Evaluate: tan^{-1} left( frac{1 - x^2}{2x} ri | vedclass

Name: Exam Paper Generator | JEE NEET Paper Generator | Vedclass
Brand: Vedclass
Rating: 5 (35 reviews)

Question

(B) Let $x = 	an 	heta$. Then $	heta = 	an^{-1} x$.The given expression is $	an^{-1} \left( \frac{1 - x^2}{2x} ight) + \cos^{-1} \left( \frac{1

Vedclass · Accepted Answer

$\frac{\pi}{2}$

Vedclass · Answer

(B) Let $x = 	an 	heta$. Then $	heta = 	an^{-1} x$.The given expression is $	an^{-1} \left( \frac{1 - x^2}{2x} ight) + \cos^{-1} \left( \frac{1 - x^2}{1 + x^2} ight)$.Substitute $x = 	an 	heta$:$= 	an^{-1} \left( \frac{1 - 	an^2 	heta}{2 	an 	heta} ight) + \cos^{-1} \left( \frac{1 - 	an^2 	heta}{1 + 	an^2 	heta} ight)$Using trigonometric identities $\cot 2	heta = \frac{1 - 	an^2 	heta}{2 	an 	heta}$ and $\cos 2	heta = \frac{1 - 	an^2 	heta}{1 + 	an^2 	heta}$:$= 	an^{-1} (\cot 2	heta) + \cos^{-1} (\cos 2	heta)$Since $	an^{-1} (\cot 2	heta) = 	an^{-1} \left( 	an \left( \frac{\pi}{2} - 2	heta ight) ight) = \frac{\pi}{2} - 2	heta$ and $\cos^{-1} (\cos 2	heta) = 2	heta$:$= \left( \frac{\pi}{2} - 2	heta ight) + 2	heta$$= \frac{\pi}{2}$.

Evaluate: $\tan^{-1} \left( \frac{1 - x^2}{2x} \right) + \cos^{-1} \left( \frac{1 - x^2}{1 + x^2} \right)$

Similar Questions

The trigonometric equation $\sin ^{-1} x = 2 \sin ^{-1} 2a$ has a real solution, if

If $\theta = \sec^{-1}(\cosh u)$,then $u =$

If $\tan ^{-1} \frac{x-1}{x-2}+\tan ^{-1} \frac{x+1}{x+2}=\frac{\pi}{4},$ then find the value of $x$.

$\operatorname{Sin}^{-1}(-\cos 2) + \operatorname{Cos}^{-1}(\sin 3) + \operatorname{Tan}^{-1}(\cot 5) = $

$\tan ^{-1} 2+\tan ^{-1} 3=$

Vedclass Test Series

Exam Paper Generator

Online Exam Module