Question

Prove that
1+ cos 0/
sin o-
sin 0/
1 + cos 0
= 2 cot 0.​

Answer 1

Question:

Prove that,

$\dfrac {1+\cos\theta}{\sin\theta}-\dfrac {\sin\theta}{1+\cos\theta}=2\cot\theta$

Solution:

$\begin {aligned}&\sf{LHS}\\\\=\ &\dfrac {1+\cos\theta}{\sin\theta}-\dfrac {\sin\theta}{1+\cos\theta}\\\\=\ &\dfrac {(1+\cos\theta)^2-\sin^2\theta}{(1+\cos\theta)\sin\theta}\\\\=\ &\dfrac {\left (\dfrac {(1+\cos\theta)^2-\sin^2\theta}{\cos^2\theta}\right)}{\left (\dfrac {(1+\cos\theta)\sin\theta}{\cos^2\theta}\right)}\\\\=\ &\dfrac {\left (\dfrac {(1+\cos\theta)^2}{\cos^2\theta}-\dfrac {\sin^2\theta}{\cos^2\theta}\right)}{\left (\dfrac {(1+\cos\theta)}{\cos\theta}\cdot\dfrac {\sin\theta}{\cos\theta}\right)}\end {aligned}$

$\begin {aligned}=\ &\dfrac {\left (\dfrac {1+\cos\theta}{\cos\theta}\right)^2-\tan^2\theta}{\left (\dfrac {1+\cos\theta}{\cos\theta}\right)\cdot\dfrac {\sin\theta}{\cos\theta}}\\\\=\ &\dfrac {(1+\sec\theta)^2-\tan^2\theta}{(1+\sec\theta)\tan\theta}\\\\=\ &\dfrac {1+2\sec\theta+\sec^2\theta-\tan^2\theta}{(1+\sec\theta)\tan\theta}\\\\=\ &\dfrac {2(1+\sec\theta)}{(1+\sec\theta)\tan\theta}\\\\=\ &\dfrac {2}{\tan\theta}\\\\=\ &2\cot\theta\\\\=\ &\sf{RHS}\\\\\end{aligned}$

Hence Proved!