The value of sin left[tan ^{-1}left(frac{1-x^ | 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$. Substituting this into the expression: $	an^{-1}\left(\frac{1-	an^2 	heta}{2 	an 	heta}

Vedclass · Accepted Answer

$1$

Vedclass · Answer

(B) Let $x = 	an 	heta$. Then $	heta = 	an^{-1} x$. Substituting this into the expression: $	an^{-1}\left(\frac{1-	an^2 	heta}{2 	an 	heta}ight) = 	an^{-1}\left(\frac{1}{	an 2	heta}ight) = 	an^{-1}(\cot 2	heta) = 	an^{-1}\left(	an\left(\frac{\pi}{2} - 2	hetaight)ight) = \frac{\pi}{2} - 2	heta$. Next,$\cos^{-1}\left(\frac{1-	an^2 	heta}{1+	an^2 	heta}ight) = \cos^{-1}(\cos 2	heta) = 2	heta$. Adding these two parts: $(\frac{\pi}{2} - 2	heta) + 2	heta = \frac{\pi}{2}$. Finally,$\sin\left(\frac{\pi}{2}ight) = 1$.

The value of $\sin \left[\tan ^{-1}\left(\frac{1-x^2}{2 x}\right)+\cos ^{-1}\left(\frac{1-x^2}{1+x^2}\right)\right]$ is

Similar Questions

If $\cot ^{-1}(\sqrt{\cos \alpha})-\tan ^{-1}(\sqrt{\cos \alpha})=x$,then the value of $\sin x$ is

If $y = \tan^{-1}\sqrt{\frac{1 + \cos x}{1 - \cos x}}$,then $\frac{dy}{dx}$ is equal to

The value of $\tan \left(\sin ^{-1}\left(\frac{3}{5}\right)+\tan ^{-1}\left(\frac{2}{3}\right)\right)$ is

If $\sum_{n=1}^k \tan ^{-1}\left(\frac{1}{n^2+3 n+3}\right)=\tan ^{-1} \alpha$, then $\alpha=$

Write the function in the simplest form: $\tan ^{-1} \left( \frac{\sqrt{1+x^{2}}-1}{x} \right), x \neq 0$

Vedclass Test Series

Exam Paper Generator

Online Exam Module