The value of frac{(tan 69^{circ} + tan 66^{ci | vedclass

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Question

(D) The given expression is in the form of the trigonometric identity $	an(A + B) = \frac{(	an A + 	an B)}{(1 - 	an A 	an B)}$.Here,$A = 69^{\cir

Vedclass · Accepted Answer

$-1$

Vedclass · Answer

(D) The given expression is in the form of the trigonometric identity $	an(A + B) = \frac{(	an A + 	an B)}{(1 - 	an A 	an B)}$.Here,$A = 69^{\circ}$ and $B = 66^{\circ}$.Therefore,the expression becomes $	an(69^{\circ} + 66^{\circ})$.$	an(69^{\circ} + 66^{\circ}) = 	an(135^{\circ})$.We can write $	an(135^{\circ})$ as $	an(180^{\circ} - 45^{\circ})$.Using the identity $	an(180^{\circ} - 	heta) = -	an 	heta$,we get $-	an 45^{\circ}$.Since $	an 45^{\circ} = 1$,the final value is $-1$.

The value of $\frac{(\tan 69^{\circ} + \tan 66^{\circ})}{(1 - \tan 69^{\circ} \tan 66^{\circ})}$ is

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