Question

If $\sin \alpha \sin \beta -\cos \alpha \cos \beta +1=0$, Prove that $1+\cot \alpha \tan \beta =0$.

Answer 1

Hint: As the given equation can be written as formula and finding out the value of $\alpha ,\beta$and further applying the trigonometric properties and formula $\tan \left( \alpha +\beta \right)=\dfrac{\tan \alpha +\tan \beta }{1-\tan \alpha \tan \beta }$, which leads to the final answer.

Complete step-by-step answer:
Given $\sin \alpha \sin \beta -\cos \alpha \cos \beta +1=0$
The equation can be further written as $\cos \alpha \cos \beta -\sin \alpha \sin \beta =1$
It looks like it is the value of a known formula.
$\cos \alpha \cos \beta -\sin \alpha \sin \beta =1$
$\cos \left( \alpha +\beta \right)=1$
The value of cosine becomes zero only when $\theta$ is ${{0}^{\circ }}$.
$\Rightarrow \left( \alpha +\beta \right)=0$ . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . (1)
Now writing the formula for $\tan \left( \alpha +\beta \right)$
$\tan \left( \alpha +\beta \right)=\dfrac{\tan \alpha +\tan \beta }{1-\tan \alpha \tan \beta }$ . . . . . . . . . . . . . . . . . . . . . . . . (2)
Substituting (1) in (2) we get,
$\tan \left( 0 \right)=\dfrac{\tan \alpha +\tan \beta }{1-\tan \alpha \tan \beta }$ . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . (a)
Since the value of $\tan \left( 0 \right)=0$
Now rewriting the above equation (a) we get,
$\tan \alpha +\tan \beta =0$. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .(b)
Now the equation (b) can be resolved to $\dfrac{1}{\cot \alpha }+\tan \beta =0$.
Further transformation gives us the $1+\cot \alpha \tan \beta =0$.
Hence proved $1+\cot \alpha \tan \beta =0$.

Note: To solve such types of problems all the related formulas should be handy. Also be careful about the signs in all the formulas. As the interval is not given we have taken the value of cosine as ${{0}^{\circ }}$ when $\cos \left( \alpha +\beta \right)=1$.