Prove that: `(sin(theta+varphi)-2sintheta+"sin"(theta-varphi))/(cos(theta+varphi)-2costheta+"cos"(theta-varphi))=t a ntheta`

Explanation

We have to proof,

`(sin(theta+varphi)-2sintheta+"sin"(theta-varphi))/(cos(theta+varphi)-2costheta+"cos"(theta-varphi))=tantheta`

Let,

L.H.S=`(sin(theta+varphi)-2sintheta+"sin"(theta-varphi))/(cos(theta+varphi)-2costheta+"cos"(theta-varphi))`

As we know the formula,
`sin(A^o)+sin(B^o)= 2sin(((A^o)+(B^o))/2)cos(((A^o)-(B^o))/2)` and `(cos A+cosB)= 2cos(((A^o)-(B^o))/2)cos(((A^o)+(B^o))/2)`

then,

L.H.S.=`(2sin(((theta+varphi)+(theta-varphi))/2)cos(((theta+varphi)-(theta-varphi))/2)-2sin(theta))/(2cos(((theta+varphi)-(theta-varphi))/2)cos(((theta+varphi)+(theta-varphi))/2)-2cos(theta)`

L.H.S.=`(2sin(theta)cos(varphi)-2sin(theta))/(2cos(varphi)cos(theta)-2cos(theta)`

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