If x = sin (2 tan^{-1} 2),y = cos (2 tan^{-1} | vedclass

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

Question

(D) We use the identities: $2 	an^{-1} (	heta) = \sin^{-1} (\frac{2 	heta}{1 + 	heta^2}) = \cos^{-1} (\frac{1 - 	heta^2}{1 + 	heta^2}) = \sec^{-

Vedclass · Accepted Answer

$z < y < x$

Vedclass · Answer

(D) We use the identities: $2 	an^{-1} (	heta) = \sin^{-1} (\frac{2 	heta}{1 + 	heta^2}) = \cos^{-1} (\frac{1 - 	heta^2}{1 + 	heta^2}) = \sec^{-1} (\frac{1 + 	heta^2}{1 - 	heta^2})$.For $x = \sin (2 	an^{-1} 2)$:$x = \sin (\sin^{-1} (\frac{2 	imes 2}{1 + 2^2})) = \frac{4}{5} = 0.8$.For $y = \cos (2 	an^{-1} 3)$:$y = \cos (\cos^{-1} (\frac{1 - 3^2}{1 + 3^2})) = \frac{1 - 9}{1 + 9} = \frac{-8}{10} = -0.8$.For $z = \sec (2 	an^{-1} 4)$:$z = \sec (\sec^{-1} (\frac{1 + 4^2}{1 - 4^2})) = \frac{1 + 16}{1 - 16} = \frac{17}{-15} \approx -1.133$.Comparing the values: $-1.133 < -0.8 < 0.8$,which means $z < y < x$.

If $x = \sin (2 \tan^{-1} 2)$,$y = \cos (2 \tan^{-1} 3)$,and $z = \sec (2 \tan^{-1} 4)$,then:

Similar Questions

$\sec ^2(\tan ^{-1} 2) + \operatorname{cosec}^2(\cot ^{-1} 3)$ is equal to

If $\cot \left(\cos^{-1} x\right) = \sec \left(\tan^{-1} \left(\frac{a}{\sqrt{b^2-a^2}}\right)\right)$,where $b > a > 0$,then $x =$

The value of $\frac{d}{dx} \tan^{-1} \left[ \frac{3a^2x - x^3}{a(a^2 - 3x^2)} \right]$ at $x = 0$ is

If $y = \cos \left(\frac{\pi}{3} + \cos^{-1} \frac{x}{2}\right)$,then $(x - y)^2 + 3y^2$ is equal to . . . . . . .

The numerical value of $\tan \left(2 \tan ^{-1}\left(\frac{1}{5}\right)+\frac{\pi}{4}\right)$ is:

Vedclass Test Series

Exam Paper Generator

Online Exam Module