Question

Solve question 4please please

Answer 1

$\dfrac{\cos\theta}{\csc\theta+1}+\dfrac{\cos\theta}{\csc\theta-1}=2\tan\theta\\\\L_s=\dfrac{\cos\theta}{\dfrac{1}{\sin\theta}+1}+\dfrac{\cos\thets}{\dfrac{1}{\sin\theta}-1}=\dfrac{\cos\theta}{\dfrac{1}{\sin\theta}+\dfrac{\sin\theta}{\sin\theta}}+\dfrac{\cos\theta}{\dfrac{1}{\sin\theta}-\dfrac{\sin\theta}{\sin\theta}}$
$=\dfrac{\cos\theta}{\dfrac{1+\sin\theta}{\sin\theta}}+\dfrac{\cos\theta}{\dfrac{1-\sin\theta}{\sin\theta}}=\dfrac{\cos\theta\sin\theta}{1+\sin\theta}+\dfrac{\cos\theta\sin\theta}{1-\sin\theta}\\\\=\dfrac{\cos\theta\sin\theta(1-\sin\theta)}{(1+\sin\theta)(1-\sin\theta)}+\dfrac{\cos\theta\sin\theta(1+\sin\theta)}{(1+\sin\theta)(1-\sin\theta)}\\\\=\dfrac{cos\theta\sin\theta-\cos\theta\sin^2\theta+\cos\theta\sin\theta+\cos\theta\sin^2\theta}{1^2-\sin^2\theta}$
$=\dfrac{2\sin\theta\cos\theta}{1-\sin^2\theta}=\dfrac{2\sin\theta\cos\theta}{\cos^2\theta}=\dfrac{2\sin\theta}{\cos\theta}=2\cdot\dfrac{\sin\theta}{\cos\theta}=2\tan\theta=R_s$

$Used:\\\csc x=\dfrac{1}{\sin x}\\\\\sin^2x+\cos^2x=1\to \cos^2x=1-\sin^2x\\\\\tan x=\dfrac{\sin x}{\cos x}$