Question

The value of the angle ${\tan }^{–1}(\tan {65}^{\circ }-2\tan {40}^{\circ })$ in degrees is equal to

• A. $-{20}^{\circ }$
• B. ${20}^{\circ }$
• C. ${25}^{\circ }$
• D. ${40}^{\circ }$

Answer 1

The correct option is C ${25}^{\circ }$

$\tan {65}^{\circ }-2\tan {40}^{\circ }=\tan ({45}^{\circ }+{20}^{\circ })-2\tan {40}^{\circ }=\frac{1+\tan {20}^{\circ }}{1-\tan {20}^{\circ }}-\frac{4\tan {20}^{\circ }}{1-{\tan }^2{20}^{\circ }}=\frac{(1+\tan {20}^{\circ }{)}^2-4\tan {20}^{\circ }}{(1-\tan {20}^{\circ })(1+\tan {20}^{\circ })}=\frac{(1-\tan {20}^{\circ })(1-\tan {20}^{\circ })}{(1-\tan {20}^{\circ })(1+\tan {20}^{\circ })}=\frac{1-\tan {20}^{\circ }}{1+\tan {20}^{\circ }}=\tan ({45}^{\circ }-{20}^{\circ })=\tan {25}^{\circ }$

$\therefore {\tan }^{-1}\left(\tan {25}^{\circ }\right)={25}^{\circ }$