If tan x = frac{b}{a}, then sqrt{frac{a + b}{ | vedclass

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

Question

(B) Given that,$	an x = \frac{b}{a}.$Consider the expression $\sqrt{\frac{a + b}{a - b}} + \sqrt{\frac{a - b}{a + b}}.$Dividing the numerator and den

Vedclass · Accepted Answer

$\frac{2\cos x}{\sqrt{\cos 2x}}$

Vedclass · Answer

(B) Given that,$	an x = \frac{b}{a}.$Consider the expression $\sqrt{\frac{a + b}{a - b}} + \sqrt{\frac{a - b}{a + b}}.$Dividing the numerator and denominator inside the square roots by $a,$ we get:$= \sqrt{\frac{1 + \frac{b}{a}}{1 - \frac{b}{a}}} + \sqrt{\frac{1 - \frac{b}{a}}{1 + \frac{b}{a}}}$$= \sqrt{\frac{1 + 	an x}{1 - 	an x}} + \sqrt{\frac{1 - 	an x}{1 + 	an x}}$$= \frac{(1 + 	an x) + (1 - 	an x)}{\sqrt{(1 - 	an x)(1 + 	an x)}} = \frac{2}{\sqrt{1 - 	an^2 x}}$$= \frac{2}{\sqrt{1 - \frac{\sin^2 x}{\cos^2 x}}} = \frac{2\cos x}{\sqrt{\cos^2 x - \sin^2 x}}$Since $\cos^2 x - \sin^2 x = \cos 2x,$ we have:$= \frac{2\cos x}{\sqrt{\cos 2x}}.$

If $\tan x = \frac{b}{a},$ then $\sqrt{\frac{a + b}{a - b}} + \sqrt{\frac{a - b}{a + b}} = $

Similar Questions

The value of $e^{\log _{10} \tan 1^{\circ}+\log _{10} \tan 2^{\circ}+\log _{10} \tan 3^{\circ}+\ldots+\log _{10} \tan 89^{\circ}}$ is

Let $a$ and $b$ be non-negative real numbers. If $\sin x + a \cos x = b$,then $|a \sin x - \cos x| = $

If $e^{\sin x}-e^{-\sin x}-4=0,$ then the number of real values of $x$ is

The value of $(1-\cos \theta)(1+\cos \theta)(1+\cot^2 \theta)$,when $\theta = \frac{\pi}{15}$ is

If $\tan \theta + \cot \theta = 4$,then $\tan^{4} \theta + \cot^{4} \theta = $

Vedclass Test Series

Exam Paper Generator

Online Exam Module