cos^{-1}left(x^2 + frac{1}{x^2} - 1right) + s | vedclass

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(B) Let $t = x^2$. Since $x \in R - \{0\}$,$t > 0$.For $\cos^{-1}(t + \frac{1}{t} - 1)$ to be defined,we must have $-1 \leq t + \frac{1}{t} - 1 \leq 1

Vedclass · Accepted Answer

$\frac{\pi}{4}$

Vedclass · Answer

(B) Let $t = x^2$. Since $x \in R - \{0\}$,$t > 0$.For $\cos^{-1}(t + \frac{1}{t} - 1)$ to be defined,we must have $-1 \leq t + \frac{1}{t} - 1 \leq 1$.Since $t + \frac{1}{t} \geq 2$ for $t > 0$,we have $t + \frac{1}{t} - 1 \geq 1$.Thus,the only possible value is $t + \frac{1}{t} - 1 = 1$,which implies $t + \frac{1}{t} = 2$.This occurs only when $t = 1$,i.e.,$x^2 = 1$.Substituting $x^2 = 1$ into the expression:$\cos^{-1}(1 + 1 - 1) + \sin^{-1}(1 - 1) + 	an^{-1}(1)$$= \cos^{-1}(1) + \sin^{-1}(0) + 	an^{-1}(1)$$= 0 + 0 + \frac{\pi}{4} = \frac{\pi}{4}$.

$\cos^{-1}\left(x^2 + \frac{1}{x^2} - 1\right) + \sin^{-1}\left(x^2 - \frac{1}{x^2}\right) + \tan^{-1}(x^2)$ is equal to (where $x \in R - \{0\}$)

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