Question
Evaluate: $\frac{\tan 65^{\circ}}{\cot 25^{\circ}}$
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Answer
$\frac{\tan 65^{\circ}}{\cot 25^{\circ}}=\frac{\tan (90-25)^{\circ}}{\cot 25^{\circ}}$
$\frac{\cot 25^{\circ}}{\cot 25^{\circ}}=1[\because \tan (90-\theta)=\cot \theta]$
$\frac{\cot 25^{\circ}}{\cot 25^{\circ}}=1[\because \tan (90-\theta)=\cot \theta]$
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