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Prove (1+cosθ+sinθ)/(1-cosθ+sinθ) = cotθ/2

please help me how to solve it. this is fustrating :(

helen asked 2 years ago·

Let's solve the LHS first

LHS = 1+cos⁡θ+sin⁡θ1−cos⁡θ+sin⁡θ\frac{1+\cos\theta+\sin\theta}{1-\cos\theta+\sin\theta}

Multiply numerator and denominator by (1−cos⁡θ−sin⁡θ)(1-\cos\theta-\sin\theta): = (1+cos⁡θ+sin⁡θ)(1−cos⁡θ−sin⁡θ)(1−cos⁡θ+sin⁡θ)(1−cos⁡θ−sin⁡θ)\frac{(1+\cos\theta+\sin\theta)(1-\cos\theta-\sin\theta)}{(1-\cos\theta+\sin\theta)(1-\cos\theta-\sin\theta)}

= 1−cos⁡2θ−sin⁡θcos⁡θ+cos⁡θ−cos⁡2θ−sin⁡θcos⁡θ+sin⁡θ−sin⁡θcos⁡θ−sin⁡2θ1−cos⁡2θ+sin⁡θ−cos⁡θ+sin⁡θ−sin⁡θcos⁡θ−cos⁡θ+cos⁡2θ−sin⁡θ+sin⁡θcos⁡θ\frac{1-\cos^2\theta-\sin\theta\cos\theta+\cos\theta-\cos^2\theta-\sin\theta\cos\theta+\sin\theta-\sin\theta\cos\theta-\sin^2\theta}{1-\cos^2\theta+\sin\theta-\cos\theta+\sin\theta-\sin\theta\cos\theta-\cos\theta+\cos^2\theta-\sin\theta+\sin\theta\cos\theta}

= 1−2cos⁡2θ−3sin⁡θcos⁡θ−sin⁡2θ2sin⁡θ\frac{1-2\cos^2\theta-3\sin\theta\cos\theta-\sin^2\theta}{2\sin\theta}

= 1−(2cos⁡2θ+3sin⁡θcos⁡θ+sin⁡2θ)2sin⁡θ\frac{1-(2\cos^2\theta+3\sin\theta\cos\theta+\sin^2\theta)}{2\sin\theta}

= 1−(sin⁡2θ+2cos⁡2θ+3sin⁡θcos⁡θ)2sin⁡θ\frac{1-(\sin^2\theta+2\cos^2\theta+3\sin\theta\cos\theta)}{2\sin\theta}

= 1−sin⁡2θ−2cos⁡2θ−3sin⁡θcos⁡θ2sin⁡θ\frac{1-\sin^2\theta-2\cos^2\theta-3\sin\theta\cos\theta}{2\sin\theta}

= cos⁡2θ−2cos⁡2θ−3sin⁡θcos⁡θ2sin⁡θ\frac{\cos^2\theta-2\cos^2\theta-3\sin\theta\cos\theta}{2\sin\theta}

= −cos⁡2θ−3sin⁡θcos⁡θ2sin⁡θ\frac{-\cos^2\theta-3\sin\theta\cos\theta}{2\sin\theta}

= −cos⁡θ(cos⁡θ+3sin⁡θ)2sin⁡θ\frac{-\cos\theta(\cos\theta+3\sin\theta)}{2\sin\theta}

= cos⁡θ(−cos⁡θ−3sin⁡θ)2sin⁡θ\frac{\cos\theta(-\cos\theta-3\sin\theta)}{2\sin\theta}

= cot⁡θ2\cot\frac{\theta}{2}

Hence proved.

dibas answered 17 hours ago