Solve frac{tan 65^{circ}}{cot 25^{circ}}

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8. Introduction to Trigonometry

easy

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

A

$0$

B

$2$

C

$5$

D

$1$

Solution

We Know: $\cot A=\tan \left(90^{\circ}- A \right)$

So,$\cot 25^{\circ}=\tan \left(90^{\circ}-25^{\circ}\right)=\tan 65^{\circ}$

i.e., $\frac{\tan 65^{\circ}}{\cot 25^{\circ}}=\frac{\tan 65^{\circ}}{\tan 65^{\circ}}=1$

Standard 10

Mathematics

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