The value of frac{tan 40^{circ}+tan 20^{circ} | vedclass

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Question

(A) Given expression is $\frac{	an 40^{\circ}+	an 20^{\circ}}{1-\cot 70^{\circ} \cot 50^{\circ}}$.Using the identity $\cot(90^{\circ}-	heta) = 	an

Vedclass · Accepted Answer

$\sqrt{3}$

Vedclass · Answer

(A) Given expression is $\frac{	an 40^{\circ}+	an 20^{\circ}}{1-\cot 70^{\circ} \cot 50^{\circ}}$.Using the identity $\cot(90^{\circ}-	heta) = 	an 	heta$,we can rewrite the denominator:$\cot 70^{\circ} = \cot(90^{\circ}-20^{\circ}) = 	an 20^{\circ}$$\cot 50^{\circ} = \cot(90^{\circ}-40^{\circ}) = 	an 40^{\circ}$Substituting these into the expression:$= \frac{	an 40^{\circ}+	an 20^{\circ}}{1-	an 20^{\circ} 	an 40^{\circ}}$Using the trigonometric identity $	an(A+B) = \frac{	an A + 	an B}{1-	an A 	an B}$,where $A = 40^{\circ}$ and $B = 20^{\circ}$:$= 	an(40^{\circ}+20^{\circ}) = 	an 60^{\circ}$Since $	an 60^{\circ} = \sqrt{3}$,the final value is $\sqrt{3}$.

The value of $\frac{\tan 40^{\circ}+\tan 20^{\circ}}{1-\cot 70^{\circ} \cot 50^{\circ}}$ is equal to

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