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Question
Prove that:
\[\frac{\cos(90^\circ - \theta)}{1 + \sin(90^\circ - \theta)} + \frac{1 + \sin(90^\circ- \theta)}{\cos(90^\circ - \theta)} = 2 cosec\theta\]
Solution
\[\begin{array}{l}(v) LHS = \frac{\cos( {90}^0 - \theta)}{1 + \sin( {90}^0 - \theta)} + \frac{1 + \sin( {90}^0 - \theta)}{\cos( {90}^0 - \theta)} \\ \end{array}\]
\[\begin{array}{l}= \frac{\sin\theta}{1 + \cos\theta} + \frac{1 + \cos\theta}{\sin\theta} \\ \end{array}\]
\[\begin{array}{l}= \frac{\sin^2 \theta + {(1 + \cos\theta)}^2}{(1 + \cos\theta)\sin\theta} \\ \end{array}\]
\[\begin{array}{l}= \frac{\sin^2 \theta + 1 + \cos^2 \theta + 2\cos\theta}{(1 + \cos\theta)\sin\theta} \\ \end{array}\]
\[\begin{array}{l}= \frac{1 + 1 + 2\cos\theta}{(1 + \cos\theta)\sin\theta} \\ \end{array}\]
\[\begin{array}{l}= \frac{2 + 2\cos\theta}{(1 + \cos\theta)\sin\theta} \\ \end{array}\]
\[\begin{array}{l}= \frac{2(1 + \cos\theta)}{(1 + \cos\theta)\sin\theta} \\ \end{array}\]
\[\begin{array}{l}= 2\frac{1}{\sin\theta} \\ \end{array}\]
\[\begin{array}{l}= 2 \ cosec\theta \\ \end{array}\]
\[\begin{array}{l}= RHS \\ \end{array}\]
= Hence proved
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