The expression frac{tan 57^{circ}+cot 37^{cir | vedclass

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Question

(B) Given expression: $\frac{	an 57^{\circ}+\cot 37^{\circ}}{	an 33^{\circ}+\cot 53^{\circ}}$Using the complementary angle identities $	an(90^{\cir

Vedclass · Accepted Answer

$	an 57^{\circ} \cot 37^{\circ}$

Vedclass · Answer

(B) Given expression: $\frac{	an 57^{\circ}+\cot 37^{\circ}}{	an 33^{\circ}+\cot 53^{\circ}}$Using the complementary angle identities $	an(90^{\circ}-	heta) = \cot 	heta$ and $\cot(90^{\circ}-	heta) = 	an 	heta$:$	an 57^{\circ} = 	an(90^{\circ}-33^{\circ}) = \cot 33^{\circ}$$\cot 37^{\circ} = \cot(90^{\circ}-53^{\circ}) = 	an 53^{\circ}$Substituting these into the expression:$= \frac{\cot 33^{\circ} + 	an 53^{\circ}}{	an 33^{\circ} + \cot 53^{\circ}}$$= \frac{\frac{1}{	an 33^{\circ}} + 	an 53^{\circ}}{	an 33^{\circ} + \frac{1}{	an 53^{\circ}}}$$= \frac{\frac{1 + 	an 33^{\circ} 	an 53^{\circ}}{	an 33^{\circ}}}{\frac{	an 33^{\circ} 	an 53^{\circ} + 1}{	an 53^{\circ}}}$$= \frac{1 + 	an 33^{\circ} 	an 53^{\circ}}{	an 33^{\circ}} 	imes \frac{	an 53^{\circ}}{1 + 	an 33^{\circ} 	an 53^{\circ}}$$= \frac{	an 53^{\circ}}{	an 33^{\circ}} = 	an 53^{\circ} \cot 33^{\circ}$Since $	an 53^{\circ} = \cot 37^{\circ}$ and $\cot 33^{\circ} = 	an 57^{\circ}$,the expression is equal to $	an 57^{\circ} \cot 37^{\circ}$.

The expression $\frac{\tan 57^{\circ}+\cot 37^{\circ}}{\tan 33^{\circ}+\cot 53^{\circ}}$ is equal to

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