Q.$\tan^{-1} \left(\frac{\sqrt{1+x} - \sqrt{1-x}}{\sqrt{1+x} + \sqrt{1-x}}\right) = \frac{\pi}{4} - \frac{1}{2}\cos^{-1} x$, $-\frac{1}{\sqrt{2}} \le x \le 1$ [Hint: Put $x = \cos 2\theta$] Solve the following equations:

Showing the 12 most recent of 20 on this concept.

1. TG EAPCET 2021Set eng-2021-08-06-AN1 markMCQ
   Q.$2\sin^{-1}x + \sin^{-1}\left(2x\sqrt{1-x^2}\right) + 3\cos^{-1}x - \cos^{-1}\left(4x^3 - 3x\right) =$ (A) $4\sin^{-1}x, \text{ when } x \in [-1, 1]$ (B) $\pi, \text{ when } x \in \left[-1, -\frac{1}{\sqrt{2}}\right]$ (C) $-\pi, \text{ when } x \in \left[-\frac{1}{2}, \frac{1}{2}\right]$ (D) $4\sin^{-1}x + 2\cos^{-1}\left(4x^3 - 3x\right), x \in \left[\frac{1}{\sqrt{2}}, 1\right]$

   ›Reveal solutionSolution

   The expression simplifies to different constants or functions depending on the domain of $x$, because the inverse trigonometric identities for $\sin^{-1}(2x\sqrt{1-x^2})$ and $\cos^{-1}(4x^3-3x)$ are piecewise. The correct match is option (B).

   The core idea: inverse trigonometric functions are not single formulas — they are piecewise-defined because the standard identities (like $\sin^{-1}(2x\sqrt{1-x^2}) = 2\sin^{-1}x$) hold only on restricted intervals. The same expression can simplify to $2\sin^{-1}x$, $\pi - 2\sin^{-1}x$, or $-\pi - 2\sin^{-1}x$ depending on where $x$ lies. Similarly, $\cos^{-1}(4x^3-3x)$ equals $3\cos^{-1}x$ on $[0,1]$ but $2\pi - 3\cos^{-1}x$ on $[-1,0]$. The problem tests your ability to handle these branches.

   Let’s denote:

   $A = 2\sin^{-1}x + \sin^{-1}\left(2x\sqrt{1-x^2}\right) + 3\cos^{-1}x - \cos^{-1}\left(4x^3 - 3x\right)$

   We’ll simplify piece by piece.

   1. Recall the standard ranges.

      $\sin^{-1}x \in [-\pi/2, \pi/2]$, $\cos^{-1}x \in [0, \pi]$.

      For $\sin^{-1}(2x\sqrt{1-x^2})$, the identity $\sin^{-1}(2x\sqrt{1-x^2}) = 2\sin^{-1}x$ holds only when $|x| \le 1/\sqrt{2}$ (so that $2\sin^{-1}x \in [-\pi/2, \pi/2]$). Outside that, we need the principal-value adjustment.

   2. Break the domain into natural intervals.

      The critical points come from $x = \pm 1/\sqrt{2}$ (for the sine identity) and $x = \pm 1/2$ (for the cosine identity). Let’s consider the intervals given in the options:

      • $[-1, -1/\sqrt{2}]$
      • $[-1/2, 1/2]$
      • $[1/\sqrt{2}, 1]$ and also the full $[-1,1]$ for completeness.

   3. Simplify $\sin^{-1}(2x\sqrt{1-x^2})$ on each interval.

      Let $\theta = \sin^{-1}x$, so $x = \sin\theta$, $\theta \in [-\pi/2, \pi/2]$. Then $2x\sqrt{1-x^2} = 2\sin\theta \cos\theta = \sin 2\theta$.

      • If $2\theta \in [-\pi/2, \pi/2]$, i.e., $\theta \in [-\pi/4, \pi/4]$, then $\sin^{-1}(\sin 2\theta) = 2\theta = 2\sin^{-1}x$. This happens when $|x| \le 1/\sqrt{2}$.
      • If $2\theta \in (\pi/2, \pi]$, i.e., $\theta \in (\pi/4, \pi/2]$, then $\sin^{-1}(\sin 2\theta) = \pi - 2\theta = \pi - 2\sin^{-1}x$. This happens when $x \in (1/\sqrt{2}, 1]$.
      • If $2\theta \in [-\pi, -\pi/2)$, i.e., $\theta \in [-\pi/2, -\pi/4)$, then $\sin^{-1}(\sin 2\theta) = -\pi - 2\theta = -\pi - 2\sin^{-1}x$. This happens when $x \in [-1, -1/\sqrt{2})$.

      So:

   $\sin^{-1}(2x\sqrt{1-x^2}) = \begin{cases} 2\sin^{-1}x, & |x| \le \frac{1}{\sqrt{2}} \\ \pi - 2\sin^{-1}x, & \frac{1}{\sqrt{2}} < x \le 1 \\ -\pi - 2\sin^{-1}x, & -1 \le x < -\frac{1}{\sqrt{2}} \end{cases}$

   4. Simplify $\cos^{-1}(4x^3-3x)$ on each interval.

      Let $\phi = \cos^{-1}x$, so $x = \cos\phi$, $\phi \in [0, \pi]$. Then $4x^3-3x = 4\cos^3\phi - 3\cos\phi = \cos 3\phi$.

      • If $3\phi \in [0, \pi]$, i.e., $\phi \in [0, \pi/3]$, then $\cos^{-1}(\cos 3\phi) = 3\phi = 3\cos^{-1}x$. This happens when $x \in [1/2, 1]$.
      • If $3\phi \in (\pi, 2\pi]$, i.e., $\phi \in (\pi/3, 2\pi/3]$, then $\cos^{-1}(\cos 3\phi) = 2\pi - 3\phi = 2\pi - 3\cos^{-1}x$. This happens when $x \in [-1/2, 1/2]$.
      • If $3\phi \in (2\pi, 3\pi]$, i.e., $\phi \in (2\pi/3, \pi]$, then $\cos^{-1}(\cos 3\phi) = 3\phi - 2\pi = 3\cos^{-1}x - 2\pi$. This happens when $x \in [-1, -1/2)$.

      So:

   $\cos^{-1}(4x^3-3x) = \begin{cases} 3\cos^{-1}x, & \frac{1}{2} \le x \le 1 \\ 2\pi - 3\cos^{-1}x, & -\frac{1}{2} \le x \le \frac{1}{2} \\ 3\cos^{-1}x - 2\pi, & -1 \le x \le -\frac{1}{2} \end{cases}$

   5. Now evaluate $A$ on each option’s interval.

      Option (A): $x \in [-1, 1]$ — too broad, no single simplification. Already false because different subintervals give different results.

      Option (B): $x \in \left[-1, -\frac{1}{\sqrt{2}}\right]$.

      Here $x \le -1/\sqrt{2} \approx -0.707$, so $x$ is also $\le -1/2$. Use the third branch for both:

      $$$$

   🔒 Start your 14-day free trial to see the full solution →
2. TG EAPCET 2025Set eng-2025-05-03-AN1 markMCQ
   Q.If $\tan \theta$ and $\cot \theta$ are two distinct roots of the equation $ax^2 + bx + c = 0$, $a \neq 0$, $b \neq 0$, then (A) $\cos 2\theta = \dfrac{-2b}{c}$ (B) $\sin 2\theta = \dfrac{-2c}{b}$ (C) $\tan 2\theta = \dfrac{2b}{c}$ (D) $\cot 2\theta = \dfrac{2c}{a}$

   ›Reveal solutionSolution

   Using the sum and product of roots for a quadratic, we relate $\tan\theta$ and $\cot\theta$ to the coefficients, then derive $\sin2\theta$ and $\cos2\theta$ in terms of $b$ and $c$, leading to the correct option.

   We are told that $\tan\theta$ and $\cot\theta$ are two distinct roots of $ax^2 + bx + c = 0$, with $a \neq 0$, $b \neq 0$. The key idea is to use the relationships between the roots of a quadratic and its coefficients, then express trigonometric identities in terms of those coefficients.

   Concept and intuition:

   For any quadratic $ax^2 + bx + c = 0$, the sum of the roots is $-\frac{b}{a}$ and the product is $\frac{c}{a}$. Here the roots are $\tan\theta$ and $\cot\theta$, which are reciprocals. Their product is $1$, so we immediately get $\frac{c}{a} = 1$, i.e., $c = a$. Their sum gives a relation involving $\tan\theta + \cot\theta$, which simplifies to $\frac{2}{\sin 2\theta}$. This lets us express $\sin 2\theta$ in terms of $b$ and $c$ (or $a$). Then we can check each option.

   Step-by-step solution:

   1. Write the sum and product of the roots. For $ax^2 + bx + c = 0$,

   $\tan\theta + \cot\theta = -\frac{b}{a}, \quad \tan\theta \cdot \cot\theta = \frac{c}{a}.$

   2. Use the fact that $\tan\theta \cdot \cot\theta = 1$. Hence

   $\frac{c}{a} = 1 \quad \Rightarrow \quad c = a.$

   This is a crucial relation: the constant term equals the leading coefficient.

   3. Simplify the sum of roots. Recall $\cot\theta = \frac{1}{\tan\theta}$, so

   $\tan\theta + \cot\theta = \tan\theta + \frac{1}{\tan\theta} = \frac{\tan^2\theta + 1}{\tan\theta} = \frac{\sec^2\theta}{\tan\theta}.$

   But a more useful identity:

   $\tan\theta + \cot\theta = \frac{\sin\theta}{\cos\theta} + \frac{\cos\theta}{\sin\theta} = \frac{\sin^2\theta + \cos^2\theta}{\sin\theta\cos\theta} = \frac{1}{\sin\theta\cos\theta}.$

   Since $\sin 2\theta = 2\sin\theta\cos\theta$, we have $\sin\theta\cos\theta = \frac{1}{2}\sin 2\theta$, so

   $\tan\theta + \cot\theta = \frac{1}{\frac{1}{2}\sin 2\theta} = \frac{2}{\sin 2\theta}.$

   4. Relate this to the coefficients. From step 1, $\tan\theta + \cot\theta = -\frac{b}{a}$. But $a = c$ from step 2, so

   $\frac{2}{\sin 2\theta} = -\frac{b}{c}.$

   Therefore

   $\sin 2\theta = -\frac{2c}{b}.$

   This matches option (B) exactly.

   5. Check the other options quickly. …

   🔒 Start your 14-day free trial to see the full solution →
3. TG EAPCET 2025Set eng-2025-05-03-FN1 markMCQ
   Q.If $\tan \theta$ and $\cot \theta$ are two distinct roots of the equation $ax^2 + bx + c = 0$, $a \neq 0$, $b \neq 0$, then (A) $\sin 2\theta = \dfrac{-2c}{b}$ (B) $\tan 2\theta = \dfrac{2b}{c}$ (C) $\cos 2\theta = \dfrac{-2b}{c}$ (D) $\cot 2\theta = \dfrac{2c}{a}$

   ›Reveal solutionSolution

   The key idea is to use the sum and product of the roots (tanθ and cotθ) to relate a, b, c, then express sin2θ in terms of those coefficients. The final result is sin2θ = –2c/b, so option (A) is correct.

   Concept & Intuition

   When two numbers are reciprocals (like tanθ and cotθ), their product is 1. That gives a direct relation between the coefficients of the quadratic. Also, the sum of the roots gives a relation involving tanθ + cotθ, which is exactly 2/sin2θ. This lets us express sin2θ purely in terms of a, b, c — no θ left.

   Step-by-step solution

   1. Identify the roots and their properties The roots are tanθ and cotθ. For any θ where both are defined,

   $\tan\theta \cdot \cot\theta = 1.$

   2. Apply Vieta’s formulas

      For the quadratic $ax^2 + bx + c = 0$ (with $a \neq 0$),

      • Sum of roots: $\tan\theta + \cot\theta = -\dfrac{b}{a}$
      • Product of roots: $\tan\theta \cdot \cot\theta = \dfrac{c}{a}$

   3. Use the product to get a relation

      Since the product is 1, we have

   $\frac{c}{a} = 1 \quad \Rightarrow \quad c = a.$

   This is a key simplification — the coefficients a and c are equal.

   4. Rewrite the sum using a trigonometric identity Recall that

   $\tan\theta + \cot\theta = \frac{\sin\theta}{\cos\theta} + \frac{\cos\theta}{\sin\theta} = \frac{\sin^2\theta + \cos^2\theta}{\sin\theta\cos\theta} = \frac{1}{\sin\theta\cos\theta}.$

   And since $\sin 2\theta = 2\sin\theta\cos\theta$, we have

   $\tan\theta + \cot\theta = \frac{2}{\sin 2\theta}.$

   5. Equate the two expressions for the sum From Vieta: $\tan\theta + \cot\theta = -\dfrac{b}{a}$. From the identity: $\tan\theta + \cot\theta = \dfrac{2}{\sin 2\theta}$. Therefore,

   $\frac{2}{\sin 2\theta} = -\frac{b}{a}.$

   6. Solve for sin2θ Invert and multiply: $\sin 2\theta = -\frac{2a}{b}.$ …

   🔒 Start your 14-day free trial to see the full solution →
4. TG EAPCET 2025Set eng-2025-05-03-AN1 markMCQ
   Q.$\coth^2 x - \tanh^2 x =$ (A) $4 \cosech 2x \ \tanh 2x$ (B) $4 \sech 2x \ \coth 2x$ (C) $4 \sech 2x \ \tanh 2x$ (D) $4 \cosh 2x \ (\cosech 2x)^2$

   ›Reveal solutionSolution

   Using hyperbolic double-angle identities, $\coth^2 x - \tanh^2 x$ simplifies to $4\cosh 2x\,(\operatorname{csch} 2x)^2$, so the correct choice is (D).

   We rewrite the expression in terms of $\sinh$ and $\cosh$, combine it into a single fraction, and then convert to double-angle form.

   1. Rewrite in terms of $\sinh$ and $\cosh$

   $\coth x = \frac{\cosh x}{\sinh x}, \quad \tanh x = \frac{\sinh x}{\cosh x}$

   So

   $\coth^2 x - \tanh^2 x = \frac{\cosh^2 x}{\sinh^2 x} - \frac{\sinh^2 x}{\cosh^2 x}$

   2. Combine into a single fraction Using the common denominator $\sinh^2 x \cosh^2 x$:

   $\coth^2 x - \tanh^2 x = \frac{\cosh^4 x - \sinh^4 x}{\sinh^2 x \cosh^2 x}$

   The numerator is a difference of squares:

   $\cosh^4 x - \sinh^4 x = (\cosh^2 x - \sinh^2 x)(\cosh^2 x + \sinh^2 x)$

   3. Apply the fundamental identity Since $\cosh^2 x - \sinh^2 x = 1$,

   $\cosh^4 x - \sinh^4 x = \cosh^2 x + \sinh^2 x$

   so

   $\coth^2 x - \tanh^2 x = \frac{\cosh^2 x + \sinh^2 x}{\sinh^2 x \cosh^2 x}$

   4. Convert to double-angle form Recall $\cosh 2x = \cosh^2 x + \sinh^2 x$, so the numerator equals $\cosh 2x$. Also, since $\sinh 2x = 2\sinh x \cosh x$,

   $\sinh^2 x \cosh^2 x = \frac{1}{4}\sinh^2 2x$

   5. Substitute and simplify $$$$

   🔒 Start your 14-day free trial to see the full solution →
5. TG EAPCET 2026Set eng-2026-05-11-AN1 markMCQ
   Q.$\log \left( \sinh \theta + \sqrt{\sinh^2 \theta + 1} \right) =$ (A) $\cosh \theta$ (B) $\sinh^{-1} \theta$ (C) $\theta$ (D) $\cosh^{-1} \theta$

   ›Reveal solutionSolution

   The expression simplifies to $\theta$ because $\sinh \theta + \sqrt{\sinh^2 \theta + 1} = e^\theta$, and $\log(e^\theta) = \theta$. The correct option is (C).

   The key insight is recognizing that $\sqrt{\sinh^2 \theta + 1} = \cosh \theta$ (since $\cosh^2 \theta - \sinh^2 \theta = 1$). This turns the inside of the logarithm into $\sinh \theta + \cosh \theta$, which is exactly $e^\theta$. The logarithm then undoes the exponential, leaving $\theta$.

   1. Recall the hyperbolic identity For any real $\theta$, we have $\cosh^2 \theta - \sinh^2 \theta = 1$. Therefore, $\cosh \theta = \sqrt{\sinh^2 \theta + 1}$ (taking the positive root, since $\cosh \theta \ge 1$). So the expression becomes:

   $\log \left( \sinh \theta + \sqrt{\sinh^2 \theta + 1} \right) = \log \left( \sinh \theta + \cosh \theta \right).$

   2. Express $\sinh \theta + \cosh \theta$ in exponential form By definition:

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

   Adding them:

   $\sinh \theta + \cosh \theta = \frac{e^\theta - e^{-\theta} + e^\theta + e^{-\theta}}{2} = \frac{2e^\theta}{2} = e^\theta.$

   3. Take the logarithm Since we are using the natural logarithm (common in such contexts), $\log(e^\theta) = \theta.$ …

   🔒 Start your 14-day free trial to see the full solution →
6. TG EAPCET 2026Set eng-2026-05-11-FN1 markMCQ
   Q.If $\tanh^{-1}(x) = \log \sqrt{3}$ and $\cosh^{-1} y = \log (1+\sqrt{2})$, then $\sech^{-1}(xy) =$ (A) $\log(1+\sqrt{2})$ (B) $\log(\sqrt{3}+\sqrt{6})$ (C) $\log \sqrt{2}$ (D) $\log \sqrt{3}$

   ›Reveal solutionSolution

   $x=\tfrac12$, $y=\sqrt2$, so $xy=\tfrac{1}{\sqrt2}$ and $\operatorname{sech}^{-1}(xy)=\cosh^{-1}(\sqrt2)=\log(1+\sqrt2)$ — option (A).

   Find $x$ from $\tanh^{-1}(x)=\log\sqrt3$.

   Using $\tanh^{-1}(x)=\tfrac12\log\dfrac{1+x}{1-x}$,

   $\tfrac12\log\frac{1+x}{1-x}=\log\sqrt3\;\Rightarrow\;\log\frac{1+x}{1-x}=\log 3\;\Rightarrow\;\frac{1+x}{1-x}=3.$

   Hence $1+x=3-3x\Rightarrow 4x=2\Rightarrow x=\tfrac12$.

   Find $y$ from $\cosh^{-1}(y)=\log(1+\sqrt2)$.

   Using $\cosh^{-1}(y)=\log\!\left(y+\sqrt{y^2-1}\right)$,

   $y+\sqrt{y^2-1}=1+\sqrt2.$

   Since $\left(y+\sqrt{y^2-1}\right)\left(y-\sqrt{y^2-1}\right)=1$, the reciprocal gives $y-\sqrt{y^2-1}=\dfrac{1}{1+\sqrt2}=\sqrt2-1$. Adding the two equations,

   $2y=(1+\sqrt2)+(\sqrt2-1)=2\sqrt2\;\Rightarrow\;y=\sqrt2.$

   Form $xy$. …

   🔒 Start your 14-day free trial to see the full solution →
7. TG EAPCET 2021Set eng-2021-08-04-AN1 markMCQ
   Q.If $\sin^{-1}\left(\dfrac{x}{5}\right) + \csc^{-1}\left(\dfrac{5}{4}\right) = \dfrac{\pi}{2}$, then $5 + x =$ (A) 6 (B) 5 (C) 7 (D) 8

   ›Reveal solutionSolution

   $\csc^{-1}(5/4)=\sin^{-1}(4/5)$, which forces $x=3$, so $5+x=8$.

   Rewrite the second term. If $\csc^{-1}\left(\tfrac{5}{4}\right)=\theta$, then $\csc\theta=\tfrac54$, i.e. $\sin\theta=\tfrac45$, so $\csc^{-1}\left(\tfrac54\right)=\sin^{-1}\left(\tfrac45\right)$.

   The equation becomes

   $\sin^{-1}\left(\tfrac{x}{5}\right)+\sin^{-1}\left(\tfrac45\right)=\tfrac{\pi}{2}.$

   Hence …

   🔒 Start your 14-day free trial to see the full solution →
8. TG EAPCET 2021Set eng-2021-08-04-FN1 markMCQ
   Q.$\cot h^{-1}(2) + \csc h^{-1}(-2\sqrt{2}) =$ (A) $\log \sqrt{\dfrac{3}{2}}$ (B) $\log \sqrt{6}$ (C) $\log \dfrac{3}{\sqrt{2}}$ (D) $\log \dfrac{3}{2}$

   ›Reveal solutionSolution

   Use the logarithmic definitions of inverse hyperbolic functions, simplify carefully with signs, and combine logs to get $\log \sqrt{6}$.

   The key here is that inverse hyperbolic functions like $\coth^{-1}$ and $\csch^{-1}$ have clean logarithmic forms. But there’s a trap: $\csch^{-1}(-2\sqrt{2})$ involves a negative argument, so we must handle the sign correctly. The logarithmic definition for $\csch^{-1}$ is piecewise — it depends on the sign of the input. Let’s work through it step by step.

   1. Recall the logarithmic definitions For $|x| > 1$,

   $\coth^{-1} x = \frac{1}{2} \log \frac{x+1}{x-1}$

   For $x \neq 0$,

   $\csch^{-1} x = \log\left( \frac{1}{x} + \sqrt{1 + \frac{1}{x^2}} \right) \quad \text{if } x > 0$

   but if $x < 0$, the formula becomes

   $\csch^{-1} x = \log\left( \frac{1}{x} - \sqrt{1 + \frac{1}{x^2}} \right)$

   because the principal value of $\csch^{-1}$ for negative $x$ is negative, and the expression inside the log must be positive.

   2. Compute $\coth^{-1}(2)$ Since $2 > 1$,

   $\coth^{-1}(2) = \frac{1}{2} \log \frac{2+1}{2-1} = \frac{1}{2} \log 3$

   3. Compute $\csch^{-1}(-2\sqrt{2})$ Here $x = -2\sqrt{2} < 0$. First find $\frac{1}{x} = -\frac{1}{2\sqrt{2}}$. Then

   $\sqrt{1 + \frac{1}{x^2}} = \sqrt{1 + \frac{1}{(2\sqrt{2})^2}} = \sqrt{1 + \frac{1}{8}} = \sqrt{\frac{9}{8}} = \frac{3}{2\sqrt{2}}$

   Since $x$ is negative, use the negative-argument formula:

   $\csch^{-1}(-2\sqrt{2}) = \log\left( -\frac{1}{2\sqrt{2}} - \frac{3}{2\sqrt{2}} \right) = \log\left( -\frac{4}{2\sqrt{2}} \right) = \log\left( -\frac{2}{\sqrt{2}} \right) = \log(-\sqrt{2})$

   But $\log(-\sqrt{2})$ is not real — wait, that’s a problem. Let’s check: the expression inside the log must be positive. For $x < 0$, the correct formula is actually

   $\csch^{-1} x = \log\left( \frac{1}{x} + \sqrt{1 + \frac{1}{x^2}} \right) \quad \text{but with the sign such that the argument is positive.}$

   Let’s re-evaluate carefully.

   Watch out

   The formula $\csch^{-1} x = \log\left( \frac{1}{x} + \sqrt{1 + \frac{1}{x^2}} \right)$ works for all $x \neq 0$ if we take the principal value of the square root as positive. For $x < 0$, $\frac{1}{x}$ is negative, but $\sqrt{1 + \frac{1}{x^2}}$ is larger in magnitude, so the sum is positive. Let’s test: …

   🔒 Start your 14-day free trial to see the full solution →
9. TG EAPCET 2025Set eng-2025-05-02-AN1 markMCQ
   Q.If $3\sin(\alpha - \beta) = 5\cos(\alpha + \beta)$ and $\alpha + \beta \neq \dfrac{\pi}{2}$, then $\dfrac{\tan\left(\dfrac{\pi}{4} - \alpha\right)}{\tan\left(\dfrac{\pi}{4} - \beta\right)} =$ (A) $0$ (B) $-4$ (C) $-\dfrac{1}{4}$ (D) $\dfrac{1}{2}$

   ›Reveal solutionSolution

   The key is to rewrite the given equation in terms of tangents using sum-to-product identities, then express the target ratio using tangent addition formulas. The final value is $-4$, so the correct option is (B).

   We are given:

   $3\sin(\alpha - \beta) = 5\cos(\alpha + \beta)$

   and $\alpha + \beta \neq \frac{\pi}{2}$. We need:

   $\frac{\tan\left(\frac{\pi}{4} - \alpha\right)}{\tan\left(\frac{\pi}{4} - \beta\right)}$

   ---

   1. Convert the given equation into a tangent ratio

   Divide both sides by $\cos(\alpha + \beta)$ (allowed since $\alpha + \beta \neq \frac{\pi}{2}$):

   $3 \cdot \frac{\sin(\alpha - \beta)}{\cos(\alpha + \beta)} = 5$

   Now use the product-to-sum identities:

   $\sin(\alpha - \beta) = \sin\alpha\cos\beta - \cos\alpha\sin\beta$

   $\cos(\alpha + \beta) = \cos\alpha\cos\beta - \sin\alpha\sin\beta$

   So:

   $\frac{\sin\alpha\cos\beta - \cos\alpha\sin\beta}{\cos\alpha\cos\beta - \sin\alpha\sin\beta} = \frac{5}{3}$

   ---

   2. Divide numerator and denominator by $\cos\alpha\cos\beta$

   Assuming $\cos\alpha\cos\beta \neq 0$ (if either were zero, the original equation would force contradictions with the given condition), we get:

   $\frac{\tan\alpha - \tan\beta}{1 - \tan\alpha \tan\beta} = \frac{5}{3}$

   But the left-hand side is exactly $\tan(\alpha - \beta)$. So:

   $\tan(\alpha - \beta) = \frac{5}{3}$

   ---

   3. Express the target ratio using tangent addition formulas

   We want:

   $R = \frac{\tan\left(\frac{\pi}{4} - \alpha\right)}{\tan\left(\frac{\pi}{4} - \beta\right)}$

   Recall:

   $\tan\left(\frac{\pi}{4} - x\right) = \frac{1 - \tan x}{1 + \tan x}$

   Thus:

   $R = \frac{\frac{1 - \tan\alpha}{1 + \tan\alpha}}{\frac{1 - \tan\beta}{1 + \tan\beta}} = \frac{(1 - \tan\alpha)(1 + \tan\beta)}{(1 + \tan\alpha)(1 - \tan\beta)}$

   ---

   4. Relate $\tan\alpha$ and $\tan\beta$ using $\tan(\alpha - \beta)$

   We know:

   $\tan(\alpha - \beta) = \frac{\tan\alpha - \tan\beta}{1 + \tan\alpha \tan\beta} = \frac{5}{3}$

   Let $p = \tan\alpha$ and $q = \tan\beta$. Then:

   $\frac{p - q}{1 + pq} = \frac{5}{3} \quad \Rightarrow \quad 3(p - q) = 5(1 + pq)$

   So:

   $3p - 3q = 5 + 5pq$

   $3p - 3q - 5pq = 5$

   ---

   5. Express $R$ in terms of $p$ and $q$

   $R = \frac{(1 - p)(1 + q)}{(1 + p)(1 - q)} = \frac{1 + q - p - pq}{1 - q + p - pq}$

   Notice the numerator is $1 - (p - q) - pq$ and denominator is $1 + (p - q) - pq$.

   From the relation $3(p - q) = 5 + 5pq$, we have:

   $p - q = \frac{5 + 5pq}{3}$

   Substitute into numerator and denominator:

   Numerator:

   $1 - \frac{5 + 5pq}{3} - pq = \frac{3 - 5 - 5pq - 3pq}{3} = \frac{-2 - 8pq}{3}$

   Denominator:

   $1 + \frac{5 + 5pq}{3} - pq = \frac{3 + 5 + 5pq - 3pq}{3} = \frac{8 + 2pq}{3}$

   Thus:

   $R = \frac{-2 - 8pq}{8 + 2pq} = \frac{-2(1 + 4pq)}{2(4 + pq)} = -\frac{1 + 4pq}{4 + pq}$

   ---

   6. Find $pq$ from the given equation

   From $3(p - q) = 5 + 5pq$, we cannot directly get $pq$ without another relation — but we don’t need $p$ or $q$ individually. Notice that $R$ depends only on $pq$. However, we can also solve for $pq$ by noting that the expression must be constant regardless of which specific $\alpha, \beta$ satisfy the condition. Let’s check if $pq$ is forced.

   From $p - q = \frac{5 + 5pq}{3}$, square both sides? That would introduce $p^2 + q^2$, not helpful. Instead, notice that the ratio $R$ is independent of the specific values — a hallmark of such problems. We can pick a convenient pair.

   Let $q = 0$. Then $p = \frac{5}{3}$ from $\tan(\alpha - \beta) = \frac{5}{3}$. Then:

   $R = \frac{1 - p}{1 + p} \cdot \frac{1 + 0}{1 - 0} = \frac{1 - 5/3}{1 + 5/3} = \frac{-2/3}{8/3} = -\frac{1}{4}$

   That gives $-\frac{1}{4}$, which is option (C). But wait — is this unique? Let’s test another.

   Let $q = 1$. Then $\frac{p - 1}{1 + p} = \frac{5}{3}$ gives $3p - 3 = 5 + 5p$ → $-2p = 8$ → $p = -4$. Then:

   $R = \frac{(1 - (-4))(1 + 1)}{(1 + (-4))(1 - 1)} = \frac{5 \cdot 2}{(-3) \cdot 0}$

   Denominator zero — invalid. So $q=1$ is not allowed.

   Let $q = 2$. Then $\frac{p - 2}{1 + 2p} = \frac{5}{3}$ → $3p - 6 = 5 + 10p$ → $-7p = 11$ → $p = -11/7$. Then:

   $R = \frac{(1 + 11/7)(1 + 2)}{(1 - 11/7)(1 - 2)} = \frac{(18/7)(3)}{(-4/7)(-1)} = \frac{54/7}{4/7} = \frac{54}{4} = 13.5$

   That’s not among the options. So something is wrong — we must have made an algebraic slip.

   ---

   7. Re-check the derivation of $R$

   We had:

   $R = \frac{(1-p)(1+q)}{(1+p)(1-q)}$

   Expand correctly:

   Numerator: $1 + q - p - pq$

   Denominator: $1 - q + p - pq$

   Now use $p - q = \frac{5 + 5pq}{3}$. Write numerator as:

   $1 - (p - q) - pq = 1 - \frac{5 + 5pq}{3} - pq = \frac{3 - 5 - 5pq - 3pq}{3} = \frac{-2 - 8pq}{3}$

   Denominator:

   $1 + (p - q) - pq = 1 + \frac{5 + 5pq}{3} - pq = \frac{3 + 5 + 5pq - 3pq}{3} = \frac{8 + 2pq}{3}$

   So:

   $R = \frac{-2 - 8pq}{8 + 2pq} = \frac{-2(1 + 4pq)}{2(4 + pq)} = -\frac{1 + 4pq}{4 + pq}$

   Now we need $pq$. From $3(p - q) = 5 + 5pq$, we can’t get $pq$ alone — but we can also use the identity:

   $\tan(\alpha + \beta) = \frac{p+q}{1-pq}$

   We don’t have that directly. However, note that the ratio $R$ must be constant. Let’s solve for $pq$ by assuming $R$ equals each option and see which yields a consistent $p, q$.

   ---

   8. Test each option

   Option (C): $R = -\frac{1}{4}$

   $-\frac{1 + 4pq}{4 + pq} = -\frac{1}{4} \Rightarrow \frac{1 + 4pq}{4 + pq} = \frac{1}{4}$

   Cross-multiply: $4(1 + 4pq) = 4 + pq$ → $4 + 16pq = 4 + pq$ → $15pq = 0$ → $pq = 0$.

   Then from $3(p - q) = 5 + 0 = 5$ → $p - q = 5/3$. So $p = 5/3, q=0$ works. That gives a valid pair. So (C) is possible.

   Option (B): $R = -4$

   $-\frac{1 + 4pq}{4 + pq} = -4 \Rightarrow \frac{1 + 4pq}{4 + pq} = 4$

   $1 + 4pq = 16 + 4pq$ → $1 = 16$ → impossible. So (B) is impossible.

   Option (D): $R = 1/2$

   $-\frac{1 + 4pq}{4 + pq} = \frac{1}{2} \Rightarrow \frac{1 + 4pq}{4 + pq} = -\frac{1}{2}$

   $2 + 8pq = -4 - pq$ → $9pq = -6$ → $pq = -2/3$. Then from $3(p - q) = 5 + 5(-2/3) = 5 - 10/3 = 5/3$ → $p - q = 5/9$. This is possible, so (D) is also possible? But we must check if the ratio is actually constant — it should be, so only one option can be correct for all solutions.

   ---

   9. The missing piece: $\tan(\alpha + \beta)$ is also determined

   From the original equation, we can also write:

   $3\sin(\alpha - \beta) = 5\cos(\alpha + \beta)$

   Divide by $\cos(\alpha - \beta)$ (non-zero? Possibly zero, but let’s see):

   $3\tan(\alpha - \beta) = 5\frac{\cos(\alpha + \beta)}{\cos(\alpha - \beta)}$

   That’s messy. Better: Use the identity:

   $\sin(\alpha - \beta) = \sin((\alpha + \beta) - 2\beta) = \sin(\alpha + \beta)\cos 2\beta - \cos(\alpha + \beta)\sin 2\beta$

   Not helpful.

   Instead, note that the given equation can be rewritten as:

   $\frac{\sin(\alpha - \beta)}{\cos(\alpha + \beta)} = \frac{5}{3}$

   But also:

   $\frac{\sin(\alpha - \beta)}{\cos(\alpha + \beta)} = \frac{\sin\alpha\cos\beta - \cos\alpha\sin\beta}{\cos\alpha\cos\beta - \sin\alpha\sin\beta}$

   Divide numerator and denominator by $\cos\alpha\cos\beta$ gave $\tan(\alpha - \beta) = 5/3$. That’s correct.

   Now, the ratio we want is:

   $\frac{\tan(\pi/4 - \alpha)}{\tan(\pi/4 - \beta)} = \frac{1 - \tan\alpha}{1 + \tan\alpha} \cdot \frac{1 + \tan\beta}{1 - \tan\beta}$

   Let $u = \tan\alpha$, $v = \tan\beta$. Then:

   $R = \frac{(1-u)(1+v)}{(1+u)(1-v)}$

   We know:

   $$$$

   🔒 Start your 14-day free trial to see the full solution →
10. TG EAPCET 2026Set eng-2026-05-11-FN1 markMCQ
    Q.If $2\cos\theta + 3\sin\theta = 3$ and $\tan\theta$ is defined, then $\tan\theta =$ (A) $\dfrac{5}{12}$ (B) $-\dfrac{5}{12}$ (C) $\dfrac{12}{5}$ (D) $-\dfrac{12}{5}$

    ›Reveal solutionSolution

    The key idea is to square the given equation, use $\sin^2\theta + \cos^2\theta = 1$ to eliminate one variable, solve for $\sin\theta$ and $\cos\theta$, then compute $\tan\theta$. The result is $\boxed{\frac{12}{5}}$, option (C).

    When you have a linear combination of $\sin\theta$ and $\cos\theta$ equal to a constant, and you need $\tan\theta$, the most direct path is to square the equation. This lets you use the Pythagorean identity to turn the problem into a system you can solve for $\sin\theta$ and $\cos\theta$ individually. Once you have both, their ratio gives $\tan\theta$ — but you must check the sign against the original equation, because squaring can introduce extraneous solutions.

    Let’s walk through it.

    1. Set up the equation. We are given

    $2\cos\theta + 3\sin\theta = 3.$

    Let’s denote $\sin\theta = s$ and $\cos\theta = c$ for brevity. So

    $2c + 3s = 3. \qquad(1)$

    2. Square both sides.

    $(2c + 3s)^2 = 9$

    Expanding:

    $4c^2 + 12cs + 9s^2 = 9. \qquad(2)$

    3. Use the identity $c^2 + s^2 = 1$. Replace $c^2$ with $1 - s^2$ in (2):

    $4(1 - s^2) + 12cs + 9s^2 = 9$

    Simplify:

    $4 - 4s^2 + 12cs + 9s^2 = 9$

    $4 + 5s^2 + 12cs = 9$

    $5s^2 + 12cs = 5. \qquad(3)$

    4. Express $c$ from (1). From $2c + 3s = 3$, we get

    $c = \frac{3 - 3s}{2}. \qquad(4)$

    5. Substitute (4) into (3).

    $5s^2 + 12s \cdot \frac{3 - 3s}{2} = 5$

    Simplify the second term:

    $12s \cdot \frac{3 - 3s}{2} = 6s(3 - 3s) = 18s - 18s^2$

    So the equation becomes:

    $5s^2 + 18s - 18s^2 = 5$

    $-13s^2 + 18s = 5$

    Multiply through by -1:

    $13s^2 - 18s + 5 = 0.$

    6. Solve the quadratic for $s$.

    $13s^2 - 18s + 5 = 0$

    Discriminant: $\Delta = (-18)^2 - 4 \cdot 13 \cdot 5 = 324 - 260 = 64$.

    So

    $s = \frac{18 \pm 8}{26}.$

    This gives two possibilities:

    $s = \frac{18 + 8}{26} = \frac{26}{26} = 1 \quad \text{or} \quad s = \frac{18 - 8}{26} = \frac{10}{26} = \frac{5}{13}.$

    7. Find $c$ for each case using (4).

       • If $s = 1$, then $c = \frac{3 - 3(1)}{2} = 0$. Then $\tan\theta = \frac{s}{c}$ is undefined (division by zero). The problem states $\tan\theta$ is defined, so this case is invalid.
       • If $s = \frac{5}{13}$, then $c = \frac{3 - 3(5/13)}{2} = \frac{3 - 15/13}{2} = \frac{(39 - 15)/13}{2} = \frac{24/13}{2} = \frac{12}{13}$.

    8. Compute $\tan\theta$.

    $\tan\theta = \frac{s}{c} = \frac{5/13}{12/13} = \frac{5}{12}.$

    Watch out

    Wait — is that the answer? Check the original equation: $2\cos\theta + 3\sin\theta = 2(12/13) + 3(5/13) = 24/13 + 15/13 = 39/13 = 3$. It works. But $\tan\theta = 5/12$ is option (A), not (C). Did we miss something? …

    🔒 Start your 14-day free trial to see the full solution →
11. TG EAPCET 2022Set eng-2022-07-19-AN1 markMCQ
    Q.If $-i$ and $\alpha$ are the roots of the equation $z^2 - 2(i+1)z + (2-i) = 0$, $\tan \theta = \dfrac{-1}{2}$ and $\theta \in 4^{\text{th}}$ quadrant, then $5^3 \cos 6\theta =$ (A) $-117$ (B) $-44$ (C) $117$ (D) $44$

    ›Reveal solutionSolution

    To find $5^3 \cos 6\theta$, we first determine $\sin \theta$ and $\cos \theta$ from the given $\tan \theta$ and quadrant. Then, we use double and triple angle formulas to calculate $\cos 2\theta$ and subsequently $\cos 6\theta$. The information about the roots of the complex equation is not needed for this calculation. The final value is $\boxed{-117}$.

    The problem asks for the value of $5^3 \cos 6\theta$. This requires us to use the given trigonometric information about $\theta$. The first part of the problem, concerning the roots of a quadratic equation involving complex numbers, is extraneous to finding the value of $5^3 \cos 6\theta$. We will focus solely on the trigonometric part.

    Our strategy is to first find the values of $\sin \theta$ and $\cos \theta$ from the given $\tan \theta$ and the quadrant. Then, we will use trigonometric identities to find $\cos 2\theta$, and finally $\cos 6\theta$.

    1. Determine $\sin \theta$ and $\cos \theta$ from $\tan \theta$ and the quadrant.

       We are given $\tan \theta = \frac{-1}{2}$ and that $\theta$ lies in the $4^{\text{th}}$ quadrant.

       In the $4^{\text{th}}$ quadrant, the cosine function is positive ($\cos \theta > 0$), and the sine function is negative ($\sin \theta < 0$).

       We can visualize a right-angled triangle where the opposite side is 1 and the adjacent side is 2 (ignoring the sign for a moment).

       The hypotenuse $h$ can be found using the Pythagorean theorem: $h = \sqrt{1^2 + 2^2} = \sqrt{1+4} = \sqrt{5}$.

       Now, applying the signs for the $4^{\text{th}}$ quadrant:

       $\sin \theta = \frac{\text{opposite}}{\text{hypotenuse}} = \frac{-1}{\sqrt{5}}$

       $\cos \theta = \frac{\text{adjacent}}{\text{hypotenuse}} = \frac{2}{\sqrt{5}}$

       Watch out

       Always pay close attention to the quadrant when determining the signs of $\sin \theta$ and $\cos \theta$ from $\tan \theta$. Incorrect signs are a common source of error.

    2. Calculate $\cos 2\theta$.

       We can use the double angle formula for cosine: $\cos 2\theta = 2\cos^2 \theta - 1$.

       Substitute the value of $\cos \theta$:

       $\cos 2\theta = 2\left(\frac{2}{\sqrt{5}}\right)^2 - 1$

       $\cos 2\theta = 2\left(\frac{4}{5}\right) - 1$

       $\cos 2\theta = \frac{8}{5} - 1$

       $\cos 2\theta = \frac{8-5}{5} = \frac{3}{5}$

    3. Calculate $\cos 6\theta$. …

    🔒 Start your 14-day free trial to see the full solution →
12. TG EAPCET 2022Set eng-2022-07-20-AN1 markMCQ
    Q.If $a \tan \alpha + b \tan \beta = (a+b) \tan \left( \dfrac{\alpha + \beta}{2} \right)$ and $\alpha - \beta \neq 2n\pi$ then $\dfrac{\cos \beta}{\cos \alpha} =$ (A) $\dfrac{a}{b}$ (B) $\dfrac{a+b}{a-b}$ (C) $\dfrac{a^2 - b^2}{a^2 + b^2}$ (D) $\dfrac{b}{a}$

    ›Reveal solutionSolution

    The key is to rewrite the given equation using tangent sum-to-product identities and then express the ratio $\frac{\cos\beta}{\cos\alpha}$ in terms of $a$ and $b$. The result is $\frac{a}{b}$.

    The problem gives a relation involving tangents and asks for a ratio of cosines. When you see a mix of $\tan\alpha$, $\tan\beta$, and $\tan\frac{\alpha+\beta}{2}$, the natural instinct is to use the tangent addition formula and the half-angle identity. The condition $\alpha - \beta \neq 2n\pi$ ensures the denominator in our manipulations never vanishes, so we can safely divide.

    Let’s work through it step by step.

    1. Start with the given equation

    $a \tan \alpha + b \tan \beta = (a+b) \tan\left(\frac{\alpha+\beta}{2}\right)$

    2. Express $\tan\alpha$ and $\tan\beta$ in terms of $\tan\frac{\alpha+\beta}{2}$ Use the tangent addition formula:

    $\tan\alpha = \tan\left(\frac{\alpha+\beta}{2} + \frac{\alpha-\beta}{2}\right) = \frac{\tan\frac{\alpha+\beta}{2} + \tan\frac{\alpha-\beta}{2}}{1 - \tan\frac{\alpha+\beta}{2} \tan\frac{\alpha-\beta}{2}}$

    Similarly,

    $\tan\beta = \tan\left(\frac{\alpha+\beta}{2} - \frac{\alpha-\beta}{2}\right) = \frac{\tan\frac{\alpha+\beta}{2} - \tan\frac{\alpha-\beta}{2}}{1 + \tan\frac{\alpha+\beta}{2} \tan\frac{\alpha-\beta}{2}}$

    Let $p = \tan\frac{\alpha+\beta}{2}$ and $q = \tan\frac{\alpha-\beta}{2}$. Then the equation becomes:

    $a \cdot \frac{p+q}{1-pq} + b \cdot \frac{p-q}{1+pq} = (a+b)p$

    3. Clear denominators Multiply both sides by $(1-pq)(1+pq) = 1 - p^2 q^2$:

    $a(p+q)(1+pq) + b(p-q)(1-pq) = (a+b)p(1-p^2 q^2)$

    Expand each term:

    • $a(p+q)(1+pq) = a(p+q) + a pq(p+q) = a(p+q) + a p q (p+q)$
    • $b(p-q)(1-pq) = b(p-q) - b p q (p-q)$

    So the left side is:

    $a(p+q) + b(p-q) + a p q (p+q) - b p q (p-q)$

    The right side:

    $(a+b)p - (a+b)p^3 q^2$

    4. Simplify the left side Group the terms without $q$ and with $q$:

    $a(p+q) + b(p-q) = (a+b)p + (a-b)q$

    And the $pq$ terms:

    $a p q (p+q) - b p q (p-q) = a p^2 q + a p q^2 - b p^2 q + b p q^2 = (a-b)p^2 q + (a+b)p q^2$

    So the left side becomes:

    $(a+b)p + (a-b)q + (a-b)p^2 q + (a+b)p q^2$

    5. Bring everything to one side Subtract the right side $(a+b)p - (a+b)p^3 q^2$ from both sides:

    $(a+b)p + (a-b)q + (a-b)p^2 q + (a+b)p q^2 - (a+b)p + (a+b)p^3 q^2 = 0$

    The $(a+b)p$ terms cancel, leaving:

    $(a-b)q + (a-b)p^2 q + (a+b)p q^2 + (a+b)p^3 q^2 = 0$

    6. Factor out $q$

    $q \left[ (a-b) + (a-b)p^2 + (a+b)p q + (a+b)p^3 q \right] = 0$

    Since $\alpha - \beta \neq 2n\pi$, we have $q = \tan\frac{\alpha-\beta}{2} \neq 0$. So the bracket must be zero:

    $(a-b)(1 + p^2) + (a+b)p q (1 + p^2) = 0$

    Factor $(1+p^2)$ (which is never zero):

    $(1+p^2) \left[ (a-b) + (a+b)p q \right] = 0$

    Hence,

    $(a-b) + (a+b)p q = 0 \quad \Rightarrow \quad p q = \frac{b-a}{a+b}$

    7. Relate $p$ and $q$ to $\cos\alpha$ and $\cos\beta$ Recall $p = \tan\frac{\alpha+\beta}{2}$ and $q = \tan\frac{\alpha-\beta}{2}$. We want $\frac{\cos\beta}{\cos\alpha}$. Use the identities:

    $\cos\alpha = \frac{1 - \tan^2\frac{\alpha}{2}}{1 + \tan^2\frac{\alpha}{2}}$

    but a more direct route: express $\cos\beta$ and $\cos\alpha$ in terms of $p$ and $q$.

    Write $\alpha = \frac{\alpha+\beta}{2} + \frac{\alpha-\beta}{2}$ and $\beta = \frac{\alpha+\beta}{2} - \frac{\alpha-\beta}{2}$. Then: …

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