Question

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

• A. $-1$
• B. $1$
• C. Zero
• D. $\pi$

Answer 1

Answer: (A)

$-1$

$\sin \alpha \sin \beta -\cos \alpha \cos \beta +1=0$
$\Rightarrow cos\alpha cos\beta -sin\alpha sin\beta =1\Rightarrow cos(\alpha +\beta )=1$
$\therefore (\alpha +\beta )=0$

$\therefore tan(\alpha +\beta )=0\Rightarrow \frac{\tan \alpha +\tan \beta }{1-\tan \alpha \tan \beta }=0$

$\Rightarrow \tan \beta =-\tan \alpha$
$\Rightarrow \frac{\tan \beta }{\tan \alpha }=-1$
$\therefore \cot \alpha \cdot \tan \beta =-1$