Question

Prove that:

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

Answer 1

\[\begin{array}{l} LHS = \frac{\cos( {90}^\circ -  \theta)\sec( {90}^\circ - \theta)\tan\theta}{\text{cosec} ( {90}^\circ- \theta)\sin( {90}^\circ - \theta)\cot( {90}^\circ - \theta)} + \frac{\tan( {90}^\circ - \theta)}{\cot\theta} \\ \end{array}\]
\[\begin{array}{l}= \frac{\sin\theta    \text           cosec\theta\tan\theta}{\sec\theta\cos\theta\tan\theta} + \frac{\cot\theta}{\cot\theta} \\ \end{array}\]

= 1 + 1

= 2

= RHS

Hence proved.