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प्रश्न
If sin α sin β − cos α cos β + 1 = 0, prove that 1 + cot α tan β = 0.
उत्तर
Given:
\[\sin\alpha \sin\beta - \cos \alpha \cos \beta + 1 = 0 \]
\[ \Rightarrow - (\cos\alpha \cos\beta - \sin\alpha \sin\beta) + 1 = 0\]
\[ \Rightarrow - \cos(\alpha + \beta) + 1 = 0\]
\[ \Rightarrow \cos(\alpha + \beta) = 1\]
\[\text{ Therefore, }\sin(\alpha + \beta) = 0 . . . . (1) (\text{ Since }\sin\theta = \sqrt{1 - \cos^2 \theta} ) \]
Hence ,
\[1 + \cot\alpha \tan\beta = 1 + \frac{\cos\alpha \sin\beta}{\sin\alpha \cos\beta} \]
\[ = \frac{\sin\alpha\cos\beta + \cos\alpha\sin\beta}{\sin\alpha \cos\beta}\]
\[ = \frac{\sin(\alpha + \beta)}{\sin\alpha\cos\beta}\]
\[ = 0 . . . \left\{\text{ From eq }(1) \right\}\]
Hence proved .
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