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

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

Question

(b) Given that, $	an x = \frac{b}{a}$

Now $\sqrt {\frac{{a + b}}{{a - b}}} + \sqrt {\frac{{a - b}}{{a + b}}}$

$= \sqrt {\frac{{1 + b/a}}{{1 - b

Vedclass · Accepted Answer

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

Vedclass · Answer

(b) Given that, $	an x = \frac{b}{a}$

Now $\sqrt {\frac{{a + b}}{{a - b}}} + \sqrt {\frac{{a - b}}{{a + b}}}$

$= \sqrt {\frac{{1 + b/a}}{{1 - b/a}}} + \sqrt {\frac{{1 - b/a}}{{1 + b/a}}} $

$ = \frac{2}{{\sqrt {1 - \frac{{{b^2}}}{{{a^2}}}} }} = \frac{2}{{\sqrt {1 - {{	an }^2}x} }} $

$= \frac{2}{{\sqrt {1 - \frac{{{{\sin }^2}x}}{{{{\cos }^2}x}}} }} $

$= \frac{{2\cos x}}{{\sqrt {\cos 2x} }}$.

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

Similar Questions

$4 \,\,sin5^o \,\,sin55^o \,\,sin65^o$ has the values equal to

Prove that $\cot 4 x(\sin 5 x+\sin 3 x)=\cot x(\sin 5 x-\sin 3 x)$

$2 \sin \left(\frac{\pi}{22}\right) \sin \left(\frac{3 \pi}{22}\right) \sin \left(\frac{5 \pi}{22}\right) \sin \left(\frac{7 \pi}{22}\right) \sin \left(\frac{9 \pi}{22}\right)$ is

Number of values of $ x \in \left[ {0,2\pi } \right]$ satisfying the equation $cotx - cosx = 1 - cotx. cosx$

$1 - 2{\sin ^2}\left( {\frac{\pi }{4} + \theta } \right) = $