Question

If $\tan y=\tan \alpha \tan h\beta$ and $\tan z=\cot \alpha \tan h\beta$ then $\tan (y+z)=$

• A. $\cot h2\beta \csc 2\alpha$
• B. $\sin h2\beta \csc 2\alpha$
• C. $\sec h2\beta \csc 2\alpha$
• D. $\tan h2\beta \csc 2\alpha$

Answer 1

The correct option is B $\sin h2\beta \csc 2\alpha$

$\tan (y+z)=\frac{\tan y+\tan z}{1-\tan y\tan z}$
From the question,
$\Rightarrow \tan (y+z)=\frac{\tan \alpha \tan h\beta +\cot \alpha \tan h\beta }{1-\tan \alpha \cot \alpha \tan h^2\beta }$
$\tan (y+z)=\frac{\tan h\beta (\tan \alpha +\cot \alpha )}{1-\tan h^2\beta }$
$=\frac{\tan h\beta (\tan \alpha +\cot \alpha )}{\sec h^2\beta }$
$=\tan h\beta \cos h^2\beta (\frac{\sin \alpha }{\cos \alpha }+\frac{\cos \alpha }{\sin \alpha })$
$=\frac{\sin h\beta }{\cos h\beta }\cos h^2\beta (\frac{\sin \alpha }{\cos \alpha }+\frac{\cos \alpha }{\sin \alpha })$
$=\sin h\beta \cos h\beta \frac{1}{\sin \alpha \cos \alpha }$
Multiply and divide the above equation by $2$ we get
$=2\sin h\beta \cos h\beta \frac{1}{2\sin \alpha \cos \alpha }$
$=\sin h2\beta \csc 2\alpha$