JEE Advanced · Mathematics · 21. ITF

The total number of real solutions of the equation $\theta=\tan ^{-1}(2 \tan \theta)-\frac{1}{2} \sin ^{-1}\left(\frac{6 \tan \theta}{9+\tan ^2 \theta}\right)$ is (Here, the inverse trigonometric functions $\sin ^{-1} x$ and $\tan ^{-1} x$ assume values in $\left[-\frac{\pi}{2}, \frac{\pi}{2}\right]$ and $\left(-\frac{\pi}{2}, \frac{\pi}{2}\right)$, respectively.)

Verified Solution

Answer & Solution

Correct Answer

(C) 3

Step-by-step Solution

Detailed explanation

Let $\alpha=\frac{1}{2} \sin ^{-1}\left(\frac{6 \tan \theta}{9+\tan ^2 \theta}\right)$
$\begin{aligned}
& \theta=\tan ^{-1}(2 \tan \theta)-\alpha \\
& \theta+\alpha=\tan ^{-1}(2 \tan \theta) \\
& \tan (\theta+\alpha)=2 \tan \theta \\
& \frac{\tan \theta+\tan \alpha}{1-\tan \alpha \tan \theta}=2 \tan \theta ....(1)\\
& \sin 2 \alpha=\frac{6 \tan \theta}{9+\tan ^2 \theta}=\frac{2 \tan \alpha}{1+\tan ^2 \alpha} \\
& 3 \tan \theta+3 \tan \theta \tan ^2 \alpha=9 \tan \alpha+\tan \alpha \tan ^2 \theta \\
& 3(\tan \theta-3 \tan \alpha)=\tan \alpha \tan \theta(\tan \theta-3 \tan \alpha) \\
& \tan \theta=\frac{3}{\tan \alpha} \text { or } \tan \theta=3 \tan \alpha
\end{aligned}$
$\begin{aligned} & \text { Case-I : } \tan \theta=3 \tan \alpha \\ & \frac{\tan \theta+\frac{\tan \theta}{3}}{1-\frac{\tan ^2 \theta}{3}}=2 \tan \theta\end{aligned}$
$\tan \theta=0, \quad \frac{2}{3}=1-\frac{\tan ^2 \theta}{3} \quad \Rightarrow \tan \theta=1,-1$
\(\tan \theta=0,-1,1

\)\begin{aligned} & \text { Case-II : } \tan \theta=\frac{3}{\tan \alpha} \\ & \frac{\tan \theta+\frac{3}{\tan \theta}}{-2}=2 \tan \theta\end{aligned}$

Apply the relevant identity from the chapter and substitute the values established in the previous step, keeping the original constraint in view.

Use the simplified expression to isolate the unknown, then plug the result back into the question setup to confirm the working is internally consistent.

Cross-check against a boundary or limiting case. Both the dimensional balance and the qualitative behaviour at the extreme should agree with the candidate answer.

Combine the steps above to identify the correct option. The remaining choices each fail at least one of the consistency, dimensional, or boundary checks.

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From JEE Advanced

Columns 1, 2 and 3 contain conics, equations of tangents to the conics and points of contact, respectively.

Column 1	Column 2	Column 3
(I) $x^{2} + y^{2} = a^{2}$	(i) $m y = m^{2} x + a$	(P) $(\frac{a}{m^{2}}, \frac{2 a}{m})$
(II) $x^{2} + a^{2} y^{2} = a^{2}$	(ii) $y = m x + a \sqrt{m^{2} + 1}$	(Q) $(\frac{- m a}{\sqrt{m^{2} + 1}}, \frac{a}{\sqrt{m^{2} + 1}})$
(III) $y^{2} = 4 a x$	(iii) $y = m x + \sqrt{a^{2} m^{2} - 1}$	(R) $(\frac{- a^{2} m}{\sqrt{a^{2} m^{2} + 1}}, \frac{1}{\sqrt{a^{2} m^{2} + 1}})$
(IV) $x^{2} - a^{2} y^{2} = a^{2}$	(iv) $y = m x + \sqrt{a^{2} m^{2} + 1}$	(S) $(\frac{- a^{2} m}{\sqrt{a^{2} m^{2} - 1}}, \frac{- 1}{\sqrt{a^{2} m^{2} - 1}})$

For

a = \sqrt{2}

, if a tangent is drawn to a suitable conic (Column 1) at the point of contact

(- 1, 1)

, then which of the following options is the only Correct combination for obtaining its equation?

JEE Advanced 2017 Hard

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