Question
Prove the following$: \tan\theta \tan(90^\circ - \theta ) = \cot\theta \cot(90^\circ - \theta )$
✓
Answer
$\text{L.H.S.}$
$= \tan\theta \tan(90^\circ - \theta )$
$= \cot(90^\circ - \theta ) \times \cot\theta$
$= \cot\theta \cot(90^\circ - \theta )$
$= \text{R.H.S.}$
$= \tan\theta \tan(90^\circ - \theta )$
$= \cot(90^\circ - \theta ) \times \cot\theta$
$= \cot\theta \cot(90^\circ - \theta )$
$= \text{R.H.S.}$
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