Question

$\tan {65}^o=?$

• A. $\tan {25}^o-\tan {40}^o$
• B. $\tan {25}^o+\tan {40}^o$
• C. $\tan {25}^o-2\tan {40}^o$
• D. $\tan {25}^o+2\tan {40}^o$

Answer 1

The correct option is D $\tan {25}^o+2\tan {40}^o$

$\sin (A-B)=\sin A\cos B-\cos A\sin B$

$\frac{\sin (A-B)}{\cos A\cos B}=\frac{\sin A\cos B}{\cos A\cos B}-\frac{\cos A\sin B}{\cos A\cos B}=\tan A-\tan B$

Let $A={65}^{\circ }$ and $B={40}^{\circ }$

$\tan {65}^{\circ }-\tan {40}^{\circ }=$$\frac{\sin {25}^{\circ }}{\cos {65}^{\circ }\cos {40}^{\circ }}$$=\frac{\sin ({90}^{\circ }-{65}^{\circ })}{\cos {65}^{\circ }\cos {40}^{\circ }}$$=\frac{\cos {65}^{\circ }}{\cos {65}^{\circ }\cos {40}^{\circ }}$$=\frac{1}{\cos {40}^{\circ }}$

$\tan {65}^{\circ }-\tan {40}^{\circ }$$=\frac{1}{\cos {40}^{\circ }}$

Subtract $\tan {40}^{\circ }$ from both sides,

$\tan {65}^{\circ }-\tan {40}^{\circ }-\tan {40}^{\circ }$$=\frac{1}{\cos {40}^{\circ }}-\tan {40}^{\circ }$

$⟹\tan {65}^{\circ }-2\tan {40}^{\circ }$$=\frac{1-\sin {40}^{\circ }}{\cos {40}^{\circ }}$

$⟹\tan {65}^{\circ }-2\tan {40}^{\circ }$$=\frac{(\cos {20}^{\circ }-\sin {20}^{\circ }{)}^2}{(\cos {20}^{\circ }-\sin {20}^{\circ })(\cos {20}^{\circ }+\sin {20}^{\circ })}$

$⟹\tan {65}^{\circ }-2\tan {40}^{\circ }$$=\frac{\tan {45}^{\circ }-\tan {20}^{\circ }}{1-\tan {45}^{\circ }\tan {20}^{\circ }}$

$⟹\tan {65}^{\circ }-2\tan {40}^{\circ }$$=\tan ({45}^{\circ }-{20}^{\circ })$

$⟹\tan {65}^{\circ }=\tan {25}^{\circ }+2\tan {40}^{\circ }$