Question

Solve question 4
please please

Answer 1

`(\cos\theta)/(\csc\theta+1)+(\cos\theta)/(\csc\theta-1)=2\tan\theta\\\\L_s=(\cos\theta)/((1)/(\sin\theta)+1)+(\cos\thets)/((1)/(\sin\theta)-1)=(\cos\theta)/((1)/(\sin\theta)+(\sin\theta)/(\sin\theta))+(\cos\theta)/((1)/(\sin\theta)-(\sin\theta)/(\sin\theta))`

`=(\cos\theta)/((1+\sin\theta)/(\sin\theta))+(\cos\theta)/((1-\sin\theta)/(\sin\theta))=(\cos\theta\sin\theta)/(1+\sin\theta)+(\cos\theta\sin\theta)/(1-\sin\theta)\\\\=(\cos\theta\sin\theta(1-\sin\theta))/((1+\sin\theta)(1-\sin\theta))+(\cos\theta\sin\theta(1+\sin\theta))/((1+\sin\theta)(1-\sin\theta))\\\\=(cos\theta\sin\theta-\cos\theta\sin^2\theta+\cos\theta\sin\theta+\cos\theta\sin^2\theta)/(1^2-\sin^2\theta)`

`=(2\sin\theta\cos\theta)/(1-\sin^2\theta)=(2\sin\theta\cos\theta)/(\cos^2\theta)=(2\sin\theta)/(\cos\theta)=2\cdot(\sin\theta)/(\cos\theta)=2\tan\theta=R_s`


`Used:\\\csc x=(1)/(\sin x)\\\\\sin^2x+\cos^2x=1\to \cos^2x=1-\sin^2x\\\\\tan x=(\sin x)/(\cos x)`