Question

Prove that:

\[\frac{\sin\theta  \cos(90^\circ - \theta)\cos\theta}{\sin(90^\circ- \theta)} + \frac{\cos\theta  \sin(90^\circ - \theta)\sin\theta}{\cos(90^\circ - \theta)}\]

Answer 1

LHS \[\begin{array}{l} = \frac{\sin\theta\cos( {90}^\circ - \theta)\cos\theta}{\sin( {90}^\circ - \theta)} + \frac{\cos\theta\sin( {90}^\circ - \theta)\sin\theta}{\cos( {90}^\circ - \theta)} \\ \end{array}\]
\[\begin{array}{l}= \frac{\sin\theta\sin\theta\cos\theta}{\cos\theta} + \frac{\cos\theta\cos\theta\sin\theta}{\sin\theta} \\ \end{array}\]
\[\begin{array}{l}= \sin^2 \theta + \cos^2 \theta \\ \end{array}\]
= 1

= RHS

Hence proved.