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

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

Question

(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 $x \in R - \{0\}$)

Similar Questions

If $\frac{(x + 1)^2}{x^3 + x} = \frac{A}{x} + \frac{Bx + C}{x^2 + 1}$,then $\sin^{-1}\left(\frac{A}{C}\right) = $

Consider the following statements:
Statement-$I$: $\operatorname{Cosh}^{-1} x = \operatorname{Tanh}^{-1} x$ has no solution.
Statement-$II$: $\operatorname{Cosh}^{-1} x = \operatorname{Coth}^{-1} x$ has only one solution.
The correct answer is:

$\sin^{-1} (\sin \frac{3\pi}{5}) = \dots \dots \dots$

$\tan \left[ \sec^{-1} \sqrt{1 + x^2} \right] = $

If ${\cos ^{ - 1}}\left( {\frac{1}{x}} \right) = \theta $,then $\tan \theta =$

Vedclass Test Series

Exam Paper Generator

Online Exam Module