Question

If $\sin \alpha \sin \beta -\cos \alpha \cos \beta +1=0$, then

• A. $\cot \alpha +\cot \beta =0$
• B. $1+\cot \alpha \cot \beta =0$
• C. $1+\tan \alpha \tan \beta =0$
• D. None of these

Answer 1

The correct option is D

None of these

Explanation for the correct option.

Given that, $\sin \alpha \sin \beta -\cos \alpha \cos \beta +1=0$

$$\begin{gathered} \Rightarrow \cos \alpha \cos \beta -\sin \alpha \sin \beta =1 \\ \Rightarrow \cos (\alpha +\beta )=1 \\ \Rightarrow \alpha +\beta =0....\left(1\right) \end{gathered}$$
Take $\tan$ on both sides in equation (1) we get:
$\begin{array}{rcl}\tan (\alpha +\beta )& =& \tan 0\\ \frac{{\displaystyle (\tan \alpha +\tan \beta )}}{(1-\tan \alpha \tan \beta )}& =& 0\\ \tan \alpha +\tan \beta & =& 0\\ \tan \beta & =& -\tan \alpha \\ \frac{{\displaystyle \tan \beta }}{\tan \alpha }& =& -1\\ \tan \beta \cot \alpha +1& =& 0\end{array}$

This doesn't match any of the options.
Hence, option D is correct.