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Question Number 43810 by gyugfeet last updated on 15/Sep/18

$$\frac{1-cos\theta +co\beta -cos(\theta +\beta )}{1+cos\theta -cos\beta -cos(\theta +\beta )}=tan\frac{\theta }{2}.cot\frac{\beta }{2}$$

Answered by tanmay.chaudhury50@gmail.com last updated on 15/Sep/18

$$=\frac{2{sin}^2\frac{\theta }{2}+2sin(\frac{\theta }{2}+\beta )sin\left(\frac{\theta }{2}\right)}{2{cos}^2\left(\frac{\theta }{2}\right)-2cos(\frac{\theta }{2}+\beta )cos\left(\frac{\theta }{2}\right)}$$$$=tan\frac{\theta }{2}\times \frac{sin\frac{\theta }{2}+sin(\frac{\theta }{2}+\beta )}{cos\left(\frac{\theta }{2}\right)-cos(\frac{\theta }{2}+\beta )}$$$$=tan\left(\frac{\theta }{2}\right)\times \frac{2sin\left(\frac{\theta +\beta }{2}\right)cos\left(\frac{\beta }{2}\right)}{2sin\left(\frac{\theta +\beta }{2}\right)sin\left(\frac{\beta }{2}\right)}$$$$=tan\left(\frac{\theta }{2}\right)cot\left(\frac{\beta }{2}\right)$$$$$$$$$$

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