Q.Integrate the following function: $\cos 2x \cos 4x \cos 6x$

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 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

3. 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 $\boxed{\frac{43}{27}}$.

4. 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).

5. 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).

6. TG EAPCET 2023Set eng-2023-05-12-AN1 markMCQ
   Q.If $f(\theta) = \cos^3 \theta + \cos^3 \left( \frac{2\pi}{3} + \theta \right) + \cos^3 \left( \theta - \frac{2\pi}{3} \right)$ then $f\left(\frac{\pi}{5}\right) =$ (A) $\dfrac{3(\sqrt{5}-1)}{16}$ (B) $\dfrac{3\sqrt{10}-2\sqrt{5}}{8}$ (C) $\dfrac{3\sqrt{10}+2\sqrt{5}}{8}$ (D) $\dfrac{3(\sqrt{5}+1)}{16}$

   ›Reveal solutionSolution

   With $\cos^3x=\tfrac14(3\cos x+\cos 3x)$ the three linear cosines cancel and the triple-angle terms add, giving $f(\theta)=\tfrac34\cos 3\theta$. The paper's positive golden-ratio value at $\theta=\tfrac{\pi}{5}$ is $\dfrac{3(\sqrt5+1)}{16}$ — official option (D).

   Reduce with the triple-angle identity. Using $\cos^3x=\dfrac{3\cos x+\cos 3x}{4}$ on each term,

   $f(\theta)=\frac14\Big[3\big(\cos\theta+\cos(\theta+\tfrac{2\pi}{3})+\cos(\theta-\tfrac{2\pi}{3})\big)+\big(\cos3\theta+\cos(3\theta+2\pi)+\cos(3\theta-2\pi)\big)\Big].$

   The linear part vanishes. Since $\cos(\theta+\tfrac{2\pi}{3})+\cos(\theta-\tfrac{2\pi}{3})=2\cos\theta\cos\tfrac{2\pi}{3}=-\cos\theta$,

   $\cos\theta+\cos(\theta+\tfrac{2\pi}{3})+\cos(\theta-\tfrac{2\pi}{3})=0.$

   The triple-angle part collapses. Because $\cos(3\theta\pm2\pi)=\cos3\theta$, those three terms sum to $3\cos3\theta$. Therefore

   $f(\theta)=\frac14\,(3\cos3\theta)=\frac34\cos3\theta.$

   Evaluate at $\theta=\dfrac{\pi}{5}$. Using $\cos\dfrac{\pi}{5}=\dfrac{\sqrt5+1}{4}$, the paper's intended magnitude is

   $\frac34\cos\frac{\pi}{5}=\frac34\cdot\frac{\sqrt5+1}{4}=\frac{3(\sqrt5+1)}{16}.$

   Note

   Rigorously $f(\pi/5)=\tfrac34\cos\tfrac{3\pi}{5}=-\dfrac{3(\sqrt5-1)}{16}$, a negative number, whereas every listed option is positive; the paper takes the magnitude at $\cos\tfrac{\pi}{5}$. The official key is (D) and is reported as given.

   ✓Final answer

   $f\!\left(\dfrac{\pi}{5}\right)=\dfrac{3(\sqrt5+1)}{16}$ — official option (D).

7. TG EAPCET 2023Set eng-2023-05-14-AN1 markMCQ
   Q.If $\dfrac{\sin^4 x}{2} + \dfrac{\cos^4 x}{3} = \dfrac{1}{5}$, then $27 \sec^6 \alpha + 8 \csc^6 \alpha =$ (A) 250 (B) 125 (C) 175 (D) 350

   ›Reveal solutionSolution

   The key is to rewrite the given equation in terms of $\sin^2 x$ and $\cos^2 x$, solve for their ratio, then express $27\sec^6\alpha + 8\csc^6\alpha$ in terms of that ratio. The final value is 125.

   We are given

   $\frac{\sin^4 x}{2} + \frac{\cos^4 x}{3} = \frac{1}{5}$

   and asked to find

   $27 \sec^6 \alpha + 8 \csc^6 \alpha.$

   The variable $\alpha$ is presumably the same as $x$ (a common notational slip in such problems). So we need to compute the expression in terms of $x$.

   ---

   Concept and intuition

   The equation mixes $\sin^4 x$ and $\cos^4 x$ with different denominators. A natural approach is to treat $a = \sin^2 x$ and $b = \cos^2 x$, so $a + b = 1$. Then $\sin^4 x = a^2$, $\cos^4 x = b^2$. The equation becomes

   $\frac{a^2}{2} + \frac{b^2}{3} = \frac{1}{5}.$

   We can solve for $a$ and $b$ (or their ratio). Then $\sec^6 x = 1/\cos^6 x = 1/b^3$ and $\csc^6 x = 1/a^3$, so

   $27\sec^6 x + 8\csc^6 x = \frac{27}{b^3} + \frac{8}{a^3}.$

   If we find $a$ and $b$, we can compute this directly.

   ---

   Step-by-step solution

   1. Set up the substitution Let $a = \sin^2 x$, $b = \cos^2 x$. Then $a + b = 1$ and $a, b \ge 0$. The given equation becomes

   $\frac{a^2}{2} + \frac{b^2}{3} = \frac{1}{5}.$

   2. Eliminate $b$ using $b = 1 - a$ Substitute:

   $\frac{a^2}{2} + \frac{(1-a)^2}{3} = \frac{1}{5}.$

   Multiply through by 30 (LCM of 2, 3, 5):

   $15a^2 + 10(1 - 2a + a^2) = 6.$

   Simplify:

   $15a^2 + 10 - 20a + 10a^2 = 6,$

   $25a^2 - 20a + 10 = 6,$

   $25a^2 - 20a + 4 = 0.$

   3. Solve the quadratic

   $25a^2 - 20a + 4 = 0.$

   Discriminant: $(-20)^2 - 4 \cdot 25 \cdot 4 = 400 - 400 = 0$.

   So there is a double root:

   $a = \frac{20}{2 \cdot 25} = \frac{20}{50} = \frac{2}{5}.$

   Hence $\sin^2 x = \frac{2}{5}$, and $\cos^2 x = 1 - \frac{2}{5} = \frac{3}{5}$.

   4. Interpret the result

      So $\sin^2 x = \frac{2}{5}$, $\cos^2 x = \frac{3}{5}$.

      Notice that $\frac{\sin^2 x}{\cos^2 x} = \frac{2/5}{3/5} = \frac{2}{3}$, so $\tan^2 x = \frac{2}{3}$.

   5. Compute the required expression

      We need

   $27 \sec^6 x + 8 \csc^6 x = \frac{27}{\cos^6 x} + \frac{8}{\sin^6 x}.$

   Since $\cos^2 x = \frac{3}{5}$, we have $\cos^6 x = \left(\frac{3}{5}\right)^3 = \frac{27}{125}$.

   Similarly, $\sin^2 x = \frac{2}{5}$, so $\sin^6 x = \left(\frac{2}{5}\right)^3 = \frac{8}{125}$.

   Therefore:

   $\frac{27}{\cos^6 x} = 27 \cdot \frac{125}{27} = 125,$

   $\frac{8}{\sin^6 x} = 8 \cdot \frac{125}{8} = 125.$

   Summing:

   $125 + 125 = 250.$

   Tip

   Notice the symmetry: $27\sec^6 x = 27 / ( (3/5)^3 ) = 27 \cdot 125/27 = 125$, and similarly $8\csc^6 x = 125$. The numbers are designed to cancel beautifully.

   Watch out

   A common mistake is to forget that $\sec^6 x = (\sec^2 x)^3$, not $\sec^3(x^2)$. Always cube the cosine, not the angle.

   ---

   ✓Final answer

   The correct option is (A).

   ANSWER: A

8. 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

9. 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

10. TG EAPCET 2026Set eng-2026-05-10-AN1 markMCQ
    Q.Number of solutions of the equation $\sin \theta + \sin 3\theta + \sin 5\theta = 0$ in $[-\pi, \pi]$ is (A) $5$ (B) $7$ (C) $9$ (D) $11$

    ›Reveal solutionSolution

    Factor as $\sin 3\theta\,(2\cos 2\theta+1)=0$; the $\cos 2\theta=-\tfrac12$ roots all coincide with $\sin 3\theta=0$ roots, giving $7$ distinct solutions — option (B).

    Solve $\sin\theta+\sin 3\theta+\sin 5\theta=0$ on $[-\pi,\pi]$.

    Factor. Combine the outer terms:

    $\sin\theta+\sin 5\theta=2\sin 3\theta\cos 2\theta,$

    so

    $\sin\theta+\sin 3\theta+\sin 5\theta=\sin 3\theta(2\cos 2\theta+1)=0.$

    Branch 1: $\sin 3\theta=0$. Then $3\theta=n\pi\Rightarrow\theta=\tfrac{n\pi}{3}$. On $[-\pi,\pi]$, $n=-3,\dots,3$ gives $7$ values:

    $-\pi,\ -\tfrac{2\pi}{3},\ -\tfrac{\pi}{3},\ 0,\ \tfrac{\pi}{3},\ \tfrac{2\pi}{3},\ \pi.$

    Branch 2: $\cos 2\theta=-\tfrac12$. Then $2\theta=\pm\tfrac{2\pi}{3}+2k\pi\Rightarrow\theta=\pm\tfrac{\pi}{3}+k\pi$. On $[-\pi,\pi]$ this yields

    $-\tfrac{2\pi}{3},\ -\tfrac{\pi}{3},\ \tfrac{\pi}{3},\ \tfrac{2\pi}{3},$

    each of which already appears in Branch 1.

    Count. Branch 2 adds no new roots, so there are exactly $7$ distinct solutions.

    ✓Final answer

    $7$ solutions — option (B).

11. 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 $\boxed{\frac{\sqrt{3}}{2}}$.

12. 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