Question

$\tan h\left(x+y\right)$ equals

• A. $\frac{\tan h\left(x\right)+\tan h\left(y\right)}{1-\tan h\left(x\right)\tan h\left(y\right)}$
• B. $\frac{\tan h\left(x\right)+\tan h\left(y\right)}{1+\tan h\left(x\right)\tan h\left(y\right)}$
• C. $\frac{\tan h\left(x\right)-\tan h\left(y\right)}{1-\tan h\left(x\right)\tan h\left(y\right)}$
• D. $\frac{\tan h\left(x\right)-\tan h\left(y\right)}{1+\tan h\left(x\right)\tan h\left(y\right)}$

Answer 1

The correct option is B

$\frac{\tan h\left(x\right)+\tan h\left(y\right)}{1+\tan h\left(x\right)\tan h\left(y\right)}$

Explanation for the correct option :

We know that $\sin h(x+y)=\sin h\left(x\right)\cos h\left(y\right)+\cos h\left(x\right)\sin h\left(y\right)$ and $\cos h(x+y)=\cos h\left(x\right)\cos h\left(y\right)+\sin h\left(x\right)\sin h\left(y\right)$.

Then,

$$\begin{gathered} \tan h(x+y)=\frac{\sin h(x+y)}{\cos h(x+y)} \\ =\frac{\sin h\left(x\right)\cos h\left(y\right)+\cos h\left(x\right)\sin h\left(y\right)}{\cos h\left(x\right)\cos h\left(y\right)+\sin h\left(x\right)\sin h\left(y\right)} \\ Numeratoranddenominatordividedby\cos h\left(x\right)\cos h\left(y\right) \\ =\frac{{\displaystyle \frac{\sin h\left(x\right)\cos h\left(y\right)}{\cos h\left(x\right)\cos h\left(y\right)}}+{\displaystyle \frac{\cos h\left(x\right)\sin h\left(y\right)}{\cos h\left(x\right)\cos h\left(y\right)}}}{1+{\displaystyle \frac{\sin h\left(x\right)\sin h\left(y\right)}{\cos h\left(x\right)\cos h\left(y\right)}}} \\ =\frac{\tan h\left(x\right)+\tan h\left(y\right)}{1+\tan h\left(x\right)\tan h\left(y\right)} \end{gathered}$$

Hence, option (B) is the correct answer.