Q.Find the value of the following: $\cot^{-1} \left(\frac{\sqrt{1+\sin x} + \sqrt{1-\sin x}}{\sqrt{1+\sin x} - \sqrt{1-\sin x}}\right) = \frac{x}{2}$, $x \in \left(0, \frac{\pi}{4}\right)$

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:

      $$$$

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2. 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 …

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3. 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 $$$$

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4. 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: …

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5. 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:

   $$$$

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6. 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. …

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

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

   [!FORMULA] $\frac{\sinh(x+y) + \sinh(x-y)}{\cosh(x+y) - \cosh(x-y)} =$

   (A) $\tan hy$ (B) $\cot hy$ (C) $\tan hx \cot hy$ (D) $\tan hy \cot hx$

   ›Reveal solutionSolution

   We simplify the expression by applying the sum and difference formulas for hyperbolic sine and cosine, leading to a cancellation of terms. The final simplified expression is $\boxed{\coth y}$.

   The problem asks us to simplify a fraction involving sums and differences of hyperbolic functions. The most direct way to approach this is to expand each term in the numerator and denominator using the standard sum and difference formulas for hyperbolic sine ($\sinh$) and hyperbolic cosine ($\cosh$). Once expanded, we can combine like terms and simplify the resulting fraction.

   1. Recall the sum and difference formulas for hyperbolic functions.

      These formulas are analogous to their trigonometric counterparts, but with some sign differences.

      $\sinh(A+B) = \sinh A \cosh B + \cosh A \sinh B$

      $\sinh(A-B) = \sinh A \cosh B - \cosh A \sinh B$

      $\cosh(A+B) = \cosh A \cosh B + \sinh A \sinh B$

      $\cosh(A-B) = \cosh A \cosh B - \sinh A \sinh B$

   2. Simplify the numerator: $\sinh(x+y) + \sinh(x-y)$.

      Substitute the formulas for $\sinh(x+y)$ and $\sinh(x-y)$:

   $\sinh(x+y) + \sinh(x-y) = (\sinh x \cosh y + \cosh x \sinh y) + (\sinh x \cosh y - \cosh x \sinh y)$

   Notice that the terms $\cosh x \sinh y$ and $-\cosh x \sinh y$ cancel each other out.
   

   $\sinh(x+y) + \sinh(x-y) = 2 \sinh x \cosh y$

   3. Simplify the denominator: $\cosh(x+y) - \cosh(x-y)$. Substitute the formulas for $\cosh(x+y)$ and $\cosh(x-y)$:

   $\cosh(x+y) - \cosh(x-y) = (\cosh x \cosh y + \sinh x \sinh y) - (\cosh x \cosh y - \sinh x \sinh y)$

   > [!WARNING]
   > Be careful with the negative sign when subtracting the second expression. It changes the sign of both terms inside the parenthesis.
   

   $\cosh(x+y) - \cosh(x-y) = \cosh x \cosh y + \sinh x \sinh y - \cosh x \cosh y + \sinh x \sinh y$

   Here, the terms $\cosh x \cosh y$ and $-\cosh x \cosh y$ cancel each other out. …
   

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9. TG EAPCET 2021Set eng-2021-08-06-FN1 markMCQ
   Q.If $\cot A = \dfrac{11}{60}$, $\cos B = \dfrac{7}{25}$ and neither $A$ nor $B$ is in the first quadrant, then $\left(A + \dfrac{B}{2}\right)$ lies in the quadrant (A) I (B) II (C) III (D) IV

   ›Reveal solutionSolution

   Both angles are pinned to a single quadrant by the sign data, and $A+\tfrac{B}{2}\approx 42.7^\circ$ lands in Quadrant I.

   Locate $A$. $\cot A=\dfrac{11}{60}>0$, so $A$ is in Q1 or Q3. Since $A$ is not in the first quadrant, $A$ is in Q3, where $\sin A,\cos A<0$. The reference triangle has legs $11,60$ and hypotenuse $61$, giving $A=180^\circ+\arctan\dfrac{60}{11}\approx 259.6^\circ$. …

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10. 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 $xy$. …

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11. TG EAPCET 2024Set eng-2024-05-10-FN1 markMCQ
    Q.If $\tan A + \tan B + \cot A + \cot B = \tan A \tan B - \cot A \cot B$ and $0^\circ < A+B < 270^\circ$, then $A+B=$ (A) $45^\circ$ (B) $135^\circ$ (C) $150^\circ$ (D) $225^\circ$

    ›Reveal solutionSolution

    The given equation simplifies to $\tan(A+B) = -1$, and with the constraint $0^\circ < A+B < 270^\circ$, the only possible value is $135^\circ$.

    We start with the equation:

    $\tan A + \tan B + \cot A + \cot B = \tan A \tan B - \cot A \cot B.$

    The key insight is to rewrite everything in terms of $\tan A$ and $\tan B$, because $\cot x = \frac{1}{\tan x}$. This lets us combine terms and eventually use the tangent addition formula.

    1. Rewrite cotangents Let $x = \tan A$, $y = \tan B$. Then $\cot A = \frac{1}{x}$, $\cot B = \frac{1}{y}$. The equation becomes:

    $x + y + \frac{1}{x} + \frac{1}{y} = xy - \frac{1}{xy}.$

    2. Combine terms on each side

       Left side: $x + y + \frac{x+y}{xy} = (x+y)\left(1 + \frac{1}{xy}\right)$.

       Right side: $xy - \frac{1}{xy} = \frac{x^2 y^2 - 1}{xy}$.

       So we have:

    $(x+y)\left(1 + \frac{1}{xy}\right) = \frac{x^2 y^2 - 1}{xy}.$

    3. Multiply through by $xy$ (valid since $A,B$ not multiples of $90^\circ$, so $x,y \neq 0$):

    $(x+y)(xy + 1) = x^2 y^2 - 1.$

    4. Expand and simplify

       Left: $(x+y)(xy+1) = x^2 y + x + x y^2 + y$.

       Right: $x^2 y^2 - 1$.

       Bring all to one side:

    $x^2 y + x + x y^2 + y - x^2 y^2 + 1 = 0.$

    Group terms: $(x^2 y + x y^2) + (x + y) - x^2 y^2 + 1 = 0$.

    Factor $xy$ from the first group: $xy(x+y) + (x+y) - x^2 y^2 + 1 = 0$.

    So $(x+y)(xy+1) - (x^2 y^2 - 1) = 0$, which is just our earlier equation rearranged — we need a different grouping.

    5. Better grouping Write as:

    $x^2 y + x y^2 + x + y - x^2 y^2 + 1 = 0.$

    Rearrange: $(x^2 y - x^2 y^2) + (x y^2 + y) + (x + 1) = 0$

    Factor $x^2 y(1 - y) + y(x y + 1) + (x+1) = 0$ — not neat.

    Instead, notice the symmetric structure: try adding $1$ to both sides of the original simplified equation? Let's go back.

    6. A cleaner algebraic path From step 3: $(x+y)(xy+1) = x^2 y^2 - 1$. Notice $x^2 y^2 - 1 = (xy - 1)(xy + 1)$. So:

    $(x+y)(xy+1) = (xy - 1)(xy + 1).$

    If $xy+1 \neq 0$, we can divide both sides by it:

    $x + y = xy - 1.$

    This is much simpler!

    7. Interpret the result Recall $x = \tan A$, $y = \tan B$. So:

    $\tan A + \tan B = \tan A \tan B - 1.$

    Rearranging:

    $$$$

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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: …

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