Evaluate frac{tan 65^{circ}}{cot 25^{circ}} | vedclass

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Question

We Know: $\cot A=	an \left(90^{\circ}- A ight)$

So,$\cot 25^{\circ}=	an \left(90^{\circ}-25^{\circ}ight)=	an 65^{\circ}$

i.e., $\frac{	a

Vedclass · Accepted Answer

$1$

Vedclass · Answer

We Know: $\cot A=	an \left(90^{\circ}- A ight)$

So,$\cot 25^{\circ}=	an \left(90^{\circ}-25^{\circ}ight)=	an 65^{\circ}$

i.e., $\frac{	an 65^{\circ}}{\cot 25^{\circ}}=\frac{	an 65^{\circ}}{	an 65^{\circ}}=1$

Evaluate $\frac{\tan 65^{\circ}}{\cot 25^{\circ}}$

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