Browse Questions

# If $\sin\alpha\sin\beta+1=0$.Find $1+\cot \alpha\tan\beta=0$

$(a)\;1\qquad(b)\;2\qquad(c)\;3\qquad(d)\;0$

Given :
$\sin\alpha\sin \beta-\cos\alpha\cos\beta+1=0$
We can change as
$\cos\alpha\cos\beta-\sin\alpha\sin\beta=1$
It is of the form
$\cos (\alpha+\beta)=1$
Now,$1+\cot \alpha\tan\beta=1+\large\frac{\cos\alpha}{\sin\alpha}.\frac{\sin \beta}{\cos\beta}$
$\qquad\quad\qquad\qquad\;\;= \large\frac{\sin\alpha\cos\beta+\cos\alpha\sin\beta}{\sin\alpha\cos\beta}$
$\qquad\quad\qquad\qquad\;\;= \large\frac{\sin(\alpha+\beta)}{\sin\alpha\cos\beta}$
We know that
$\sin^2(\alpha+\beta)=1-\cos^2(\alpha+\beta)$
$\qquad\quad\quad\;\;= 1-1$
$\qquad\quad\quad\;\;= 0$
Hence $\large\frac{0}{\sin\alpha\cos\beta}$$=0$
Hence (d) is the correct option.