Question

Find x from the following equations:
\[cosec\left( \frac{\pi}{2} + \theta \right) + x \cos \theta \cot\left( \frac{\pi}{2} + \theta \right) = \sin\left( \frac{\pi}{2} + \theta \right)\]

Answer 1

\[90^\circ = \frac{\pi}{2}\]
 We have: 
\[ cosec\left( 90^\circ + \theta \right) + x \cos \theta \cot\left( 90^\circ + \theta \right) = \sin\left( 90^\circ + \theta \right)\]
\[ \Rightarrow \sec \theta + x \cos \theta \left[ - \tan \theta \right] = \cos \theta\]
\[ \Rightarrow \sec \theta - x cos\theta tan\theta = \cos \theta\]
\[ \Rightarrow \sec \theta - x cos\theta \times \frac{\sin \theta}{\cos \theta} = \cos \theta\]
\[ \Rightarrow \sec \theta - x \sin\theta = \cos \theta\]
\[ \Rightarrow \sec \theta - \cos \theta = x \sin\theta$\]
\[ \Rightarrow \frac{1}{\cos \theta} - cos\theta = x \sin\theta\]
\[ \Rightarrow \frac{1 - \cos^2 \theta}{\cos \theta} = x \sin\theta$\]
\[ \Rightarrow \frac{\sin^2 \theta}{cos\theta} = x \sin\theta\]
\[ \Rightarrow \frac{\sin^2 \theta}{\cos \theta \sin \theta} = x\]
\[ \Rightarrow \frac{\sin \theta}{\cos \theta} = x\]
\[ \Rightarrow \tan\theta = x\]
\[ \therefore x = \tan\theta\]