We need to prove thatsin⁡θ−cos⁡θ+1sin⁡θ+cos⁡θ−1=1sec⁡θ−tan⁡θ

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dagars5nx · Accepted Answer

sin⁡θ−cos⁡θ+1sin⁡θ+cos⁡θ−1=tan⁡θ−1+sec⁡θtan⁡θ+1−sec⁡θ (dividing the numerator and the denominator by cos⁡θ)=tan⁡θ−1+sec⁡θtan⁡θ−sec⁡θ+(sec2θ−tan2θ) (putting 1=sec2θ−tan2θ)=tan⁡θ+sec⁡θ−1tan⁡θ−sec⁡θ−(tan⁡θ−sec⁡θ)(tan⁡θ+sec⁡θ)=tan⁡θ+sec⁡θ−1−(tan⁡θ−sec⁡θ)(tan⁡θ+sec⁡θ−1)=1sec⁡θ−tan⁡θ

Spencer Murillo · Accepted Answer

sin⁡θ−cos⁡θ+1sin⁡θ+cos⁡θ−1=1sec⁡θ−tan⁡θBy takingL.H.S (Left hand side)=sin⁡θ−cos⁡θ+1sin⁡θ+cos⁡θ−1=2sin⁡θ2cos⁡θ2+2sin2θ22sin⁡θ2cos⁡θ2−2sin2θ2[By applying 1−cos⁡θ=2sin2θ2]Which after simplification gives :sin⁡θ2+cos⁡θ2cos⁡θ2−sin⁡θ2Now taking R.H.S we get:1sec⁡θ−tan⁡θ=cos⁡θ1−sin⁡θ=cos2θ−sin2θcos2θ2+sin2θ2−2sin⁡θ2cos⁡θ2=(cos2θ2−sin⁡θ2)(cos⁡θ2−sin⁡θ2)2=sin⁡θ2+cos⁡θ2cos⁡θ2−sin⁡θ2=L.H.Swhere the second equality comes from applying cos⁡θ=cos2θ2−sin2θ2 and sin⁡θ=2sin⁡θ2cos⁡θ2

We need to prove that \frac{\sin\theta-\cos\theta+1}{\sin\theta+\cos\theta-1}=\frac{1}{\sec\theta-\tan\theta}

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