Q.Evaluate: $\int \dfrac{\cos 5x+\cos 4x}{1-2\cos 3x}\,dx$

Showing the 12 most recent of 42 on this concept.

1. 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 $k=5n$. …

   🔒 Start your 14-day free trial to see the full solution →
2. 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 $x\to0$,

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

   🔒 Start your 14-day free trial to see the full solution →
3. 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 $\beta=4\pi$. …

   🔒 Start your 14-day free trial to see the full solution →
4. 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 $3A = 6\theta$.

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

   🔒 Start your 14-day free trial to see the full solution →
5. 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

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

   🔒 Start your 14-day free trial to see the full solution →
6. 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 $\sin^2 x = \frac{2}{5}$, $\cos^2 x = \frac{3}{5}$. …

   🔒 Start your 14-day free trial to see the full solution →
7. 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, $\cos 2\theta$. …

   🔒 Start your 14-day free trial to see the full solution →
8. 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 $$$$

   🔒 Start your 14-day free trial to see the full solution →
9. TG EAPCET 2022Set eng-2022-07-20-AN1 markMCQ
   Q.In a triangle ABC, if $\dfrac{a}{\tan A} = \dfrac{b}{\tan B} = \dfrac{c}{\tan C}$ then $\cos^2 A + \cos^2 B + \cos^2 C =$ (A) $\sqrt{2}$ (B) $\dfrac{3}{4}$ (C) $\dfrac{\sqrt{3} + 1}{2}$ (D) $\dfrac{2\sqrt{3} - 1}{2}$

   ›Reveal solutionSolution

   The condition $\frac{a}{\tan A} = \frac{b}{\tan B} = \frac{c}{\tan C}$ forces the triangle to be equilateral, so each angle is $60^\circ$, and the sum of squares of cosines is $3 \cdot \frac{1}{4} = \frac{3}{4}$.

   The key is to connect the given ratio condition to the angles of the triangle. In any triangle, sides are proportional to sines of opposite angles (the Law of Sines). So we can rewrite the condition entirely in terms of angles, and see what restriction it imposes.

   1. Rewrite using the Law of Sines. In $\triangle ABC$, we have $\frac{a}{\sin A} = \frac{b}{\sin B} = \frac{c}{\sin C} = 2R$ (where $R$ is the circumradius). So $a = 2R \sin A$, $b = 2R \sin B$, $c = 2R \sin C$. The given condition becomes:

   $\frac{2R \sin A}{\tan A} = \frac{2R \sin B}{\tan B} = \frac{2R \sin C}{\tan C}.$

   Cancel $2R$ (non-zero), and recall $\tan A = \frac{\sin A}{\cos A}$, so $\frac{\sin A}{\tan A} = \sin A \cdot \frac{\cos A}{\sin A} = \cos A$.

   Hence the condition simplifies beautifully to:

   $\cos A = \cos B = \cos C.$

   2. Interpret $\cos A = \cos B = \cos C$ in a triangle. In a triangle, angles are between $0^\circ$ and $180^\circ$, and cosine is strictly decreasing on $(0^\circ, 180^\circ)$. So equal cosines imply equal angles. Therefore: $A = B = C = 60^\circ.$ …

   🔒 Start your 14-day free trial to see the full solution →
10. 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 $e^{2x}$ $$$$

    🔒 Start your 14-day free trial to see the full solution →
11. 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 $\tan 2A = \frac{12}{5}$ $\sin 4A = \frac{2 \cdot \frac{12}{5}}{1 + \left(\frac{12}{5}\right)^2} = \frac{\frac{24}{5}}{1 + \frac{144}{25}} = \frac{\frac{24}{5}}{\frac{25}{25} + \frac{144}{25}} = \frac{\frac{24}{5}}{\frac{169}{25}} = \frac{24}{5} \cdot \frac{25}{169} = \frac{24 \cdot 5}{169} = \frac{120}{169}.$ …

    🔒 Start your 14-day free trial to see the full solution →
12. TG EAPCET 2023Set eng-2023-05-13-AN1 markMCQ
    Q.If $\tan \alpha = \dfrac{-12}{5}$, $\cot \beta = \dfrac{7}{24}$, $\alpha$ does not belong to second quadrant and $\beta$ does not belong to first quadrant, then $\sqrt{13} \sin \dfrac{\alpha}{2} + \cos \dfrac{\beta}{2} + \tan \dfrac{\alpha}{2} \cot \dfrac{\beta}{2} =$ (A) $\dfrac{31}{10}$ (B) $\dfrac{19}{10}$ (C) $\dfrac{21}{10}$ (D) $\dfrac{-9}{10}$

    ›Reveal solutionSolution

    Quadrant analysis puts $\alpha$ in Q4 and $\beta$ in Q3; the half-angle values give $2-\tfrac{3}{5}+\tfrac12=\tfrac{19}{10}$.

    Given $\tan\alpha=-\dfrac{12}{5}$ with $\alpha$ not in Q2, the only quadrant with $\tan<0$ left is Q4, so $\cos\alpha=\dfrac{5}{13}$ and $\alpha\in(\tfrac{3\pi}{2},2\pi)\Rightarrow\dfrac{\alpha}{2}\in(\tfrac{3\pi}{4},\pi)$ (Q2).

    $\sin\dfrac{\alpha}{2}=\sqrt{\dfrac{1-\cos\alpha}{2}}=\sqrt{\dfrac{4}{13}}=\dfrac{2}{\sqrt{13}},\quad \cos\dfrac{\alpha}{2}=-\sqrt{\dfrac{1+\cos\alpha}{2}}=-\dfrac{3}{\sqrt{13}},\quad \tan\dfrac{\alpha}{2}=-\dfrac{2}{3}.$

    Given $\cot\beta=\dfrac{7}{24}$ (so $\tan\beta=\dfrac{24}{7}$) with $\beta$ not in Q1, take $\beta$ in Q3, so $\cos\beta=-\dfrac{7}{25}$ and $\beta\in(\pi,\tfrac{3\pi}{2})\Rightarrow\dfrac{\beta}{2}\in(\tfrac{\pi}{2},\tfrac{3\pi}{4})$ (Q2). …

    🔒 Start your 14-day free trial to see the full solution →