Q.Integrate the following function: $\sin 3x \cos 4x$

Showing the 12 most recent of 42 on this concept.

1. TG EAPCET 2022Set eng-2022-07-19-FN1 markMCQ
   Q.$\lim_{x \to 0} \left( \frac{4!}{x^8} \left( 1 - \cos \frac{x^2}{3} - \cos \frac{x^2}{4} + \cos \frac{x^2}{3} \cos \frac{x^2}{4} \right) \right) =$ (A) $8$ (B) $\frac{1}{6}$ (C) $\frac{1}{24}$ (D) $\frac{2}{3}$

   ›Reveal solutionSolution

   The bracket factors as $(1-\cos\tfrac{x^2}{3})(1-\cos\tfrac{x^2}{4})$; each behaves like $\tfrac{t^2}{2}$, giving $\dfrac{4!}{x^8}\cdot\dfrac{x^8}{576}=\dfrac{1}{24}$ — option (C).

   Factor the expression

   With $A=\dfrac{x^2}{3}$ and $B=\dfrac{x^2}{4}$,

   $1-\cos A-\cos B+\cos A\cos B=(1-\cos A)(1-\cos B).$

   Use $1-\cos t\sim\tfrac{t^2}{2}$

   As $x\to0$,

   $1-\cos\frac{x^2}{3}\sim\frac12\left(\frac{x^2}{3}\right)^2=\frac{x^4}{18},\qquad 1-\cos\frac{x^2}{4}\sim\frac12\left(\frac{x^2}{4}\right)^2=\frac{x^4}{32}.$

   Their product is

   $\frac{x^4}{18}\cdot\frac{x^4}{32}=\frac{x^8}{576}.$

   Assemble the limit

   $\lim_{x\to0}\frac{4!}{x^8}\cdot\frac{x^8}{576}=\frac{24}{576}=\frac{1}{24}.$

   ✓Final answer

   The limit is $\dfrac{1}{24}$ — option (C).

2. TG EAPCET 2022Set eng-2022-07-19-AN1 markMCQ
   Q.Let $\alpha$ be the period of $3 \sin \frac{\pi x}{3} - \cos \frac{\pi x}{2} + \tan \frac{\pi x}{4}$, $\beta$ be the period of $\sin^2 \left( \frac{\pi}{7} + \frac{x}{4} \right) - \sin^2 \left( \frac{\pi}{7} - \frac{x}{4} \right)$, and $\gamma$ be the period of $\cos^4 x + \sin^4 x$. Then $\frac{\alpha \gamma}{\beta} =$ (A) $\frac{3}{2}$ (B) $\frac{3}{4}$ (C) $3$ (D) $6$

   ›Reveal solutionSolution

   $\alpha=12,\ \beta=4\pi,\ \gamma=\tfrac{\pi}{2}$, so $\dfrac{\alpha\gamma}{\beta}=\dfrac{3}{2}$.

   Period $\alpha$: for $3\sin\frac{\pi x}{3}-\cos\frac{\pi x}{2}+\tan\frac{\pi x}{4}$ the individual periods are $\frac{2\pi}{\pi/3}=6$, $\frac{2\pi}{\pi/2}=4$, and $\frac{\pi}{\pi/4}=4$. Their LCM is $\alpha=12$.

   Period $\beta$: using $\sin^2A-\sin^2B=\sin(A+B)\sin(A-B)$ with $A=\frac{\pi}{7}+\frac{x}{4},\ B=\frac{\pi}{7}-\frac{x}{4}$:

   $\sin^2\!\left(\tfrac{\pi}{7}+\tfrac{x}{4}\right)-\sin^2\!\left(\tfrac{\pi}{7}-\tfrac{x}{4}\right)=\sin\tfrac{2\pi}{7}\,\sin\tfrac{x}{2}.$

   The period of $\sin\frac{x}{2}$ is $\frac{2\pi}{1/2}=4\pi$, so $\beta=4\pi$.

   Period $\gamma$: $\cos^4x+\sin^4x = 1-\tfrac12\sin^2 2x = \tfrac34+\tfrac14\cos 4x$, whose period is $\frac{2\pi}{4}=\frac{\pi}{2}$, so $\gamma=\frac{\pi}{2}$.

   $\frac{\alpha\gamma}{\beta}=\frac{12\cdot\frac{\pi}{2}}{4\pi}=\frac{6\pi}{4\pi}=\frac{3}{2}.$

   ✓Final answer

   $\dfrac{\alpha\gamma}{\beta}=\dfrac{3}{2}$ — option (A).

3. TG EAPCET 2021Set eng-2021-08-05-AN1 markMCQ
   Q.

   [!FORMULA] $\frac{\cot^2 15^\circ - 1}{\cot^2 15^\circ + 1} =$

   (A) $\frac{1}{2}$ (B) $\frac{\sqrt{3}}{2}$ (C) $\frac{3\sqrt{3}}{4}$ (D) $\frac{\sqrt{3}}{4}$

   ›Reveal solutionSolution

   The expression simplifies by converting $\cot^2 15^\circ$ to $\frac{\cos^2 15^\circ}{\sin^2 15^\circ}$, which transforms the expression into the double angle identity for cosine, $\cos(2 \times 15^\circ)$, resulting in $\cos 30^\circ = \frac{\sqrt{3}}{2}$.

   The problem asks us to evaluate a trigonometric expression that involves $\cot^2 15^\circ$. The structure of the expression, $\frac{X - 1}{X + 1}$, is a strong indicator that we should try to relate it to one of the double angle identities for cosine.

   We know the fundamental trigonometric identity $\sin^2 \theta + \cos^2 \theta = 1$.

   We also know the double angle identity for cosine:

   $\cos 2\theta = \cos^2 \theta - \sin^2 \theta$

   Our strategy is to rewrite $\cot^2 15^\circ$ in terms of $\sin^2 15^\circ$ and $\cos^2 15^\circ$. This will allow us to simplify the given expression into a form that directly matches the $\cos 2\theta$ identity.

   1. Rewrite $\cot^2 15^\circ$ using sine and cosine: The definition of the cotangent function is $\cot \theta = \frac{\cos \theta}{\sin \theta}$. Therefore, for $\theta = 15^\circ$, we have:

   $\cot^2 15^\circ = \frac{\cos^2 15^\circ}{\sin^2 15^\circ}$

   2. Substitute this into the given expression: Now, substitute this equivalent form of $\cot^2 15^\circ$ into the original expression:

   $\frac{\cot^2 15^\circ - 1}{\cot^2 15^\circ + 1} = \frac{\frac{\cos^2 15^\circ}{\sin^2 15^\circ} - 1}{\frac{\cos^2 15^\circ}{\sin^2 15^\circ} + 1}$

   To simplify this complex fraction, we multiply both the numerator and the denominator by $\sin^2 15^\circ$. This clears the inner denominators:
   

   $\frac{\left(\frac{\cos^2 15^\circ}{\sin^2 15^\circ} - 1\right) \times \sin^2 15^\circ}{\left(\frac{\cos^2 15^\circ}{\sin^2 15^\circ} + 1\right) \times \sin^2 15^\circ} = \frac{\cos^2 15^\circ - \sin^2 15^\circ}{\cos^2 15^\circ + \sin^2 15^\circ}$

   3. Apply fundamental trigonometric identities: We can now recognize the numerator and denominator as standard trigonometric identities:
      • The numerator, $\cos^2 15^\circ - \sin^2 15^\circ$, is the double angle identity for cosine, $\cos 2\theta$.
      • The denominator, $\cos^2 15^\circ + \sin^2 15^\circ$, is the Pythagorean identity, which equals $1$. Applying these identities with $\theta = 15^\circ$:

   $\frac{\cos^2 15^\circ - \sin^2 15^\circ}{\cos^2 15^\circ + \sin^2 15^\circ} = \frac{\cos(2 \times 15^\circ)}{1} = \cos 30^\circ$

   4. Evaluate $\cos 30^\circ$: The value of $\cos 30^\circ$ is a standard trigonometric value:

   $\cos 30^\circ = \frac{\sqrt{3}}{2}$

   Tip

   An alternative approach is to use the identity $\cot \theta = \frac{1}{\tan \theta}$. Substituting this into the expression gives $\frac{\frac{1}{\tan^2 15^\circ} - 1}{\frac{1}{\tan^2 15^\circ} + 1}$. Multiplying the numerator and denominator by $\tan^2 15^\circ$ yields $\frac{1 - \tan^2 15^\circ}{1 + \tan^2 15^\circ}$, which is another form of the double angle identity for cosine, $\cos 2\theta$. This also leads to $\cos 30^\circ = \frac{\sqrt{3}}{2}$.

   ✓Final answer

   The value of the expression is $\boxed{\frac{\sqrt{3}}{2}}$.

4. TG EAPCET 2024Set eng-2024-05-09-AN1 markMCQ
   Q.The solution set of the equation $\cos^2 2x + \sin^2 3x = 1$ is (A) $\left\{ x / x = n\pi + \frac{\pi}{2}, n \in \mathbb{Z} \right\}$ (B) $\left\{ x / x = 2n\pi \pm \frac{\pi}{4}, n \in \mathbb{Z} \right\}$ (C) $\left\{ x / x = \frac{n\pi}{5}, n \in \mathbb{Z} \right\}$ (D) $\left\{ x / x = n\pi + (-1)^n \frac{\pi}{6}, n \in \mathbb{Z} \right\}$

   ›Reveal solutionSolution

   The equation $\cos^2 2x + \sin^2 3x = 1$ simplifies to $\sin^2 3x = \sin^2 2x$, which leads to two families of solutions; the union of these gives $x = \frac{n\pi}{5}$, so the correct option is (C).

   We start with the equation

   $\cos^2 2x + \sin^2 3x = 1.$

   A natural first thought is to use the identity $\cos^2 \theta = 1 - \sin^2 \theta$, but here the angles are different ($2x$ and $3x$). Instead, recall the Pythagorean identity: $\cos^2 \alpha + \sin^2 \alpha = 1$. Our equation looks similar, but the angles don’t match. That mismatch is the key: we can rewrite $\cos^2 2x$ as $1 - \sin^2 2x$, then the equation becomes

   $1 - \sin^2 2x + \sin^2 3x = 1 \quad\Rightarrow\quad \sin^2 3x - \sin^2 2x = 0.$

   So we have $\sin^2 3x = \sin^2 2x$. This is a clean, symmetric condition. Taking square roots gives $\sin 3x = \pm \sin 2x$, which is equivalent to two cases: $\sin 3x = \sin 2x$ or $\sin 3x = -\sin 2x$. But we can handle both elegantly using the identity $\sin A = \sin B$ or the difference-of-squares factorization.

   1. Rewrite using difference of squares

   $\sin^2 3x - \sin^2 2x = 0 \quad\Rightarrow\quad (\sin 3x - \sin 2x)(\sin 3x + \sin 2x) = 0.$

   So either $\sin 3x = \sin 2x$ or $\sin 3x = -\sin 2x$.

   2. Solve $\sin 3x = \sin 2x$ The general solution for $\sin A = \sin B$ is

   $A = B + 2n\pi \quad\text{or}\quad A = \pi - B + 2n\pi,\quad n\in\mathbb{Z}.$

   • First branch: $3x = 2x + 2n\pi \Rightarrow x = 2n\pi$.
   • Second branch: $3x = \pi - 2x + 2n\pi \Rightarrow 5x = \pi + 2n\pi \Rightarrow x = \frac{\pi}{5} + \frac{2n\pi}{5} = \frac{(2n+1)\pi}{5}$.

   3. Solve $\sin 3x = -\sin 2x$

      Note $-\sin 2x = \sin(-2x)$. So we have $\sin 3x = \sin(-2x)$. Again apply the same formula:

      • First branch: $3x = -2x + 2n\pi \Rightarrow 5x = 2n\pi \Rightarrow x = \frac{2n\pi}{5}$.
      • Second branch: $3x = \pi - (-2x) + 2n\pi = \pi + 2x + 2n\pi \Rightarrow x = \pi + 2n\pi$, i.e., $x = (2n+1)\pi$.

   4. Combine all solutions

      From step 2: $x = 2n\pi$ and $x = \frac{(2n+1)\pi}{5}$.

      From step 3: $x = \frac{2n\pi}{5}$ and $x = (2n+1)\pi$.

      Notice that $2n\pi$ and $(2n+1)\pi$ are just multiples of $\pi$, which are already included in $\frac{2n\pi}{5}$ when $n$ is a multiple of 5? Actually, let’s check:

      • $x = 2n\pi$ is $\frac{10n\pi}{5}$, which is of the form $\frac{2k\pi}{5}$ with $k=5n$.
      • $x = (2n+1)\pi$ is $\frac{5(2n+1)\pi}{5}$, which is of the form $\frac{(2m+1)\pi}{5}$ with $m=5n+2$? Not exactly — better to see the pattern directly.

      The two families from steps 2 and 3 are:

      • $x = \frac{2n\pi}{5}$ (from step 3, first branch)
      • $x = \frac{(2n+1)\pi}{5}$ (from step 2, second branch)

      These two together cover all integer multiples of $\frac{\pi}{5}$: any integer $k$ can be written as either $2n$ or $2n+1$. So the complete solution set is

   $x = \frac{n\pi}{5},\quad n\in\mathbb{Z}.$

   Tip

   The key insight: $\sin^2 3x = \sin^2 2x$ is equivalent to $\sin 3x = \pm \sin 2x$, and both cases together yield all multiples of $\frac{\pi}{5}$. No need to check extraneous solutions because squaring didn’t introduce any here — we started from squares.

   Watch out

   A common mistake is to stop after solving only $\sin 3x = \sin 2x$ and miss the $\sin 3x = -\sin 2x$ case. That would give only half the solutions, like $x = \frac{(2n+1)\pi}{5}$ and $x=2n\pi$, missing the $\frac{2n\pi}{5}$ family.

   Thus the solution set matches option (C).

   ✓Final answer

   The correct option is (C).

   ANSWER: C

5. TG EAPCET 2022Set eng-2022-07-20-AN1 markMCQ
   Q.If $\sin \theta - \cos \theta = \dfrac{1}{\sqrt{3}}$, then $\sin (2\theta) + \cos (4\theta) + \sin (6\theta) =$ (A) $\dfrac{37}{27}$ (B) $-\dfrac{37}{27}$ (C) $-\dfrac{43}{27}$ (D) $\dfrac{43}{27}$

   ›Reveal solutionSolution

   We first determine $\sin(2\theta)$ from the given equation by squaring it. Then, we use double and triple angle formulas to find $\cos(4\theta)$ and $\sin(6\theta)$ in terms of $\sin(2\theta)$, and sum these values to get the final result $\boxed{\frac{43}{27}}$.

   The core idea here is to simplify the given expression $\sin \theta - \cos \theta = \frac{1}{\sqrt{3}}$ to find a value for $\sin(2\theta)$. Once $\sin(2\theta)$ is known, we can use standard trigonometric identities (specifically, double and triple angle formulas) to express $\cos(4\theta)$ and $\sin(6\theta)$ in terms of $\sin(2\theta)$. This strategy allows us to evaluate the entire expression without needing to find the value of $\theta$ itself.

   Let's break down the solution step-by-step.

   1. Find $\sin(2\theta)$ from the given equation. We are given the equation $\sin \theta - \cos \theta = \frac{1}{\sqrt{3}}$. To introduce $\sin(2\theta)$, which is $2\sin\theta\cos\theta$, we can square both sides of the equation:

   $(\sin \theta - \cos \theta)^2 = \left(\frac{1}{\sqrt{3}}\right)^2$

   Expand the left side using $(a-b)^2 = a^2 - 2ab + b^2$:
   

   $\sin^2 \theta + \cos^2 \theta - 2\sin \theta \cos \theta = \frac{1}{3}$

   Recall the fundamental trigonometric identity $\sin^2 \theta + \cos^2 \theta = 1$ and the double angle formula $\sin(2\theta) = 2\sin \theta \cos \theta$. Substitute these into the equation:
   

   $1 - \sin(2\theta) = \frac{1}{3}$

   Now, solve for $\sin(2\theta)$:
   

   $\sin(2\theta) = 1 - \frac{1}{3}$

   $\sin(2\theta) = \frac{2}{3}$

   2. Find $\cos(4\theta)$ using $\sin(2\theta)$. We need to express $\cos(4\theta)$ in terms of $\sin(2\theta)$. We can use the double angle formula for cosine:

      $\cos(2A) = 1 - 2\sin^2 A$

      Let $A = 2\theta$. Then $2A = 4\theta$.

   $\cos(4\theta) = 1 - 2\sin^2(2\theta)$

   Substitute the value of $\sin(2\theta) = \frac{2}{3}$ we found in Step 1:
   

   $\cos(4\theta) = 1 - 2\left(\frac{2}{3}\right)^2$

   $\cos(4\theta) = 1 - 2\left(\frac{4}{9}\right)$

   $\cos(4\theta) = 1 - \frac{8}{9}$

   $\cos(4\theta) = \frac{1}{9}$

   3. Find $\sin(6\theta)$ using $\sin(2\theta)$. We need to express $\sin(6\theta)$ in terms of $\sin(2\theta)$. We can use the triple angle formula for sine:

      $\sin(3A) = 3\sin A - 4\sin^3 A$

      Let $A = 2\theta$. Then $3A = 6\theta$.

   $\sin(6\theta) = \sin(3 \cdot 2\theta) = 3\sin(2\theta) - 4\sin^3(2\theta)$

   Substitute the value of $\sin(2\theta) = \frac{2}{3}$:
   

   $\sin(6\theta) = 3\left(\frac{2}{3}\right) - 4\left(\frac{2}{3}\right)^3$

   $\sin(6\theta) = 2 - 4\left(\frac{8}{27}\right)$

   $\sin(6\theta) = 2 - \frac{32}{27}$

   To combine these, find a common denominator:
   

   $\sin(6\theta) = \frac{2 \cdot 27}{27} - \frac{32}{27}$

   $\sin(6\theta) = \frac{54 - 32}{27}$

   $\sin(6\theta) = \frac{22}{27}$

   4. Calculate the final expression. Now we have all the components:
      • $\sin(2\theta) = \frac{2}{3}$
      • $\cos(4\theta) = \frac{1}{9}$
      • $\sin(6\theta) = \frac{22}{27}$ Add these values together:

   $\sin(2\theta) + \cos(4\theta) + \sin(6\theta) = \frac{2}{3} + \frac{1}{9} + \frac{22}{27}$

   To sum these fractions, find a common denominator, which is 27:
   

   $= \frac{2 \cdot 9}{27} + \frac{1 \cdot 3}{27} + \frac{22}{27}$

   $= \frac{18}{27} + \frac{3}{27} + \frac{22}{27}$

   $= \frac{18 + 3 + 22}{27}$

   $= \frac{43}{27}$

   ✓Final answer

   The value of $\sin (2\theta) + \cos (4\theta) + \sin (6\theta)$ is $\boxed{\frac{43}{27}}$.

6. TG EAPCET 2022Set eng-2022-07-20-FN1 markMCQ
   Q.

   [!FORMULA] $\frac{1}{\sin 250^\circ} + \frac{\sqrt{3}}{\cos 290^\circ} =$

   (A) $\frac{1}{\sqrt{3}}$ (B) $4$ (C) $\frac{4}{\sqrt{3}}$ (D) $1$

   ›Reveal solutionSolution

   We simplify the given expression by first reducing the angles to acute angles using quadrant rules, then combining the fractions, and finally applying trigonometric identities for sum/difference of angles and double angles. The expression simplifies to $\boxed{4}$.

   The problem asks us to evaluate a trigonometric expression involving angles outside the first quadrant. The core idea is to first simplify these angles to their acute equivalents using quadrant rules and then use standard trigonometric identities to simplify the resulting expression.

   Here's the breakdown of the approach:

   1. Reduce angles: Convert $250^\circ$ and $290^\circ$ to equivalent acute angles ($0^\circ$ to $90^\circ$) using the properties of trigonometric functions in different quadrants. This involves determining the sign of the function in that quadrant and using the appropriate reduction formula (e.g., $180^\circ \pm \theta$ or $360^\circ - \theta$).
   2. Combine fractions: Once the angles are simplified, the expression will be in terms of acute angles. Combine the two fractions into a single fraction.
   3. Apply identities: The numerator and denominator of the combined fraction will likely simplify further using standard trigonometric identities, such as the sum/difference formulas for sine or cosine, or double angle formulas.

   Let's work through the steps.

   1. Reduce $\sin 250^\circ$: The angle $250^\circ$ lies in the third quadrant ($180^\circ < 250^\circ < 270^\circ$). In the third quadrant, the sine function is negative. We can write $250^\circ$ as $180^\circ + 70^\circ$. Using the reduction formula $\sin(180^\circ + \theta) = -\sin \theta$:

   $\sin 250^\circ = \sin(180^\circ + 70^\circ) = -\sin 70^\circ$

   2. Reduce $\cos 290^\circ$: The angle $290^\circ$ lies in the fourth quadrant ($270^\circ < 290^\circ < 360^\circ$). In the fourth quadrant, the cosine function is positive. We can write $290^\circ$ as $360^\circ - 70^\circ$. Using the reduction formula $\cos(360^\circ - \theta) = \cos \theta$:

   $\cos 290^\circ = \cos(360^\circ - 70^\circ) = \cos 70^\circ$

   3. Substitute the reduced values into the expression: Now, substitute these simplified terms back into the original expression:

   $\frac{1}{\sin 250^\circ} + \frac{\sqrt{3}}{\cos 290^\circ} = \frac{1}{-\sin 70^\circ} + \frac{\sqrt{3}}{\cos 70^\circ}$

   This can be rewritten as:
   

   $\frac{\sqrt{3}}{\cos 70^\circ} - \frac{1}{\sin 70^\circ}$

   4. Combine the fractions: To combine these, we find a common denominator, which is $\sin 70^\circ \cos 70^\circ$:

   $\frac{\sqrt{3} \sin 70^\circ - \cos 70^\circ}{\sin 70^\circ \cos 70^\circ}$

   5. Simplify the numerator: The numerator is in the form $A \sin \theta - B \cos \theta$. We can express this in the form $R \sin(\theta - \alpha)$. Factor out $2$ from the numerator:

   $2 \left( \frac{\sqrt{3}}{2} \sin 70^\circ - \frac{1}{2} \cos 70^\circ \right)$

   We know that $\cos 30^\circ = \frac{\sqrt{3}}{2}$ and $\sin 30^\circ = \frac{1}{2}$. Substitute these values:
   

   $2 (\cos 30^\circ \sin 70^\circ - \sin 30^\circ \cos 70^\circ)$

   This matches the sine difference identity: $\sin(A - B) = \sin A \cos B - \cos A \sin B$.
   So, the numerator becomes:
   

   $2 \sin(70^\circ - 30^\circ) = 2 \sin 40^\circ$

   6. Simplify the denominator: The denominator is $\sin 70^\circ \cos 70^\circ$. This can be simplified using the double angle identity for sine:

      $\sin 2A = 2 \sin A \cos A \implies \sin A \cos A = \frac{1}{2} \sin 2A$

      Applying this, the denominator becomes:

   $\frac{1}{2} \sin(2 \times 70^\circ) = \frac{1}{2} \sin 140^\circ$

   7. Substitute simplified numerator and denominator back into the expression: The expression now is:

   $\frac{2 \sin 40^\circ}{\frac{1}{2} \sin 140^\circ} = \frac{4 \sin 40^\circ}{\sin 140^\circ}$

   8. Reduce $\sin 140^\circ$: The angle $140^\circ$ lies in the second quadrant ($90^\circ < 140^\circ < 180^\circ$). In the second quadrant, the sine function is positive. We can write $140^\circ$ as $180^\circ - 40^\circ$. Using the reduction formula $\sin(180^\circ - \theta) = \sin \theta$:

   $\sin 140^\circ = \sin(180^\circ - 40^\circ) = \sin 40^\circ$

   9. Final simplification: Substitute $\sin 140^\circ = \sin 40^\circ$ back into the expression:

   $\frac{4 \sin 40^\circ}{\sin 40^\circ}$

   Since $\sin 40^\circ \neq 0$, we can cancel $\sin 40^\circ$ from the numerator and denominator.
   The final value is $4$.
   

   ✓Final answer

   The value of the expression is $\boxed{4}$.

7. TG EAPCET 2022Set eng-2022-07-18-FN1 markMCQ
   Q.If $\tan A = \dfrac{2}{3}$, then $\sin 4A =$ (A) $\dfrac{8}{27}$ (B) $\dfrac{120}{169}$ (C) $\dfrac{144}{169}$ (D) $\dfrac{16}{27}$

   ›Reveal solutionSolution

   Use the double-angle identity for tangent to find $\tan 2A$, then use the sine double-angle formula in terms of tangent to get $\sin 4A$ directly. The result is $\frac{120}{169}$, which corresponds to option (B).

   We are given $\tan A = \frac{2}{3}$ and need $\sin 4A$. The direct approach: find $\sin 2A$ and $\cos 2A$ using tangent double-angle formulas, then use $\sin 4A = 2 \sin 2A \cos 2A$. Alternatively, we can compute $\tan 2A$ first and then express $\sin 4A$ in terms of $\tan 2A$ — that’s often cleaner because we avoid square roots.

   Concept & Intuition

   The key is that $\sin 4A$ can be written as $2 \sin 2A \cos 2A$, and both $\sin 2A$ and $\cos 2A$ can be expressed rationally in terms of $\tan A$ (or $\tan 2A$). Since $\tan A$ is given as a simple fraction, we can compute $\tan 2A$ exactly, then use the identity $\sin \theta = \frac{2 \tan(\theta/2)}{1+\tan^2(\theta/2)}$ with $\theta = 4A$ — but more directly, we use $\sin 4A = \frac{2 \tan 2A}{1+\tan^2 2A}$.

   Let’s go step by step.

   1. Find $\tan 2A$ using the double-angle formula

   $\tan 2A = \frac{2 \tan A}{1 - \tan^2 A} = \frac{2 \cdot \frac{2}{3}}{1 - \left(\frac{2}{3}\right)^2} = \frac{\frac{4}{3}}{1 - \frac{4}{9}} = \frac{\frac{4}{3}}{\frac{5}{9}} = \frac{4}{3} \cdot \frac{9}{5} = \frac{36}{15} = \frac{12}{5}.$

   2. Express $\sin 4A$ in terms of $\tan 2A$ Recall the identity: for any angle $\theta$,

   $\sin \theta = \frac{2 \tan(\theta/2)}{1 + \tan^2(\theta/2)}.$

   Here $\theta = 4A$, so $\theta/2 = 2A$. Thus

   $\sin 4A = \frac{2 \tan 2A}{1 + \tan^2 2A}.$

   3. Substitute $\tan 2A = \frac{12}{5}$

   $\sin 4A = \frac{2 \cdot \frac{12}{5}}{1 + \left(\frac{12}{5}\right)^2} = \frac{\frac{24}{5}}{1 + \frac{144}{25}} = \frac{\frac{24}{5}}{\frac{25}{25} + \frac{144}{25}} = \frac{\frac{24}{5}}{\frac{169}{25}} = \frac{24}{5} \cdot \frac{25}{169} = \frac{24 \cdot 5}{169} = \frac{120}{169}.$

   4. Check against the options The value $\frac{120}{169}$ matches option (B).

   Tip

   A common pitfall is to compute $\sin 2A$ and $\cos 2A$ separately using right-triangle methods, which can introduce sign ambiguity. Using the tangent-based identity avoids that entirely — it’s always valid as long as the denominator isn’t zero.

   Watch out

   If you instead tried to find $\sin A$ and $\cos A$ directly from $\tan A = 2/3$, you’d get $\sin A = \frac{2}{\sqrt{13}}$, $\cos A = \frac{3}{\sqrt{13}}$. Then $\sin 2A = 2 \cdot \frac{2}{\sqrt{13}} \cdot \frac{3}{\sqrt{13}} = \frac{12}{13}$, $\cos 2A = \cos^2 A - \sin^2 A = \frac{9}{13} - \frac{4}{13} = \frac{5}{13}$, and $\sin 4A = 2 \cdot \frac{12}{13} \cdot \frac{5}{13} = \frac{120}{169}$. That works too, but the tangent method is faster and avoids square roots.

   ✓Final answer

   The correct option is (B).

   ANSWER: B

8. TG EAPCET 2023Set eng-2023-05-13-AN1 markMCQ
   Q.If $\sinh x = -\dfrac{4}{3}$ then $\sinh 2x + \cosh 2x =$ (A) $-\dfrac{31}{41}$ (B) $-\dfrac{20}{9}$ (C) $\dfrac{49}{41}$ (D) $\dfrac{1}{9}$

   ›Reveal solutionSolution

   Using the definitions of hyperbolic functions and the identity $\sinh 2x + \cosh 2x = e^{2x}$, we find $e^x$ from $\sinh x = -4/3$, then square to get $e^{2x}$, yielding the result $1/9$, which corresponds to option (D).

   We are given $\sinh x = -\dfrac{4}{3}$ and need $\sinh 2x + \cosh 2x$.

   The key insight: recall that $\sinh 2x + \cosh 2x = e^{2x}$ because

   $\sinh t = \frac{e^t - e^{-t}}{2}, \quad \cosh t = \frac{e^t + e^{-t}}{2}$

   so

   $\sinh t + \cosh t = e^t.$

   Thus the problem reduces to finding $e^{2x}$ from $\sinh x = -\dfrac{4}{3}$.

   1. Express $\sinh x$ in terms of $e^x$

   $\sinh x = \frac{e^x - e^{-x}}{2} = -\frac{4}{3}.$

   Multiply by 2:

   $e^x - e^{-x} = -\frac{8}{3}.$

   2. Let $u = e^x$ (so $u > 0$). Then $e^{-x} = 1/u$, and the equation becomes

   $u - \frac{1}{u} = -\frac{8}{3}.$

   Multiply through by $u$:

   $u^2 - 1 = -\frac{8}{3}u \quad\Rightarrow\quad u^2 + \frac{8}{3}u - 1 = 0.$

   3. Solve the quadratic Multiply by 3:

   $3u^2 + 8u - 3 = 0.$

   Discriminant: $\Delta = 8^2 - 4\cdot3\cdot(-3) = 64 + 36 = 100$.

   So

   $u = \frac{-8 \pm 10}{6}.$

   This gives $u = \frac{2}{6} = \frac{1}{3}$ or $u = \frac{-18}{6} = -3$.

   Since $u = e^x > 0$, we take $u = \frac{1}{3}$.

   4. Find $e^{2x}$

   $e^{2x} = (e^x)^2 = \left(\frac{1}{3}\right)^2 = \frac{1}{9}.$

   Therefore,

   $\sinh 2x + \cosh 2x = e^{2x} = \frac{1}{9}.$

   Watch out

   A common mistake is to compute $\sinh 2x$ and $\cosh 2x$ separately using double-angle formulas, then add — this is fine but slower. The fastest path is the identity $\sinh t + \cosh t = e^t$, which collapses the sum to a single exponential.

   Tip

   Notice that $\sinh x$ negative gives $e^x < 1$, consistent with $e^x = 1/3$. The positive root $u = 1/3$ is the only physically valid one.

   ✓Final answer

   The correct option is (D).

   ANSWER: D

9. TG EAPCET 2024Set eng-2024-05-10-AN1 markMCQ
   Q.$\sin 20^\circ (4 + \sec 20^\circ) =$ (A) $\sqrt{3}$ (B) $-\sqrt{3}$ (C) $1$ (D) $-1$

   ›Reveal solutionSolution

   The key idea is to rewrite the expression using sine and cosine, then apply triple-angle identities to simplify it to a known constant. The final result is $\sqrt{3}$.

   We start with the expression

   $\sin 20^\circ (4 + \sec 20^\circ).$

   The presence of $\sec 20^\circ$ suggests rewriting everything in terms of sine and cosine, since $\sec \theta = 1/\cos \theta$. Then we look for a way to combine terms into a known trigonometric identity — the triple-angle formulas for sine and cosine are natural here because $20^\circ$ is one-third of $60^\circ$.

   ---

   1. Rewrite in terms of sine and cosine

   $\sin 20^\circ \left(4 + \frac{1}{\cos 20^\circ}\right) = 4\sin 20^\circ + \frac{\sin 20^\circ}{\cos 20^\circ} = 4\sin 20^\circ + \tan 20^\circ.$

   2. Express $\tan 20^\circ$ as $\frac{\sin 20^\circ}{\cos 20^\circ}$ and combine over a common denominator

   $4\sin 20^\circ + \frac{\sin 20^\circ}{\cos 20^\circ} = \frac{4\sin 20^\circ \cos 20^\circ + \sin 20^\circ}{\cos 20^\circ} = \frac{\sin 20^\circ (4\cos 20^\circ + 1)}{\cos 20^\circ}.$

   3. Use the double-angle identity $2\sin 20^\circ \cos 20^\circ = \sin 40^\circ$, so $4\sin 20^\circ \cos 20^\circ = 2\sin 40^\circ$. Then the numerator becomes

   $2\sin 40^\circ + \sin 20^\circ.$

   4. Apply the triple-angle identity for sine Recall: $\sin 3\theta = 3\sin\theta - 4\sin^3\theta$. For $\theta = 20^\circ$, $\sin 60^\circ = \frac{\sqrt{3}}{2}$, so

   $\frac{\sqrt{3}}{2} = 3\sin 20^\circ - 4\sin^3 20^\circ.$

   This doesn't directly match our numerator, but we can also use the identity for $\sin 3\theta$ in terms of products:

   $\sin 3\theta = 4\sin\theta \sin(60^\circ - \theta) \sin(60^\circ + \theta).$

   For $\theta = 20^\circ$, this gives

   $\sin 60^\circ = 4\sin 20^\circ \sin 40^\circ \sin 80^\circ.$

   So $\sin 20^\circ \sin 40^\circ \sin 80^\circ = \frac{\sqrt{3}}{8}$.

   But we have $2\sin 40^\circ + \sin 20^\circ$ — not a product. So let's try a different path.

   5. Better approach: Use the triple-angle identity for cosine $\cos 3\theta = 4\cos^3\theta - 3\cos\theta$. For $\theta = 20^\circ$, $\cos 60^\circ = \frac{1}{2}$, so

   $\frac{1}{2} = 4\cos^3 20^\circ - 3\cos 20^\circ.$

   Multiply both sides by 2:

   $1 = 8\cos^3 20^\circ - 6\cos 20^\circ.$

   Rearranging:

   $8\cos^3 20^\circ - 6\cos 20^\circ - 1 = 0.$

   6. Relate this to our expression We have $\frac{2\sin 40^\circ + \sin 20^\circ}{\cos 20^\circ}$. Note $\sin 40^\circ = 2\sin 20^\circ \cos 20^\circ$, so

   $2\sin 40^\circ = 4\sin 20^\circ \cos 20^\circ.$

   Then numerator = $4\sin 20^\circ \cos 20^\circ + \sin 20^\circ = \sin 20^\circ (4\cos 20^\circ + 1)$.

   So the whole expression is

   $\frac{\sin 20^\circ (4\cos 20^\circ + 1)}{\cos 20^\circ} = \tan 20^\circ (4\cos 20^\circ + 1).$

   7. Now use the triple-angle cosine identity From $8\cos^3 20^\circ - 6\cos 20^\circ = 1$, divide by $\cos 20^\circ$ (nonzero):

   $8\cos^2 20^\circ - 6 = \frac{1}{\cos 20^\circ}.$

   So $\frac{1}{\cos 20^\circ} = 8\cos^2 20^\circ - 6$.

   But we have $4\cos 20^\circ + 1$, not $1/\cos 20^\circ$. Let's instead multiply the triple-angle identity by something clever.

   8. A cleaner trick: Multiply numerator and denominator by something Consider the original expression:

   $\sin 20^\circ (4 + \sec 20^\circ) = \frac{\sin 20^\circ (4\cos 20^\circ + 1)}{\cos 20^\circ}.$

   Now use the identity $\sin 20^\circ = \frac{\sqrt{3}}{2} \cdot \frac{1}{2\cos 20^\circ + 1}$? Not directly.

   Instead, recall the triple-angle formula for sine in product form:

   $\sin 60^\circ = 4\sin 20^\circ \sin(60^\circ - 20^\circ) \sin(60^\circ + 20^\circ) = 4\sin 20^\circ \sin 40^\circ \sin 80^\circ.$

   So $\sin 20^\circ \sin 40^\circ \sin 80^\circ = \frac{\sqrt{3}}{8}$.

   Also note $\sin 80^\circ = \cos 10^\circ$, but maybe we can relate $4\cos 20^\circ + 1$ to something like $\frac{\sin 60^\circ}{\sin 20^\circ \sin 40^\circ}$?

   Actually, $\sin 40^\circ = 2\sin 20^\circ \cos 20^\circ$, so

   $\frac{\sqrt{3}}{8} = \sin 20^\circ \cdot (2\sin 20^\circ \cos 20^\circ) \cdot \sin 80^\circ = 2\sin^2 20^\circ \cos 20^\circ \sin 80^\circ.$

   This is getting messy.

   9. Let’s go back to the simplest algebraic manipulation We have $E = 4\sin 20^\circ + \tan 20^\circ$. Write $\tan 20^\circ = \frac{\sin 20^\circ}{\cos 20^\circ}$, so

   $E = \frac{4\sin 20^\circ \cos 20^\circ + \sin 20^\circ}{\cos 20^\circ} = \frac{2\sin 40^\circ + \sin 20^\circ}{\cos 20^\circ}.$

   Now use the identity $\sin 40^\circ = \sin(60^\circ - 20^\circ) = \sin 60^\circ \cos 20^\circ - \cos 60^\circ \sin 20^\circ = \frac{\sqrt{3}}{2}\cos 20^\circ - \frac{1}{2}\sin 20^\circ$.

   Then

   $2\sin 40^\circ = \sqrt{3}\cos 20^\circ - \sin 20^\circ.$

   So numerator becomes

   $(\sqrt{3}\cos 20^\circ - \sin 20^\circ) + \sin 20^\circ = \sqrt{3}\cos 20^\circ.$

   Therefore

   $E = \frac{\sqrt{3}\cos 20^\circ}{\cos 20^\circ} = \sqrt{3}.$

   10. Conclusion The expression simplifies exactly to $\sqrt{3}$, independent of any approximations.

   Watch out

   A common mistake is to try to evaluate numerically and guess, but $20^\circ$ is not a standard angle with a simple exact sine/cosine. The triple-angle trick or the sine subtraction identity is essential.

   Tip

   The step $2\sin 40^\circ = \sqrt{3}\cos 20^\circ - \sin 20^\circ$ comes from $\sin(60^\circ - 20^\circ)$ expansion — a neat way to eliminate the $\sin 20^\circ$ term.

   ✓Final answer

   The correct option is (A).

   ANSWER: A

10. TG EAPCET 2025Set eng-2025-05-02-AN1 markMCQ
    Q.

    [!FORMULA] $\frac{\sin 1^\circ + \sin 2^\circ + \ldots + \sin 89^\circ}{2(\cos 1^\circ + \cos 2^\circ + \ldots + \cos 44^\circ) + 1} =$

    (A) $2$ (B) $\frac{1}{\sqrt{2}}$ (C) $\frac{1}{2}$ (D) $\sqrt{2}$

    ›Reveal solutionSolution

    The key idea is to pair symmetric sine terms and use sum-to-product identities, leading to a telescoping simplification. The final value is $\frac{1}{\sqrt{2}}$, so the correct option is (B).

    ---

    Concept and intuition:

    We have a sum of sines from $1^\circ$ to $89^\circ$ in the numerator, and a sum of cosines from $1^\circ$ to $44^\circ$ (doubled, plus one) in the denominator. The symmetry $\sin(90^\circ - x) = \cos x$ suggests pairing terms. Also, the classic identity $\sin x + \sin(90^\circ - x) = \sqrt{2} \sin(x + 45^\circ)$ or, more directly, using sum-to-product, will collapse the numerator into something involving cosines. The denominator’s “$+1$” is a hint: $\cos 0^\circ = 1$, so we can think of it as $2\sum_{k=1}^{44} \cos k^\circ + \cos 0^\circ$, making a symmetric sum from $0^\circ$ to $44^\circ$ that pairs with the numerator’s structure.

    ---

    Step-by-step solution:

    1. Pair the sine terms symmetrically Notice $\sin 89^\circ = \cos 1^\circ$, $\sin 88^\circ = \cos 2^\circ$, …, $\sin 46^\circ = \cos 44^\circ$, and the middle term $\sin 45^\circ = \frac{1}{\sqrt{2}}$. So the numerator $S = \sin 1^\circ + \sin 2^\circ + \cdots + \sin 89^\circ$ becomes

    $S = (\sin 1^\circ + \sin 89^\circ) + (\sin 2^\circ + \sin 88^\circ) + \cdots + (\sin 44^\circ + \sin 46^\circ) + \sin 45^\circ.$

    2. Apply sum-to-product identity For any $x$, $\sin x + \sin(90^\circ - x) = 2 \sin 45^\circ \cos(45^\circ - x) = \sqrt{2} \cos(45^\circ - x)$. Thus each pair gives $\sqrt{2} \cos(45^\circ - k^\circ)$ for $k = 1, 2, \dots, 44$. So

    $S = \sqrt{2} \sum_{k=1}^{44} \cos(45^\circ - k^\circ) + \frac{1}{\sqrt{2}}.$

    3. Simplify the cosine sum As $k$ runs from $1$ to $44$, $(45^\circ - k^\circ)$ runs from $44^\circ$ down to $1^\circ$. So

    $\sum_{k=1}^{44} \cos(45^\circ - k^\circ) = \cos 44^\circ + \cos 43^\circ + \cdots + \cos 1^\circ = \sum_{k=1}^{44} \cos k^\circ.$

    Hence

    $S = \sqrt{2} \sum_{k=1}^{44} \cos k^\circ + \frac{1}{\sqrt{2}}.$

    4. Rewrite the denominator The denominator is $D = 2(\cos 1^\circ + \cdots + \cos 44^\circ) + 1$. Notice $1 = \cos 0^\circ$, so

    $D = 2\sum_{k=1}^{44} \cos k^\circ + \cos 0^\circ.$

    5. Form the ratio

    $\frac{S}{D} = \frac{\sqrt{2} \sum_{k=1}^{44} \cos k^\circ + \frac{1}{\sqrt{2}}}{2\sum_{k=1}^{44} \cos k^\circ + 1}.$

    Let $T = \sum_{k=1}^{44} \cos k^\circ$. Then

    $\frac{S}{D} = \frac{\sqrt{2} T + \frac{1}{\sqrt{2}}}{2T + 1}.$

    6. Factor and cancel Multiply numerator and denominator by $\sqrt{2}$ to clear the fraction:

    $\frac{S}{D} = \frac{2T + 1}{\sqrt{2}(2T + 1)} = \frac{1}{\sqrt{2}},$

    provided $2T + 1 \neq 0$ (which it isn’t, since all cosines are positive for angles $1^\circ$ to $44^\circ$).

    Thus the expression simplifies to $\frac{1}{\sqrt{2}}$.

    Tip

    The “+1” in the denominator is actually $\cos 0^\circ$, making the denominator exactly $\sqrt{2}$ times the numerator’s core structure after pairing. Spotting this turns a messy sum into a clean cancellation.

    Watch out

    A common mistake is to forget the middle term $\sin 45^\circ$ or to mis-pair the 44 pairs (there are exactly 44 pairs, not 44.5). Always check the count: 89 terms, 44 pairs + 1 middle term.

    ✓Final answer

    The correct option is (B).

    ANSWER: B

11. TG EAPCET 2021Set eng-2021-08-04-FN1 markMCQ
    Q.$\cot \dfrac{\pi}{16} \cdot \cot \dfrac{2\pi}{16} \cdot \cot \dfrac{3\pi}{16} \cdot \cot \dfrac{4\pi}{16} \cdot \cot \dfrac{5\pi}{16} \cdot \cot \dfrac{6\pi}{16} \cdot \cot \dfrac{7\pi}{16} =$ (A) $0$ (B) $1$ (C) $\dfrac{1}{2}$ (D) $2$

    ›Reveal solutionSolution

    The product of cotangents of complementary angles simplifies to 1; pairing each angle with its complement gives the result 1.

    The key insight is that cotangent has a beautiful symmetry: $\cot(\frac{\pi}{2} - \theta) = \tan \theta$, and $\cot \theta \cdot \tan \theta = 1$. So if we can pair each angle in the product with its complement (adding to $\frac{\pi}{2}$), the whole product collapses to 1 — except possibly for the middle term where the angle is exactly $\frac{\pi}{4}$, but $\cot \frac{\pi}{4} = 1$, so it doesn't break the pattern.

    Let's see this in action.

    1. Write the angles in radians:

       $\frac{\pi}{16}, \frac{2\pi}{16}, \frac{3\pi}{16}, \frac{4\pi}{16}, \frac{5\pi}{16}, \frac{6\pi}{16}, \frac{7\pi}{16}$.

    2. Notice that $\frac{\pi}{2} = \frac{8\pi}{16}$. So the complement of $\frac{k\pi}{16}$ is $\frac{8\pi}{16} - \frac{k\pi}{16} = \frac{(8-k)\pi}{16}$.

    3. Pair the terms:

       • $\frac{\pi}{16}$ pairs with $\frac{7\pi}{16}$ (since $1+7=8$)
       • $\frac{2\pi}{16}$ pairs with $\frac{6\pi}{16}$
       • $\frac{3\pi}{16}$ pairs with $\frac{5\pi}{16}$
       • $\frac{4\pi}{16}$ is its own complement (since $4+4=8$), i.e., $\frac{\pi}{4}$.

    4. For any pair $(\theta, \frac{\pi}{2} - \theta)$:

    $\cot \theta \cdot \cot\left(\frac{\pi}{2} - \theta\right) = \cot \theta \cdot \tan \theta = 1.$

    So each pair contributes a factor of 1.

    5. The middle term is $\cot \frac{4\pi}{16} = \cot \frac{\pi}{4} = 1$.

    6. Therefore the entire product is:

    $(1) \times (1) \times (1) \times 1 = 1.$

    Watch out

    A common mistake is to forget that $\cot(\frac{\pi}{2} - \theta) = \tan \theta$, not $\cot \theta$. If you mistakenly think it's $\cot \theta$, you'd get the product wrong. Always recall the complementary angle identities.

    Tip

    For any product of cotangents of angles that are symmetric about $\frac{\pi}{4}$, the result is 1 — provided no angle is 0 or $\frac{\pi}{2}$ (where cot is undefined). This trick saves huge computation.

    ✓Final answer

    The value is $\boxed{1}$, which corresponds to option (B).

12. TG EAPCET 2026Set eng-2026-05-11-AN1 markMCQ
    Q.The transformed equation of $4x^2 - 4xy + 7y^2 - 24 = 0$ in new coordinate system when the axes are rotated through an angle $\theta$ about the origin in the positive direction is $\dfrac{x^2}{a^2} + \dfrac{y^2}{b^2} = 1$. If $0 < \theta < \dfrac{\pi}{4}$, then $1 + b^2 \sin^2 \theta =$ (A) $a^2 \cos^2 \theta$ (B) $a^2 + \cos^2 \theta$ (C) $a^2 \sin^2 \theta$ (D) $\dfrac{a^2}{\sin^2 \theta}$

    ›Reveal solutionSolution

    By rotating the axes to eliminate the $xy$-term, we find the ellipse’s semi-axes $a^2$ and $b^2$ in terms of $\theta$, then compute $1 + b^2 \sin^2 \theta$ to match one of the options — it simplifies to $a^2 \cos^2 \theta$.

    ---

    We start with the conic

    $4x^2 - 4xy + 7y^2 - 24 = 0.$

    The presence of the $xy$-term tells us the axes are not aligned with the coordinate axes. Rotating the coordinate system by an angle $\theta$ will eliminate the $xy$-term, leaving an ellipse in standard position:

    $\frac{x'^2}{a^2} + \frac{y'^2}{b^2} = 1,$

    where $x', y'$ are the new coordinates. The problem states that after rotation by $\theta$ (with $0<\theta<\pi/4$), the equation becomes exactly that. Our job: find $1 + b^2 \sin^2 \theta$ in terms of $a^2$ and $\theta$.

    ---

    1. Rotation transformation formulas

    When we rotate axes by $\theta$ counterclockwise, the old coordinates $(x,y)$ relate to the new coordinates $(X,Y)$ (I’ll use $X,Y$ for the new ones) by:

    $x = X\cos\theta - Y\sin\theta, \quad y = X\sin\theta + Y\cos\theta.$

    We substitute these into the original equation.

    2. Substitute and collect terms

    Plug in:

    $4(X\cos\theta - Y\sin\theta)^2 - 4(X\cos\theta - Y\sin\theta)(X\sin\theta + Y\cos\theta) + 7(X\sin\theta + Y\cos\theta)^2 - 24 = 0.$

    Expand each piece:

    • $4(X\cos\theta - Y\sin\theta)^2 = 4(X^2\cos^2\theta - 2XY\cos\theta\sin\theta + Y^2\sin^2\theta)$
    • $-4(X\cos\theta - Y\sin\theta)(X\sin\theta + Y\cos\theta) = -4( X^2\cos\theta\sin\theta + XY\cos^2\theta - XY\sin^2\theta - Y^2\sin\theta\cos\theta )$
    • $7(X\sin\theta + Y\cos\theta)^2 = 7(X^2\sin^2\theta + 2XY\sin\theta\cos\theta + Y^2\cos^2\theta)$

    3. Collect $X^2$, $Y^2$, and $XY$ coefficients

    • $X^2$ coefficient:

      From first: $4\cos^2\theta$

      From second: $-4\cos\theta\sin\theta$

      From third: $7\sin^2\theta$

      So $A = 4\cos^2\theta - 4\sin\theta\cos\theta + 7\sin^2\theta$.

    • $Y^2$ coefficient:

      From first: $4\sin^2\theta$

      From second: $+4\sin\theta\cos\theta$ (since $-4 \times (-\sin\theta\cos\theta) = +4\sin\theta\cos\theta$)

      From third: $7\cos^2\theta$

      So $C = 4\sin^2\theta + 4\sin\theta\cos\theta + 7\cos^2\theta$.

    • $XY$ coefficient:

      From first: $2 \times 4 \times (-\cos\theta\sin\theta) = -8\cos\theta\sin\theta$

      From second: $-4(\cos^2\theta - \sin^2\theta) = -4\cos^2\theta + 4\sin^2\theta$

      From third: $2 \times 7 \times \sin\theta\cos\theta = 14\sin\theta\cos\theta$

      Sum: $(-8 + 14)\sin\theta\cos\theta + (-4\cos^2\theta + 4\sin^2\theta) = 6\sin\theta\cos\theta - 4(\cos^2\theta - \sin^2\theta)$.

    So $B = 6\sin\theta\cos\theta - 4\cos2\theta$.

    4. Eliminate the $XY$-term

    We want $B=0$:

    $6\sin\theta\cos\theta - 4\cos2\theta = 0 \quad\Rightarrow\quad 3\sin2\theta - 4\cos2\theta = 0.$

    Thus $\tan 2\theta = \frac{4}{3}$. Since $0<\theta<\pi/4$, $0<2\theta<\pi/2$, so this is valid.

    5. Find $\sin2\theta$ and $\cos2\theta$

    From $\tan2\theta = 4/3$, we have a 3-4-5 triangle:

    $\sin2\theta = \frac{4}{5}, \quad \cos2\theta = \frac{3}{5}.$

    6. Simplify $A$ and $C$ using double-angle identities

    Recall:

    $\cos^2\theta = \frac{1+\cos2\theta}{2}$, $\sin^2\theta = \frac{1-\cos2\theta}{2}$, $\sin\theta\cos\theta = \frac{1}{2}\sin2\theta$.

    • For $A$: $A = 4\cos^2\theta - 4\sin\theta\cos\theta + 7\sin^2\theta$ $= 4\cdot\frac{1+\cos2\theta}{2} - 4\cdot\frac{1}{2}\sin2\theta + 7\cdot\frac{1-\cos2\theta}{2}$ $= 2(1+\cos2\theta) - 2\sin2\theta + \frac{7}{2}(1-\cos2\theta)$ $= 2 + 2\cos2\theta - 2\sin2\theta + \frac{7}{2} - \frac{7}{2}\cos2\theta$ $= \left(2 + \frac{7}{2}\right) + \left(2 - \frac{7}{2}\right)\cos2\theta - 2\sin2\theta$ $= \frac{11}{2} - \frac{3}{2}\cos2\theta - 2\sin2\theta$.

    Plug $\cos2\theta = 3/5$, $\sin2\theta = 4/5$:

    $A = \frac{11}{2} - \frac{3}{2}\cdot\frac{3}{5} - 2\cdot\frac{4}{5} = \frac{11}{2} - \frac{9}{10} - \frac{8}{5}$

    $= \frac{55}{10} - \frac{9}{10} - \frac{16}{10} = \frac{30}{10} = 3$.

    • For $C$: $C = 4\sin^2\theta + 4\sin\theta\cos\theta + 7\cos^2\theta$ $= 4\cdot\frac{1-\cos2\theta}{2} + 4\cdot\frac{1}{2}\sin2\theta + 7\cdot\frac{1+\cos2\theta}{2}$ $= 2(1-\cos2\theta) + 2\sin2\theta + \frac{7}{2}(1+\cos2\theta)$ $= 2 - 2\cos2\theta + 2\sin2\theta + \frac{7}{2} + \frac{7}{2}\cos2\theta$ $= \left(2 + \frac{7}{2}\right) + \left(-2 + \frac{7}{2}\right)\cos2\theta + 2\sin2\theta$ $= \frac{11}{2} + \frac{3}{2}\cos2\theta + 2\sin2\theta$.

    Plug values:

    $C = \frac{11}{2} + \frac{3}{2}\cdot\frac{3}{5} + 2\cdot\frac{4}{5} = \frac{11}{2} + \frac{9}{10} + \frac{8}{5}$

    $= \frac{55}{10} + \frac{9}{10} + \frac{16}{10} = \frac{80}{10} = 8$.

    7. The rotated equation

    So after rotation, the equation becomes

    $3X^2 + 8Y^2 - 24 = 0 \quad\Rightarrow\quad \frac{X^2}{8} + \frac{Y^2}{3} = 1.$

    Thus $a^2 = 8$, $b^2 = 3$ (since $a > b$ for an ellipse with major axis along $X$).

    8. Compute $1 + b^2 \sin^2 \theta$

    We have $b^2 = 3$. Need $\sin^2\theta$.

    From $\cos2\theta = 1 - 2\sin^2\theta = \frac{3}{5}$, we get

    $2\sin^2\theta = 1 - \frac{3}{5} = \frac{2}{5}$, so $\sin^2\theta = \frac{1}{5}$.

    Thus

    $1 + b^2 \sin^2\theta = 1 + 3\cdot\frac{1}{5} = 1 + \frac{3}{5} = \frac{8}{5}.$

    9. Compare with options

    We know $a^2 = 8$, so $a^2 \cos^2\theta = 8 \cdot \frac{4}{5} = \frac{32}{5}$ — not that.

    Check $a^2 \cos^2\theta$? Wait, $\cos^2\theta = 1 - \sin^2\theta = 4/5$, so $a^2 \cos^2\theta = 8 \cdot 4/5 = 32/5$, no.

    But $a^2 \sin^2\theta = 8 \cdot 1/5 = 8/5$ — that matches!

    So $1 + b^2 \sin^2\theta = a^2 \sin^2\theta$.

    Tip

    Notice we didn’t need to compute $a^2$ and $b^2$ separately — we could have derived $1 + b^2 \sin^2\theta = a^2 \sin^2\theta$ directly from the rotation invariants, but the full calculation confirms it cleanly.

    ---

    ✓Final answer

    The correct option is (C).

    ANSWER: C