Q.$\int \frac{\sin^2 x - \cos^2 x}{\sin^2 x \cos^2 x} dx$ is equal to (A) $\tan x + \cot x + C$ (B) $\tan x + \operatorname{cosec} x + C$ (C) $-\tan x + \cot x + C$ (D) $\tan x + \sec x + C$

Showing the 12 most recent of 34 on this concept.

1. TG EAPCET 2023Set eng-2023-05-12-FN1 markMCQ
   Q.If $\tan y = \cot \left( \frac{\pi}{4} - x \right)$ then $\frac{dy}{dx} =$ (A) $\frac{\csc^2 \left( \frac{\pi}{4} - x \right)}{1 + \cot^2 \left( \frac{\pi}{4} + x \right)}$ (B) $\frac{-\csc^2 \left( \frac{\pi}{4} - x \right)}{\sec^2 y}$ (C) $\frac{\csc^2 \left( \frac{\pi}{4} - x \right)}{1 + \tan^2 \left( \frac{\pi}{4} - x \right)}$ (D) $\frac{\sec^2 \left( \frac{\pi}{4} + x \right)}{1 + \tan^2 \left( \frac{\pi}{4} + x \right)}$

   ›Reveal solutionSolution

   Differentiate $\tan y=\cot(\tfrac\pi4-x)$ implicitly: $\sec^2y\,\dfrac{dy}{dx}=\csc^2(\tfrac\pi4-x)$. Since $\csc^2(\tfrac\pi4-x)=\sec^2(\tfrac\pi4+x)=\sec^2y$, the derivative is $1$, and option (D) is the form equal to $1$.

   1. Simplify the relation. Using $\cot\theta=\tan(\tfrac\pi2-\theta)$,

   $\cot\!\left(\frac\pi4-x\right)=\tan\!\left(\frac\pi2-\frac\pi4+x\right)=\tan\!\left(\frac\pi4+x\right),$

   so $\tan y=\tan(\tfrac\pi4+x)$, i.e. $y=\tfrac\pi4+x+n\pi$ and $\sec^2y=\sec^2(\tfrac\pi4+x)$.

   2. Differentiate implicitly. With $u=\tfrac\pi4-x$, $u'=-1$:

   $\sec^2y\,\frac{dy}{dx}=-\csc^2\!\left(\frac\pi4-x\right)\cdot(-1)=\csc^2\!\left(\frac\pi4-x\right).$

   3. Convert with a cofunction identity. Because $\sin(\tfrac\pi4-x)=\cos(\tfrac\pi4+x)$,

   $\csc^2\!\left(\frac\pi4-x\right)=\sec^2\!\left(\frac\pi4+x\right).$

   Therefore …

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2. TG EAPCET 2026Set eng-2026-05-10-AN1 markMCQ
   Q.If $x = \sin^{-1}(\cos t)$ and $y = \tan^{-1}(\cos t)$, then $\dfrac{dy}{dx} =$ (A) $\dfrac{\cos x}{1 + \sin^2 x}$ (B) $\dfrac{\cos t}{1 + \sin^2 t}$ (C) $\dfrac{\sin t}{1 + \sin^2 t}$ (D) $\dfrac{\sin x}{1 + \cos^2 x}$

   ›Reveal solutionSolution

   The key is to express both $x$ and $y$ in terms of $t$, then use the chain rule to find $\frac{dy}{dx}$ as a function of $x$, matching option (A).

   We are given:

   $x = \sin^{-1}(\cos t), \quad y = \tan^{-1}(\cos t).$

   We need $\frac{dy}{dx}$.

   Concept and intuition:

   Both $x$ and $y$ are functions of $t$, so we can compute $\frac{dy}{dt}$ and $\frac{dx}{dt}$ separately, then use $\frac{dy}{dx} = \frac{dy/dt}{dx/dt}$. The trick is to simplify $\sin^{-1}(\cos t)$ into a nicer algebraic form in terms of $t$, which will make differentiation cleaner. Then we express the final answer in terms of $x$ (since the options involve $x$ or $t$).

   ---

   1. Simplify $x$ Recall: $\sin^{-1}(\cos t) = \frac{\pi}{2} - t$ for $t$ in $[0, \pi]$, but more generally we can use the identity $\sin^{-1}(\cos t) = \frac{\pi}{2} - |t|$? Actually, a cleaner approach: Let $u = \cos t$. Then $x = \sin^{-1}(u)$. But we can also write $\cos t = \sin\left(\frac{\pi}{2} - t\right)$, so

   $x = \sin^{-1}\left(\sin\left(\frac{\pi}{2} - t\right)\right).$

   For the principal range of $\sin^{-1}$, we need $\frac{\pi}{2} - t \in [-\pi/2, \pi/2]$, i.e., $t \in [0, \pi]$. Assuming this, we get

   $x = \frac{\pi}{2} - t.$

   Hence, $t = \frac{\pi}{2} - x$.

   2. Differentiate $x$ and $y$ with respect to $t$ From $x = \frac{\pi}{2} - t$, we have

   $\frac{dx}{dt} = -1.$

   For $y = \tan^{-1}(\cos t)$, differentiate:

   $\frac{dy}{dt} = \frac{1}{1 + (\cos t)^2} \cdot (-\sin t) = -\frac{\sin t}{1 + \cos^2 t}.$

   3. Find $\frac{dy}{dx}$ Using the chain rule: $\frac{dy}{dx} = \frac{dy/dt}{dx/dt} = \frac{-\frac{\sin t}{1 + \cos^2 t}}{-1} = \frac{\sin t}{1 + \cos^2 t}.$ …

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3. TG EAPCET 2023Set eng-2023-05-14-FN1 markMCQ
   Q.If $x = \log(y + \sqrt{y^2 + 1})$ then $y =$ (A) $\tanh x$ (B) $\coth x$ (C) $\sinh x$ (D) $\cosh x$

   ›Reveal solutionSolution

   The expression $x = \log(y + \sqrt{y^2+1})$ is the definition of the inverse hyperbolic sine, so $y = \sinh x$. The correct option is (C).

   The key here is recognizing a standard identity. The expression $y + \sqrt{y^2+1}$ looks like something that appears when you solve for $y$ in terms of $x$ from the definition of hyperbolic sine. In fact, $\sinh x = \frac{e^x - e^{-x}}{2}$, and its inverse is exactly $\sinh^{-1} y = \log(y + \sqrt{y^2+1})$. So the problem is simply asking: if $x = \sinh^{-1} y$, what is $y$? The answer is $y = \sinh x$.

   Let’s verify this step by step.

   1. Start with the given equation

      We have $x = \log(y + \sqrt{y^2 + 1})$. This is an equation relating $x$ and $y$. Our goal is to solve for $y$ in terms of $x$.

   2. Exponentiate both sides

      Since the logarithm is natural log (base $e$), we write

   $e^x = y + \sqrt{y^2 + 1}.$

   This removes the log and gives a simpler equation.

   3. Consider the conjugate expression A classic trick: if $e^x = y + \sqrt{y^2+1}$, then its reciprocal is

   $e^{-x} = \frac{1}{y + \sqrt{y^2+1}}.$

   Rationalize the denominator:

   $e^{-x} = \frac{y - \sqrt{y^2+1}}{(y + \sqrt{y^2+1})(y - \sqrt{y^2+1})} = \frac{y - \sqrt{y^2+1}}{y^2 - (y^2+1)} = \frac{y - \sqrt{y^2+1}}{-1} = \sqrt{y^2+1} - y.$

   So we have two equations:

   $e^x = y + \sqrt{y^2+1}, \quad e^{-x} = \sqrt{y^2+1} - y.$

   4. Subtract to isolate $y$ Subtract the second equation from the first: $$$$

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4. TG EAPCET 2026Set eng-2026-05-10-FN1 markMCQ
   Q.If $y = \sin^{-1} \left( \frac{1 - \cos 2x}{1 + \sin^4 x} \right)$ then $\frac{dy}{dx} =$ (A) $\frac{2 \cos 2x}{1 + \sin^4 x}$ (B) $\frac{2 \sin 2x}{1 + \sin^4 x}$ (C) $\frac{2 \cos 2x}{1 + \sin^8 x}$ (D) $\frac{2 \sin 2x}{1 + \sin^8 x}$

   ›Reveal solutionSolution

   The key is to simplify the argument of the inverse sine using trigonometric identities before differentiating. After simplification, the derivative becomes $\frac{2 \sin 2x}{1 + \sin^4 x}$, which matches option (B).

   ---

   We start with

   $y = \sin^{-1}\left( \frac{1 - \cos 2x}{1 + \sin^4 x} \right).$

   Concept and intuition

   The expression inside the inverse sine looks messy, but the numerator $1 - \cos 2x$ is a classic form that simplifies to $2\sin^2 x$. That’s a huge clue: the whole fraction might simplify to something like $\frac{2\sin^2 x}{1 + \sin^4 x}$, which is a perfect candidate for the identity $\sin^{-1}\left(\frac{2t}{1+t^2}\right) = 2\tan^{-1} t$ (for $|t|\le 1$). That substitution makes differentiation straightforward.

   ---

   Step-by-step

   1. Simplify the numerator Using the double-angle identity:

   $1 - \cos 2x = 2\sin^2 x.$

   So the argument becomes

   $\frac{2\sin^2 x}{1 + \sin^4 x}.$

   2. Recognize a standard form Let $t = \sin^2 x$. Then the argument is

   $\frac{2t}{1 + t^2}.$

   This is exactly the form $\frac{2t}{1+t^2}$, which appears in the identity

   $\sin^{-1}\left( \frac{2t}{1+t^2} \right) = 2\tan^{-1} t \quad \text{for } |t| \le 1.$

   Since $t = \sin^2 x \in [0,1]$, the condition holds. Therefore

   $y = 2\tan^{-1}(\sin^2 x).$

   3. Differentiate Differentiate $y = 2\tan^{-1}(\sin^2 x)$:

   $\frac{dy}{dx} = 2 \cdot \frac{1}{1 + (\sin^2 x)^2} \cdot \frac{d}{dx}(\sin^2 x).$

   The derivative of $\sin^2 x$ is $2\sin x \cos x = \sin 2x$. So …

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5. TG EAPCET 2022Set eng-2022-07-18-FN1 markMCQ
   Q.If $x = \csc \theta - \sin \theta$, $y = \csc^{2022} \theta - \sin^{2022} \theta$ and $\left( \dfrac{dy}{dx} \right)^2 = \dfrac{k(y^2 + 4)}{g(x)}$ where $k \in \mathbb{R}$, then $10 + k - g(2022) =$ (A) 0 (B) 6 (C) 10 (D) 14

   ›Reveal solutionSolution

   Using $\csc\theta\cdot\sin\theta=1$ one gets the standard identity $\left(\dfrac{dy}{dx}\right)^2=\dfrac{n^2\,(y^2+4)}{x^2+4}$ with $n=2022$, so $k=2022^2$ and $g(x)=x^2+4$; then $10+k-g(2022)=6$, option (B).

   Write $c=\csc\theta$ and $s=\sin\theta$, noting the key relation $cs=\csc\theta\sin\theta=1$.

   Setting up $x^2+4$ and $y^2+4$. With $x=c-s$ and $n=2022$, $y=c^{n}-s^{n}$:

   $x^2+4=(c-s)^2+4=c^2+s^2-2cs+4=c^2+s^2+2=(c+s)^2,$

   $y^2+4=(c^{n}-s^{n})^2+4=c^{2n}+s^{2n}-2(cs)^n+4=c^{2n}+s^{2n}+2=(c^{n}+s^{n})^2,$

   where $cs=1$ was used so $2cs=2$ and $(cs)^n=1$.

   The derivative. Differentiating with respect to $\theta$ and forming $\dfrac{dy}{dx}=\dfrac{dy/d\theta}{dx/d\theta}$, the algebra collapses to the compact identity

   $\left(\frac{dy}{dx}\right)^2=\frac{n^2\,(y^2+4)}{x^2+4}.$ …

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6. TG EAPCET 2021Set eng-2021-08-04-AN1 markMCQ
   Q.$\lim_{x \to 0} \frac{x^2 2^x - x^2 \sin x - x^2}{3^x + \cos x - 3^x \cos x - 1} =$ (A) $\frac{1}{\log 3} (\log 2 - 1)$ (B) $\frac{4}{\log 3} (1 - \log 2)$ (C) $\frac{4}{\log 3} (\log 2 - 1)$ (D) $\frac{2}{\log 3} (\log 2 - 1)$

   ›Reveal solutionSolution

   Factor the denominator as $(1-\cos x)(3^x-1)$ and expand the numerator to leading order $x^3(\log2-1)$; the limit is $\dfrac{2}{\log3}(\log2-1)$ — option (D).

   Factor the denominator.

   $3^x+\cos x-3^x\cos x-1=3^x(1-\cos x)-(1-\cos x)=(1-\cos x)(3^x-1).$

   As $x\to0$: $1-\cos x\sim\dfrac{x^2}{2}$ and $3^x-1\sim x\log3$, so the denominator $\sim\dfrac{x^3\log3}{2}$.

   Expand the numerator $x^2\big(2^x-\sin x-1\big)$, using $2^x=1+x\log2+O(x^2)$ and $\sin x=x+O(x^3)$: …

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7. TG EAPCET 2021Set eng-2021-08-04-AN1 markMCQ
   Q.$\lim_{x \to 0} \frac{(1 - \cos 2x)}{x \tan 2x + \frac{2x}{3} \tan 3x} =$ (A) $-6$ (B) $\frac{1}{2}$ (C) $0$ (D) $\frac{-6}{5}$

   ›Reveal solutionSolution

   This problem involves evaluating a limit that results in an indeterminate form $0/0$. We resolve this by using the trigonometric identity $1 - \cos 2x = 2 \sin^2 x$ and then applying standard limits for $\sin x / x$ and $\tan x / x$ as $x \to 0$. The final value of the limit is $\boxed{\frac{1}{2}}$.

   When evaluating limits, the first step is always to try direct substitution. If this yields a finite number, that's your limit. However, if it results in an indeterminate form like $0/0$ or $\infty/\infty$, it means the function's behavior near that point is not immediately obvious, and further manipulation is required. For expressions involving trigonometric functions as $x \to 0$, we often rely on a set of fundamental limits.

   The core idea here is to transform the given expression into a form where these standard limits can be directly applied. This usually involves using trigonometric identities to simplify terms and then dividing the numerator and denominator by appropriate powers of $x$ to create terms like $\frac{\sin kx}{kx}$ or $\frac{\tan kx}{kx}$, which approach $1$ as $x \to 0$.

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

   1. Check for Indeterminate Form

      First, substitute $x=0$ into the expression:

      Numerator: $1 - \cos(2 \cdot 0) = 1 - \cos(0) = 1 - 1 = 0$.

      Denominator: $0 \cdot \tan(2 \cdot 0) + \frac{2 \cdot 0}{3} \tan(3 \cdot 0) = 0 \cdot \tan(0) + 0 \cdot \tan(0) = 0 \cdot 0 + 0 \cdot 0 = 0$.

      Since we get the form $\frac{0}{0}$, the limit is indeterminate, and we need to simplify the expression.

   2. Apply Trigonometric Identity

      The term $1 - \cos 2x$ in the numerator is a common form that can be simplified using the double-angle identity for cosine: $\cos 2x = 1 - 2 \sin^2 x$.

      Rearranging this, we get:

      $1 - \cos 2x = 2 \sin^2 x$.

      Substituting this into the limit expression:

   $\lim_{x \to 0} \frac{2 \sin^2 x}{x \tan 2x + \frac{2x}{3} \tan 3x}$

   3. Prepare for Standard Limits We know the standard limits:

      $\lim_{x \to 0} \frac{\sin x}{x} = 1$

      $\lim_{x \to 0} \frac{\tan x}{x} = 1$

      To use these, we need to divide the numerator and denominator by an appropriate power of $x$.

      The numerator has $\sin^2 x$, which suggests dividing by $x^2$. The denominator has terms like $x \tan 2x$ and $x \tan 3x$. If we divide by $x^2$, these become $\frac{\tan 2x}{x}$ and $\frac{\tan 3x}{x}$, which are suitable for the standard limit form.

      So, divide both the numerator and the denominator by $x^2$:

   $\lim_{x \to 0} \frac{\frac{2 \sin^2 x}{x^2}}{\frac{x \tan 2x + \frac{2x}{3} \tan 3x}{x^2}}$

   $\lim_{x \to 0} \frac{2 \left(\frac{\sin x}{x}\right)^2}{\frac{\tan 2x}{x} + \frac{2}{3} \frac{\tan 3x}{x}}$

   4. Manipulate Terms to Match Standard Forms Now, let's adjust the terms in the denominator to perfectly match the standard limit form $\frac{\tan kx}{kx}$: For $\frac{\tan 2x}{x}$, multiply and divide by $2$: $\frac{\tan 2x}{x} = \frac{\tan 2x}{x} \cdot \frac{2}{2} = 2 \left(\frac{\tan 2x}{2x}\right)$ …

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8. TG EAPCET 2025Set eng-2025-05-04-FN1 markMCQ
   Q.$\lim_{x \to 0} \frac{\sqrt{\cos x} - \sqrt[3]{\cos x}}{\sin^2 x} =$ (A) $\frac{2}{3}$ (B) $-\frac{1}{6}$ (C) $-\frac{1}{12}$ (D) $\frac{1}{2}$

   ›Reveal solutionSolution

   The limit simplifies by factoring out a common power of $\cos x$ and using the series expansions for $\cos x$ and $(1+u)^\alpha$; the final value is $-\frac{1}{12}$, which corresponds to option (C).

   We want

   $L = \lim_{x \to 0} \frac{\sqrt{\cos x} - \sqrt[3]{\cos x}}{\sin^2 x}.$

   Both numerator and denominator vanish as $x \to 0$, so this is a $\frac{0}{0}$ form. The key is to rewrite the numerator in terms of a common factor and then use expansions near $x=0$.

   ---

   1. Factor out the smallest power of $\cos x$.

   Write $\sqrt{\cos x} = (\cos x)^{1/2}$ and $\sqrt[3]{\cos x} = (\cos x)^{1/3}$. The smaller exponent is $\frac{1}{3}$, so factor $(\cos x)^{1/3}$ out:

   $(\cos x)^{1/2} - (\cos x)^{1/3} = (\cos x)^{1/3}\left[(\cos x)^{1/6} - 1\right].$

   Thus

   $L = \lim_{x \to 0} \frac{(\cos x)^{1/3}\left[(\cos x)^{1/6} - 1\right]}{\sin^2 x}.$

   As $x \to 0$, $\cos x \to 1$, so $(\cos x)^{1/3} \to 1$. That factor is harmless; the interesting part is the bracket.

   ---

   2. Expand $\cos x$ near $0$.

   We know

   $\cos x = 1 - \frac{x^2}{2} + \frac{x^4}{24} + O(x^6).$

   Also $\sin^2 x = x^2 - \frac{x^4}{3} + O(x^6)$.

   ---

   3. Expand $(\cos x)^{1/6}$ using $(1+u)^\alpha$.

   Let $u = \cos x - 1 = -\frac{x^2}{2} + \frac{x^4}{24} + \cdots$. Then

   $(\cos x)^{1/6} = (1+u)^{1/6} = 1 + \frac{1}{6}u + \frac{(1/6)(1/6 - 1)}{2}u^2 + \cdots$

   $= 1 + \frac{1}{6}\left(-\frac{x^2}{2} + \frac{x^4}{24}\right) + \frac{1}{6}\left(-\frac{5}{6}\right)\frac{1}{2}\left(-\frac{x^2}{2}\right)^2 + O(x^6).$

   Compute term by term:

   • Linear in $u$: $\frac{1}{6}\left(-\frac{x^2}{2}\right) = -\frac{x^2}{12}$, and the $\frac{x^4}{24}$ part gives $+\frac{x^4}{144}$.
   • Quadratic in $u$: $\frac{1}{6}\cdot\left(-\frac{5}{6}\right)\cdot\frac{1}{2} = -\frac{5}{72}$, times $u^2 = \left(-\frac{x^2}{2}\right)^2 = \frac{x^4}{4}$, gives $-\frac{5}{72}\cdot\frac{x^4}{4} = -\frac{5x^4}{288}$.

   So

   $(\cos x)^{1/6} - 1 = -\frac{x^2}{12} + \left(\frac{1}{144} - \frac{5}{288}\right)x^4 + O(x^6).$

   The $x^4$ coefficient: $\frac{1}{144} = \frac{2}{288}$, so $\frac{2}{288} - \frac{5}{288} = -\frac{3}{288} = -\frac{1}{96}$.

   Thus

   $(\cos x)^{1/6} - 1 = -\frac{x^2}{12} - \frac{x^4}{96} + O(x^6).$

   ---

   4. Assemble the limit. …

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9. TG EAPCET 2022Set eng-2022-07-19-FN1 markMCQ
   Q.If the minimum value of $\cos(\sinh(\log x) + \cosh(\log x))$ is $k$, then $\cosh(k+1) =$ (A) $\dfrac{e + e^{-1}}{2}$ (B) $\dfrac{e^2 + e^{-2}}{2}$ (C) $e$ (D) $1$

   ›Reveal solutionSolution

   The expression simplifies using hyperbolic identities to $\cos(x + 1/x)$, whose minimum is $-1$, so $k = -1$, and $\cosh(k+1) = \cosh 0 = 1$.

   The problem looks messy at first — $\cos(\sinh(\log x) + \cosh(\log x))$ — but the key is to notice that $\sinh$ and $\cosh$ are built to combine neatly. Their sum is simply $e^{\log x}$, which is $x$. That turns the whole thing into $\cos(x + 1/x)$, and then it's just a matter of finding the minimum of a cosine function.

   Let's go step by step.

   1. Simplify the hyperbolic sum.

      Recall the definitions:

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

      Adding them:

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

      Here $t = \log x$, so $\sinh(\log x) + \cosh(\log x) = e^{\log x} = x$.

   2. Rewrite the original expression. …

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10. TG EAPCET 2021Set eng-2021-08-06-AN1 markMCQ
    Q.If $\theta$ is the acute angle between the curves $x^2 + y^2 = 2020\sqrt{2}$ and $x^2 - y^2 = 2020$, then

    [!FORMULA] $\frac{\sin \theta + \cos \theta}{\tan \theta} =$

    (A) $\sqrt{2}$ (B) $\frac{3 + \sqrt{3}}{2}$ (C) $\frac{3 + \sqrt{3}}{4}$ (D) $\frac{3 + \sqrt{3}}{6}$

    ›Reveal solutionSolution

    The curves cut at $45^\circ$, so $\dfrac{\sin\theta+\cos\theta}{\tan\theta}=\dfrac{\sqrt2}{1}=\sqrt2$.

    Slopes at a point of intersection.

    Circle $x^2+y^2=2020\sqrt2$: $\;2x+2y\,y'=0\Rightarrow m_1=-\dfrac{x}{y}$.

    Hyperbola $x^2-y^2=2020$: $\;2x-2y\,y'=0\Rightarrow m_2=\dfrac{x}{y}$.

    $\tan\theta=\left|\frac{m_1-m_2}{1+m_1m_2}\right| =\left|\frac{-2x/y}{1-x^2/y^2}\right| =\left|\frac{-2xy}{y^2-x^2}\right|.$

    Point of intersection. Subtracting the equations, $2y^2=2020(\sqrt2-1)$ and $2x^2=2020(\sqrt2+1)$, so

    $y^2-x^2=1010\big[(\sqrt2-1)-(\sqrt2+1)\big]=-2020,$

    $x^2y^2=1010^2(\sqrt2+1)(\sqrt2-1)=1010^2\Rightarrow |xy|=1010.$

    Hence …

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11. TG EAPCET 2022Set eng-2022-07-19-AN1 markMCQ
    Q.If $e_1$ is the eccentricity of the hyperbola $x = \sec \theta, y = \sqrt{2} \tan \theta$ and $e_2$ is the eccentricity of the hyperbola $x = \sqrt{2} \sec \theta, y = \tan \theta$, then $\dfrac{e_2^2}{e_1^2} =$ (A) $1$ (B) $2$ (C) $\dfrac{1}{2}$ (D) $\dfrac{1}{4}$

    ›Reveal solutionSolution

    The key idea is to convert each parametric form to its standard Cartesian equation, then use the formula $e^2 = 1 + \frac{b^2}{a^2}$ for hyperbolas. The ratio $\frac{e_2^2}{e_1^2}$ simplifies to $2$.

    We are given two hyperbolas in parametric form. The first step is always to eliminate the parameter $\theta$ to get the standard Cartesian equation. For a hyperbola, the standard forms are $\frac{x^2}{a^2} - \frac{y^2}{b^2} = 1$ (transverse axis along $x$) or $\frac{y^2}{a^2} - \frac{x^2}{b^2} = 1$ (transverse axis along $y$). The eccentricity $e$ is then given by $e^2 = 1 + \frac{b^2}{a^2}$ when the transverse axis is along $x$, and $e^2 = 1 + \frac{a^2}{b^2}$ when it is along $y$ — but the formula $e^2 = 1 + \frac{b^2}{a^2}$ works if we always take $a$ as the denominator under the positive term. We'll be careful.

    1. First hyperbola: $x = \sec\theta$, $y = \sqrt{2}\tan\theta$. Recall the identity $\sec^2\theta - \tan^2\theta = 1$. Here $x = \sec\theta$ and $\frac{y}{\sqrt{2}} = \tan\theta$. Substituting into the identity:

    $x^2 - \left(\frac{y}{\sqrt{2}}\right)^2 = 1 \quad\Rightarrow\quad x^2 - \frac{y^2}{2} = 1.$

    This is $\frac{x^2}{1^2} - \frac{y^2}{(\sqrt{2})^2} = 1$, so $a_1^2 = 1$, $b_1^2 = 2$. The transverse axis is along $x$. Hence

    $e_1^2 = 1 + \frac{b_1^2}{a_1^2} = 1 + \frac{2}{1} = 3.$

    2. Second hyperbola: $x = \sqrt{2}\sec\theta$, $y = \tan\theta$. Here $\frac{x}{\sqrt{2}} = \sec\theta$ and $y = \tan\theta$. Using the same identity: $\left(\frac{x}{\sqrt{2}}\right)^2 - y^2 = 1 \quad\Rightarrow\quad \frac{x^2}{2} - y^2 = 1.$ …

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12. TG EAPCET 2025Set eng-2025-05-02-FN1 markMCQ
    Q.$\tan^{-1} \dfrac{3}{5} + \tan^{-1} \dfrac{6}{41} + \tan^{-1} \dfrac{9}{191} =$ (A) $\tan^{-1} \dfrac{9}{10}$ (B) $\tan^{-1} \dfrac{18}{19}$ (C) $\tan^{-1} \dfrac{3}{191}$ (D) $\tan^{-1} \dfrac{6}{205}$

    ›Reveal solutionSolution

    Adding the arctangents two at a time gives $\tan^{-1}\dfrac{9}{10}$.

    Solution

    Use $\tan^{-1}p+\tan^{-1}q=\tan^{-1}\dfrac{p+q}{1-pq}$.

    First two terms:

    $\tan^{-1}\frac{3}{5}+\tan^{-1}\frac{6}{41}=\tan^{-1}\frac{\frac{3}{5}+\frac{6}{41}}{1-\frac{3}{5}\cdot\frac{6}{41}}=\tan^{-1}\frac{123+30}{205-18}=\tan^{-1}\frac{153}{187}=\tan^{-1}\frac{9}{11}.$

    Add the third term: …

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