If the value of x satisfying the equation sin | vedclass

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

Question

(B) Let $\sin^{-1} \sqrt{1-x^2} = 	heta$. Then $\sin 	heta = \sqrt{1-x^2}$.This implies $\cos 	heta = \sqrt{1 - (1-x^2)} = \sqrt{x^2} = |x|$. Since

Vedclass · Accepted Answer

$5$

Vedclass · Answer

(B) Let $\sin^{-1} \sqrt{1-x^2} = 	heta$. Then $\sin 	heta = \sqrt{1-x^2}$.This implies $\cos 	heta = \sqrt{1 - (1-x^2)} = \sqrt{x^2} = |x|$. Since the domain of $\sqrt{\frac{2}{x}-1}$ requires $x > 0$,we have $\cos 	heta = x$.Thus,$	an 	heta = \frac{\sin 	heta}{\cos 	heta} = \frac{\sqrt{1-x^2}}{x} = \sqrt{\frac{1-x^2}{x^2}} = \sqrt{\frac{1}{x^2}-1}$.Given $	an 	heta = \sqrt{\frac{2}{x}-1}$,we equate the two expressions:$\sqrt{\frac{1}{x^2}-1} = \sqrt{\frac{2}{x}-1}$.Squaring both sides: $\frac{1}{x^2} - 1 = \frac{2}{x} - 1$.$\frac{1}{x^2} = \frac{2}{x} \implies x^2 = \frac{x}{2} \implies x(x - \frac{1}{2}) = 0$.Since $x > 0$,we have $x = \frac{1}{2}$.Here,$a = 1$ and $b = 2$,which are coprime.Therefore,$a^2 + b^2 = 1^2 + 2^2 = 1 + 4 = 5$.

If the value of $x$ satisfying the equation $\sin^{-1} \sqrt{1-x^2} = \tan^{-1} \sqrt{\frac{2}{x}-1}$ is $\frac{a}{b}$ (where $a$ and $b$ are coprime),then the value of $a^2 + b^2$ is

Similar Questions

Let $f : R \to R$,$f(x) = \max\{|\tan^{-1}x|, \cot^{-1}x\}$. Consider the following statements:
$I.$ The function is continuous and derivable $\forall x \in R$.
$II.$ The range of the function is $\left[ \frac{\pi}{4}, \pi \right]$.
$III.$ $f(x)$ is many-one into.
Identify the correct option.

$50 \tan \left(3 \tan ^{-1}\left(\frac{1}{2}\right)+2 \cos ^{-1}\left(\frac{1}{\sqrt{5}}\right)\right)+4 \sqrt{2} \tan \left(\frac{1}{2} \tan ^{-1}(2 \sqrt{2})\right)$ is equal to

$\cos ^{-1}(\cos (-5))+\sin ^{-1}(\sin (6))-\tan ^{-1}(\tan (12))$ is equal to :
(The inverse trigonometric functions take the principal values)

If $2 \tan^{-1} x = 3 \sin^{-1} x$ and $x \neq 0$,then $8x^2 + 1 =$

Let $f(x) = \cos \left(2 \tan ^{-1} \sin \left(\cot ^{-1} \sqrt{\frac{1-x}{x}}\right)\right)$ for $0 < x < 1$. Then :

Vedclass Test Series

Exam Paper Generator

Online Exam Module