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  4. The value of tan 20°+2 tan 50°- tan 70° is equal to

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Q. The value of $\tan \,20^{\circ}+2 \tan \,50^{\circ}-\tan \,70^{\circ}$ is equal to

Trigonometric Functions

Solution:

$\tan 20^{\circ}+2 \tan 50^{\circ}-\tan 70^{\circ}$
$=\frac{\sin 20^{\circ}}{\cos 20^{\circ}}-\frac{\sin 70^{\circ}}{\cos 70^{\circ}}+2 \tan 50^{\circ}$
$=\frac{\sin 20^{\circ} \cos 70^{\circ}-\cos 20^{\circ} \sin 70^{\circ}}{\cos 20^{\circ} \cos 70^{\circ}}+2 \tan 50^{\circ}$
$=\frac{\sin \left(20^{\circ}-70^{\circ}\right)}{\frac{1}{2}\left[\cos \left(70^{\circ}+20^{\circ}\right)+\cos \left(70^{\circ}-20^{\circ}\right)\right]}+2 \tan 50^{\circ}$
$=\frac{2 \sin \left(-50^{\circ}\right)}{\cos 90^{\circ}+\cos 50^{\circ}}+2 \tan 50^{\circ}=\frac{-2 \sin 50^{\circ}}{0+\cos 50^{\circ}}+2 \tan 50^{\circ}$
$=-2 \tan 50^{\circ}+2 \tan 50^{\circ}=0$