Question
If $\sin\alpha\sin\beta-\cos\alpha\cos\beta+1=0$ prove that $1+\cot\alpha\tan\beta$

Answer

We have, $\sin\alpha\sin\beta-\cos\alpha\cos\beta+1=0$ $\Rightarrow-(\cos\alpha\cos\beta-\sin\alpha\sin\beta)=-1$ $\Rightarrow-\cos(\alpha+\beta)=1\ ...(1)$ $\therefore\sin(\alpha+\beta)=\sqrt{1-\cos^2(\alpha+\beta)}$ $=\sqrt{1-1^2}=0$ $\Rightarrow\sin(\alpha+\beta)=0\ ...(2)$ Now, $1+\cot\alpha\tan\beta=1+\frac{\cos\alpha}{\sin\alpha}\times\frac{\sin\beta}{\cos\beta}$ $=\frac{\sin\alpha\times\cos\beta+\cos\alpha\times\sin\beta}{\sin\alpha\times\cos\beta}$ $=\frac{\sin(\alpha+\beta)}{\sin\alpha\times\cos\beta}=\frac{0}{\sin\alpha\times\cos\beta}$ [Using equation (2)] $\therefore1+\cot\alpha\tan\beta=0$ Hence proved.

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