Question
Prove that $\cot \theta . \tan (90^\circ - \theta ) - \sec (90^\circ - \theta ). cosec \theta + 1 = 0$.
✓
Answer
$LHS = \cot \theta . \tan (90^\circ - \theta ) - \sec (90^\circ - \theta ). cosec \theta + 1= \cot \theta . \cot \theta - cosec \theta . cosec \theta + 1$
$= (\cot^2\theta - cosec^2\theta ) + 1$
$= - 1 + 1 = 0$
$= RHS$
Hence proved.
$= (\cot^2\theta - cosec^2\theta ) + 1$
$= - 1 + 1 = 0$
$= RHS$
Hence proved.
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