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.
$\sin\alpha\sin\beta-\cos\alpha\cos\beta+1=0$
$\Rightarrow-(\cos\alpha\cos\beta-\sin\alpha\sin\beta)=-1$
$\Rightarrow-\cos(\alpha+\beta)=1\ ...(1)$
$\therefore\sin(\alpha+\beta)=\sqrt{1-\cos^2(\alpha+\beta)}$
$=\sqrt{1-1^2}=0$
$\Rightarrow\sin(\alpha+\beta)=0\ ...(2)$
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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