Question:
$\frac{\tan 47^{\circ}}{\cot 43^{\circ}}=1$
Solution:
True
$\frac{\tan 47^{\circ}}{\cot 43^{\circ}}=\frac{\tan \left(90^{\circ}-43^{\circ}\right)}{\cot 43^{\circ}}=\frac{\cot 43^{\circ}}{\cot 43^{\circ}}=1$
$\left[\because \tan \left(90^{\circ}-\theta\right)=\cot \theta\right]$
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