Question

Prove that:

\[\frac{\sin\theta}{\cos(90° - \theta)} + \frac{\cos\theta}{\sin(90° - \theta)} = 2\]

Answer 1

\[\begin{array}{l}(ii) LHS=\frac{\sin\theta}{\cos( {90}^0 - \theta)} + \frac{\cos\theta}{\sin( {90}^0 -  \theta)} \\ \end{array}\]
\[\begin{array}{l}= \frac{\sin\theta}{\sin\theta} + \frac{\cos\theta}{\cos\theta} \\ \end{array}\]
\[\begin{array}{l}= 1 + 1 \\ \end{array}\]
\[\begin{array}{l}= 2 \\ \end{array}\]
\[\begin{array}{l}= \text{RHS} \\ \end{array}\]
\[\begin{array}{l}\text{Hence proved} . \\ \end{array}\]