Answer:
Step-by-step explanation:
Given the expression
[tex]csc\:x\:-\:csc\:x\:cos^2x=sin\:x[/tex]
Let us verify whether the L.H.S is equal to the R.H.S or not
[tex]\csc \left(x\right)-\csc \left(x\right)\cos ^2\left(x\right)[/tex]
[tex]\mathrm{Use\:the\:following\:identity}:\quad \cos ^2\left(x\right)=1-\sin ^2\left(x\right)[/tex]
[tex]=\csc \left(x\right)-\left(1-\sin ^2\left(x\right)\right)\csc \left(x\right)[/tex]
[tex]=\csc \left(x\right)-\csc \left(x\right)\left(1-\sin ^2\left(x\right)\right)[/tex]
[tex]=\csc \left(x\right)-\csc \left(x\right)+\sin ^2\left(x\right)\csc \left(x\right)[/tex]
[tex]\mathrm{Add\:similar\:elements:}\:\csc \left(x\right)-\csc \left(x\right)=0[/tex]
[tex]=\sin ^2\left(x\right)\csc \left(x\right)[/tex]
[tex]\mathrm{Use\:the\:following\:identity}:\quad \csc \left(x\right)=\frac{1}{\sin \left(x\right)}[/tex]
[tex]=\frac{1}{\sin \left(x\right)}\sin ^2\left(x\right)[/tex]
[tex]\mathrm{Cancel\:the\:common\:factor:}\:\sin \left(x\right)[/tex]
[tex]=\sin \left(x\right)[/tex]
Thus,
