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Question Number 18797 by Tinkutara last updated on 29/Jul/17

$$\mathrm{If}A,B,C\mathrm{are}\mathrm{the}\mathrm{angles}\mathrm{of}\mathrm{a}\mathrm{triangle},$$$$\mathrm{then}2\sin \frac{A}{2}\mathrm{cosec}\frac{B}{2}\sin \frac{C}{2}-\sin A\cot \frac{B}{2}$$$$-\cos A\mathrm{is}$$$$\left(1\right)\mathrm{Independent}\mathrm{of}A,B,C$$$$\left(2\right)\mathrm{Function}\mathrm{of}A,B,C$$$$\left(3\right)\mathrm{Function}\mathrm{of}A,B$$$$\left(4\right)\mathrm{Function}\mathrm{of}B,C$$

Answered by behi.8.3.4.1.7@gmail.com last updated on 30/Jul/17

$$\frac{2sin\frac{A}{2}.sin\frac{C}{2}}{sin\frac{B}{2}}-\frac{2sin\frac{A}{2}cos\frac{A}{2}cos\frac{B}{2}}{sin\frac{B}{2}}-cosA=$$$$=\frac{2sin\frac{A}{2}(sin\frac{C}{2}-cos\frac{A}{2}cos\frac{B}{2})}{sin\frac{B}{2}}-cosA=$$$$-2{sin}^2\frac{A}{2}-(1-2{sin}^2\frac{A}{2})=-1.◼$$$$\Rightarrow answer\left(1\right).$$$$note:sin\frac{C}{2}=sin(90-\frac{A+B}{2})=cos\frac{A+B}{2}.$$

Commented by Tinkutara last updated on 30/Jul/17

$$\mathrm{Thank}\mathrm{you}\mathrm{very}\mathrm{much}\mathrm{behi}\mathrm{Sir}!$$

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