If sin(tan^{-1}(xsqrt{2})) = cot(sin^{-1}sqrt | vedclass

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

Question

(A) Let $	an^{-1}(x\sqrt{2}) = 	heta$. Then $	an 	heta = x\sqrt{2}$.Using the identity $\sin 	heta = \frac{	an 	heta}{\sqrt{1+	an^2 	heta}}$,

Vedclass · Accepted Answer

$1/2$

Vedclass · Answer

(A) Let $	an^{-1}(x\sqrt{2}) = 	heta$. Then $	an 	heta = x\sqrt{2}$.Using the identity $\sin 	heta = \frac{	an 	heta}{\sqrt{1+	an^2 	heta}}$, we get $\sin 	heta = \frac{x\sqrt{2}}{\sqrt{1+2x^2}}$.Let $\sin^{-1}(\sqrt{1-x^2}) = \phi$. Then $\sin \phi = \sqrt{1-x^2}$.Using the identity $\cot \phi = \frac{\cos \phi}{\sin \phi} = \frac{\sqrt{1-\sin^2 \phi}}{\sin \phi}$, we get $\cot \phi = \frac{\sqrt{1-(1-x^2)}}{\sqrt{1-x^2}} = \frac{x}{\sqrt{1-x^2}}$.Equating the two sides: $\frac{x\sqrt{2}}{\sqrt{1+2x^2}} = \frac{x}{\sqrt{1-x^2}}$.Since $x \in (0, 1)$, $x 
eq 0$, we can divide by $x$: $\frac{\sqrt{2}}{\sqrt{1+2x^2}} = \frac{1}{\sqrt{1-x^2}}$.Squaring both sides: $\frac{2}{1+2x^2} = \frac{1}{1-x^2}$.$2(1-x^2) = 1+2x^2 \implies 2-2x^2 = 1+2x^2 \implies 4x^2 = 1 \implies x^2 = 1/4$.Since $x \in (0, 1)$, $x = 1/2$.

If $\sin(\tan^{-1}(x\sqrt{2})) = \cot(\sin^{-1}\sqrt{1-x^2})$ for $x \in (0, 1)$, then the value of $x$ is:

Similar Questions

If the equation $\sin^{-1} \sqrt{x} + \cos^{-1} \sqrt{x^2 - 1} + \tan^{-1} (\tan y) = a$ has at least one solution,then the number of integral values of $a$ is:

Let $S = \{ x : \cos^{-1} x = \pi + \sin^{-1} x + \sin^{-1}(2x + 1) \}$. Then $\sum_{x \in S} (2x - 1)^2$ is equal to . . . . . .

If $\alpha=3 \sin ^{-1} \frac{6}{11}$ and $\beta=3 \cos ^{-1}\left(\frac{4}{9}\right)$,where the inverse trigonometric functions take only the principal values,then the incorrect option is

If $x=\cos ^{-1}\left(\frac{1}{\sqrt{1+t^2}}\right)$ and $y=\sin ^{-1}\left(\frac{t}{\sqrt{1+t^2}}\right)$,then $\frac{dy}{dx}$ is

Find the maximum value of the expression $\lfloor \tan^{-1} x - \tan^{-1} y \rfloor - \lfloor \sin^{-1} u - \sin^{-1} v \rfloor$,where $\lfloor . \rfloor$ denotes the greatest integer function,and $x, y, u, v$ are independent real variables.

Vedclass Test Series

Exam Paper Generator

Online Exam Module