Question

√1+cos theta/√1-cos theta+ √1- cos theta/√ 1+ cos theta = 2 cosec theta​

Answer 1

Answer:

Hey!

Pls refer to the attatched image for answer.

Answer 2

Answer:

Its true.

i hope this helps!!

Step-by-step explanation:

let theta be A (or you could keep it as theta, im only keeping it as A because i cant type it in equation form :D )

LHS:

$\sqrt{\frac{1+cosA}{1-cosA}} } + \sqrt{\frac{1-cosA}{1+cosA}} }$

now lets rationalise these;

$\sqrt{(\frac{1+cosA}{1-cosA})(\frac{1+cosA}{1+cosA}) } + \sqrt{(\frac{1-cosA}{1+cosA}) (\frac{1-cosA}{1-cosA}) }$

$\frac{\sqrt{1+cosA+cosA+cos^{2}A}} {\sqrt{1+cosA-cosA-cos^{2}A}} + \frac{\sqrt{1-cosA-cosA+cos^{2}A}} {\sqrt{1-cosA+cosA-cos^{2}A}}$

$\frac{\sqrt{1^{2} +2cosA+cos^{2}A}} {\sqrt{1-cos^{2}A}} + \frac{\sqrt{1^{2} -2cosA+cos^{2}A}} {\sqrt{1-cos^{2}A}}$

using the formula; $(a+b)^{2} = a^{2} +2ab + b^{2}$ and $(a-b)^{2} = a^{2} -2ab + b^{2}$

we can factorise it as;

$\frac{\sqrt{(1+cosA)^{2}}} {\sqrt{1-cos^{2}A}} + \frac{\sqrt{(1-cosA)^{2}}} {\sqrt{1-cos^{2}A}}$

and since $sinA = \sqrt{1-cos^{2}A }$

$\frac{\sqrt{(1+cosA)^{2}}} {\sqrt{sin^{2} A}} + \frac{\sqrt{(1-cosA)^{2}}} {\sqrt{sin^{2} A}}$

$\frac{\sqrt{(1+cosA)^{2}}} {\sqrt{(sinA)^{2} }} + \frac{\sqrt{(1-cosA)^{2}}} {\sqrt{(sinA)^{2} }}$

the square roots and squares get cancelled

$\frac{{(1+cosA)}} {(sinA)}} + \frac{{(1-cosA)}} {(sinA)}}$

$\frac{{(1+cosA+1-cosA)}} {(sinA)}}$

$\frac{2}{sinA}$

$2$×$\frac{1}{sinA}$

since $\frac{1}{sinA}$ = $cosecA$

2 × cosec A

L.H.S = R.H.S

hence, proved