If sinα sinβ − cosα cosβ + 1 = 0, Prove that 1 + cotα tanβ = 0
Question
If
\[ \sin\alpha\sin\beta-\cos\alpha\cos\beta+1=0 \]
prove that:
\[ 1+\cot\alpha\tan\beta=0 \]
Proof
Given,
\[ \sin\alpha\sin\beta-\cos\alpha\cos\beta+1=0 \]
\[ \sin\alpha\sin\beta-\cos\alpha\cos\beta=-1 \]
Using
\[ \cos(\alpha+\beta) = \cos\alpha\cos\beta-\sin\alpha\sin\beta \]
\[ -\cos(\alpha+\beta)=-1 \]
\[ \cos(\alpha+\beta)=1 \]
\[ \alpha+\beta=0 \]
Therefore,
\[ \tan(\alpha+\beta)=0 \]
Using
\[ \tan(\alpha+\beta) = \frac{\tan\alpha+\tan\beta} {1-\tan\alpha\tan\beta} \]
\[ \tan\alpha+\tan\beta=0 \]
\[ \frac{\sin\alpha}{\cos\alpha} + \frac{\sin\beta}{\cos\beta} =0 \]
\[ \sin\alpha\cos\beta+\cos\alpha\sin\beta=0 \]
\[ \cos\alpha\sin\beta = -\sin\alpha\cos\beta \]
\[ \frac{\cos\alpha\sin\beta} {\sin\alpha\cos\beta} =-1 \]
\[ \cot\alpha\tan\beta=-1 \]
\[ 1+\cot\alpha\tan\beta=0 \]
Hence proved.