Question

Csc x - csc x cos^2x = sin x verify

Answer 1

• 
  `\csc \left(x\right)-\csc \left(x\right)\cos ^2\left(x\right)=\sin \left(x\right)`

Explanation:

Given the expression

`csc\:x\:-\:csc\:x\:cos^2x=sin\:x`

Let us verify whether the L.H.S is equal to the R.H.S or not

`\csc \left(x\right)-\csc \left(x\right)\cos ^2\left(x\right)`

`\mathrm{Use\:the\:following\:identity}:\quad \cos ^2\left(x\right)=1-\sin ^2\left(x\right)`

`=\csc \left(x\right)-\left(1-\sin ^2\left(x\right)\right)\csc \left(x\right)`

`=\csc \left(x\right)-\csc \left(x\right)\left(1-\sin ^2\left(x\right)\right)`

`=\csc \left(x\right)-\csc \left(x\right)+\sin ^2\left(x\right)\csc \left(x\right)`

`\mathrm{Add\:similar\:elements:}\:\csc \left(x\right)-\csc \left(x\right)=0`

`=\sin ^2\left(x\right)\csc \left(x\right)`

`\mathrm{Use\:the\:following\:identity}:\quad \csc \left(x\right)=(1)/(\sin \left(x\right))`

`=(1)/(\sin \left(x\right))\sin ^2\left(x\right)`

`\mathrm{Cancel\:the\:common\:factor:}\:\sin \left(x\right)`

`=\sin \left(x\right)`

Thus,

• 
  `\csc \left(x\right)-\csc \left(x\right)\cos ^2\left(x\right)=\sin \left(x\right)`