Test Index

TS EAMCET 11 May 2024 Shift 1 Solved Paper

Section: Mathematics

© examsnet.com

Question : 26 of 160

Marks: +1, -0

If

\sin^{-1} x - \cos^{-1} 2x = \sin^{-1} \left( \frac{\sqrt{3}}{2} \right) - \cos^{-1} \left( \frac{\sqrt{3}}{2} \right)

\tan^{-1} x + \tan^{-1} \left( \frac{x}{x+1} \right) =

[TG EAPCET 11 May 2024 Shift 1]

$\frac{\pi}{6}$
$\frac{\pi}{4}$
$\frac{\pi}{3}$
$\frac{\pi}{2}$

Solution:

\sin^{-1} x - \cos^{-1} 2x = \frac{\pi}{3} - \frac{\pi}{6} = \frac{\pi}{6}

\Rightarrow \frac{\pi}{2} - \cos^{-1} x - \cos^{-1} 2x = \frac{\pi}{6}

\Rightarrow \cos^{-1} x + \cos^{-1} 2x = \frac{\pi}{2} - \frac{\pi}{6} = \frac{\pi}{3}

\Rightarrow \cos^{-1} \left( x \cdot 2x - \sqrt{1-x^2} \sqrt{1-4x^2} \right) = \frac{\pi}{3}

\Rightarrow 2x^2 - \sqrt{1-x^2} \sqrt{1-4x^2} = \cos \frac{\pi}{3} = \frac{1}{2}

\Rightarrow \sqrt{1-x^2} \sqrt{1-4x^2} = \left( 2x^2 - \frac{1}{2} \right)^2

On solving, we get

x = \frac{1}{2}

\tan^{-1} x + \tan^{-1} \left( \frac{x}{x+1} \right) = \tan^{-1} \frac{1}{2} + \tan^{-1} \frac{1}{3}

= \tan^{-1} \left( \frac{ \frac{1}{2} + \frac{1}{3} }{ 1 - \frac{1}{6} } \right)

= \tan^{-1} \left( \frac{ \frac{5}{6} }{ \frac{5}{6} } \right) = \tan^{-1} 1 = \frac{\pi}{4}

© examsnet.com

Go to Question:

Prev Question

Next Question