Question

if $\tan {25}^{\circ }=x,$ then $\frac{\tan {155}^{\circ }-\tan {115}^{\circ }}{1+\tan {155}^{\circ }\tan {115}^{\circ }}$ is equal to

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

Answer 1

Answer: (A)

$\frac{1-x^2}{2x}$

Given, $\tan 25=x$
Then, $\frac{\tan {155}^{\circ }-\tan {115}^{\circ }}{1+\tan {155}^{\circ }\tan {115}^{\circ }}$
= $\frac{\tan (180-25{)}^{\circ }-\tan (90+25{)}^{\circ }}{1+\tan (180-25{)}^{\circ }\tan (90+25{)}^{\circ }}$
= $\frac{-\tan {25}^{\circ }+\cot {25}^{\circ }}{1+\tan {25}^{\circ }\cot {25}^{\circ }}$
= $\frac{\frac{1}{x}-x}{2}$
= $\frac{1-x^2}{2x}$