The value of tan left{frac{1}{2} cos ^{-1} fr | vedclass

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

Question

(A) Let $	heta = \frac{1}{2} \cos ^{-1} \frac{\sqrt{5}}{3}$.Then $2	heta = \cos ^{-1} \frac{\sqrt{5}}{3}$,which implies $\cos 2	heta = \frac{\sqrt{

Vedclass · Accepted Answer

$\frac{3-\sqrt{5}}{2}$

Vedclass · Answer

(A) Let $	heta = \frac{1}{2} \cos ^{-1} \frac{\sqrt{5}}{3}$.Then $2	heta = \cos ^{-1} \frac{\sqrt{5}}{3}$,which implies $\cos 2	heta = \frac{\sqrt{5}}{3}$.We need to find $	an 	heta$.Using the formula $	an 	heta = \sqrt{\frac{1-\cos 2	heta}{1+\cos 2	heta}}$:$	an 	heta = \sqrt{\frac{1-\frac{\sqrt{5}}{3}}{1+\frac{\sqrt{5}}{3}}} = \sqrt{\frac{3-\sqrt{5}}{3+\sqrt{5}}}$.Rationalizing the denominator:$	an 	heta = \sqrt{\frac{(3-\sqrt{5})(3-\sqrt{5})}{(3+\sqrt{5})(3-\sqrt{5})}} = \sqrt{\frac{(3-\sqrt{5})^2}{9-5}} = \sqrt{\frac{(3-\sqrt{5})^2}{4}} = \frac{3-\sqrt{5}}{2}$.

The value of $\tan \left\{\frac{1}{2} \cos ^{-1} \frac{\sqrt{5}}{3}\right\}$ is

Similar Questions

If $\cos \left(\cos ^{-1} \frac{\sqrt{3}}{2}+\sin ^{-1} x\right)=1$,then find the value of $x$.

Find the value of $\tan \frac{1}{2} \left[ \sin^{-1} \frac{2x}{1+x^2} + \cos^{-1} \frac{1-y^2}{1+y^2} \right]$,where $|x| <1, y>0$ and $xy < 1$.

If $(\tan ^{-1} x)^2+(\cot ^{-1} x)^2=\frac{5 \pi^2}{8}$,then $x^2+1=$

$\sinh^{-1}\left(\frac{x}{\sqrt{1-x^2}}\right)$ is equal to

Find the number of solutions for the equation $\sin^{-1} x = 2 \tan^{-1} x$ (in principal values).

Vedclass Test Series

Exam Paper Generator

Online Exam Module