Match List I with List II and select the corr | vedclass

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

Question

(D) $(P)$ $\left(\frac{1}{y^2}\left(\frac{\cos (	an ^{-1} y)+y \sin (	an ^{-1} y)}{\cot (\sin ^{-1} y)+	an (\sin ^{-1} y)}ight)^2+y^4ight)^{1 /

Vedclass · Accepted Answer

$3 \quad 4 \quad 1 \quad 2$

Vedclass · Answer

(D) $(P)$ $\left(\frac{1}{y^2}\left(\frac{\cos (	an ^{-1} y)+y \sin (	an ^{-1} y)}{\cot (\sin ^{-1} y)+	an (\sin ^{-1} y)}ight)^2+y^4ight)^{1 / 2}$$= \left[\frac{1}{y^2}\left[\frac{\left(\frac{1}{\sqrt{1+y^2}}+\frac{y \cdot y}{\sqrt{1+y^2}}ight)}{\left(\frac{\sqrt{1-y^2}}{y}+\frac{y}{\sqrt{1-y^2}}ight)}ight]^2+y^4ight]^{1 / 2}$$= \left(\frac{1}{y^2} \cdot y^2(1-y^4)+y^4ight)^{1 / 2} = 1$$(Q)$ $\cos x+\cos y=-\cos z$ and $\sin x+\sin y=-\sin z$.Squaring and adding: $(\cos x+\cos y)^2 + (\sin x+\sin y)^2 = \cos^2 z + \sin^2 z = 1$.$2 + 2\cos(x-y) = 1 \Rightarrow \cos(x-y) = -1/2$.Using $\cos(x-y) = 2\cos^2(\frac{x-y}{2}) - 1 = -1/2 \Rightarrow 2\cos^2(\frac{x-y}{2}) = 1/2 \Rightarrow \cos^2(\frac{x-y}{2}) = 1/4 \Rightarrow \cos(\frac{x-y}{2}) = 1/2$.$(R)$ $\cos 2 x(\cos (\frac{\pi}{4}-x)-\cos (\frac{\pi}{4}+x)) = 2 \sin x \cos x - 2 \sin^2 x = 2 \sin x (\cos x - \sin x)$.$\cos 2 x(\sqrt{2} \sin x) = 2 \sin x (\cos x - \sin x)$.$\sqrt{2} \sin x [\cos 2 x - \sqrt{2}(\cos x - \sin x)] = 0$.$\sin x = 0$ or $\cos^2 x - \sin^2 x = \sqrt{2}(\cos x - \sin x) \Rightarrow (\cos x - \sin x)(\cos x + \sin x - \sqrt{2}) = 0$.$\sec x = \sqrt{2}$.$(S)$ $\cot (\sin ^{-1} \sqrt{1-x^2}) = \frac{x}{\sqrt{1-x^2}}$.$\sin (	an ^{-1}(x \sqrt{6})) = \frac{x \sqrt{6}}{\sqrt{1+6 x^2}}$.$\frac{x}{\sqrt{1-x^2}} = \frac{x \sqrt{6}}{\sqrt{1+6 x^2}} \Rightarrow 1+6x^2 = 6(1-x^2) = 6-6x^2 \Rightarrow 12x^2 = 5 \Rightarrow x = \frac{1}{2} \sqrt{\frac{5}{3}}$.

List $I$	List $II$
$P$. $\left(\frac{1}{y^2}\left(\frac{\cos (\tan ^{-1} y)+y \sin (\tan ^{-1} y)}{\cot (\sin ^{-1} y)+\tan (\sin ^{-1} y)}\right)^2+y^4\right)^{1 / 2}$ takes value	$1$. $\frac{1}{2} \sqrt{\frac{5}{3}}$
$Q$. If $\cos x+\cos y+\cos z=0=\sin x+\sin y+\sin z$ then possible value of $\cos \frac{x-y}{2}$ is	$2$. $\sqrt{2}$
$R$. If $\cos (\frac{\pi}{4}-x) \cos 2 x+\sin x \sin 2 x \sec x=\cos x \sin 2 x \sec x+\cos (\frac{\pi}{4}+x) \cos 2 x$ then possible value of $\sec x$ is	$3$. $\frac{1}{2}$
$S$. If $\cot (\sin ^{-1} \sqrt{1-x^2})=\sin (\tan ^{-1}(x \sqrt{6})), x \neq 0$,then possible value of $x$ is	$4$. $1$

Similar Questions

$\sin \left(\tan ^{-1} \frac{4}{5}+\tan ^{-1} \frac{4}{3}+\tan ^{-1} \frac{1}{9}-\tan ^{-1} \frac{1}{7}\right) = $

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

If $a_1, a_2, a_3, \dots, a_n$ is an $A.P.$ with common difference $d$,then $\tan \left[ \tan^{-1} \left( \frac{d}{1 + a_1 a_2} \right) + \tan^{-1} \left( \frac{d}{1 + a_2 a_3} \right) + \dots + \tan^{-1} \left( \frac{d}{1 + a_{n-1} a_n} \right) \right] = $

For the equation $\cos^{-1} x + \cos^{-1} 2x + \pi = 0$,the number of real solutions is:

If $\sin ^{-1} x + \sin ^{-1}(1-x) = \cos ^{-1} x$,then $x \in$

Vedclass Test Series

Exam Paper Generator

Online Exam Module