cos left[ 2cos^{-1}frac{1}{5} + sin^{-1}frac{ | vedclass

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

Question

(B) We know that $\cos^{-1}x + \sin^{-1}x = \frac{\pi}{2}$ for $x \in [-1, 1]$.Given expression: $\cos \left[ \cos^{-1}\frac{1}{5} + \cos^{-1}\frac{1}

Vedclass · Accepted Answer

$-\frac{2\sqrt{6}}{5}$

Vedclass · Answer

(B) We know that $\cos^{-1}x + \sin^{-1}x = \frac{\pi}{2}$ for $x \in [-1, 1]$.Given expression: $\cos \left[ \cos^{-1}\frac{1}{5} + \cos^{-1}\frac{1}{5} + \sin^{-1}\frac{1}{5} ight]$.Using the property,this simplifies to: $\cos \left[ \frac{\pi}{2} + \cos^{-1}\frac{1}{5} ight]$.Since $\cos(\frac{\pi}{2} + 	heta) = -\sin	heta$,we have: $-\sin \left( \cos^{-1}\frac{1}{5} ight)$.Let $\cos^{-1}\frac{1}{5} = \alpha$,then $\cos\alpha = \frac{1}{5}$.Then $\sin\alpha = \sqrt{1 - \cos^2\alpha} = \sqrt{1 - (\frac{1}{5})^2} = \sqrt{1 - \frac{1}{25}} = \sqrt{\frac{24}{25}} = \frac{2\sqrt{6}}{5}$.Thus,the expression equals $-\frac{2\sqrt{6}}{5}$.

$\cos \left[ 2\cos^{-1}\frac{1}{5} + \sin^{-1}\frac{1}{5} \right] = $

Similar Questions

$\cos \left[\cos ^{-1}\left(-\frac{1}{7}\right)+\sin ^{-1}\left(-\frac{1}{7}\right)\right]$ is equal to

Let $S_{n} = \cot^{-1} 2 + \cot^{-1} 8 + \cot^{-1} 18 + \cot^{-1} 32 + \dots$ to $n^{\text{th}}$ term. Then $\lim_{n \rightarrow \infty} S_{n}$ is

Consider the following statements.
$I$. $\sin ^{-1}(y^2-4y+6)+\cos ^{-1}(y^2-4y+6) = \frac{\pi}{2}, \forall y \in R$
$II$. $\sec ^{-1}(y^2-4y+6)+\operatorname{cosec}^{-1}(y^2-4y+6) = \frac{\pi}{2}, \forall y \in R$
Which of the above statement$(s)$ is/are true?

Prove $3 \cos ^{-1} x = \cos ^{-1} (4 x^{3} - 3 x)$,where $x \in [\frac{1}{2}, 1]$.

$A$ solution of the equation $\tan^{-1}(1 + x) + \tan^{-1}(1 - x) = \frac{\pi}{2}$ is

Vedclass Test Series

Exam Paper Generator

Online Exam Module