Question

If $\tan {35}^{\circ }=k$, then the value of $\frac{\tan {145}^{\circ }-\tan {125}^{\circ }}{1+\tan {145}^{\circ }\tan {125}^{\circ }}$ is:

• A. $\frac{2k}{1-k^2}$
• B. $\frac{2k}{1+k^2}$
• C. $\frac{1-k^2}{2k}$
• D. $\frac{1-k^2}{1+k^2}$

Answer 1

The correct option is C $\frac{1-k^2}{2k}$

$\frac{\tan {145}^o-\tan {125}^o}{1+\tan {145}^o\tan {125}^o}=\tan ({145}^o-{125}^o)=\tan {20}^o$

$\tan {20}^o=\cot {70}^o=\frac{1}{\tan {70}^o}=\frac{1}{\frac{2\tan {35}^o}{1-{\tan }^2{35}^o}}=\frac{1}{\frac{2k}{1-k^2}}=\frac{1-k^2}{2k}$

Hence, the answer is $\frac{1-k^2}{2k}$