Question

Find x from the following equations:
\[x \cot\left( \frac{\pi}{2} + \theta \right) + \tan\left( \frac{\pi}{2} + \theta \right)\sin \theta + cosec\left( \frac{\pi}{2} + \theta \right) = 0\]

Answer 1

\[90^\circ = \frac{\pi}{2}\]

 We have: 

\[ x \cot\left( 90^\circ + \theta \right) + \tan\left( 90^\circ + \theta \right) \sin \theta + cosec\left( 90^\circ + \theta \right) = 0\]

\[ \Rightarrow x \left[ - \tan \theta \right] + \left[ - \cot \theta \right] \sin \theta + \sec \theta = 0\]

\[ \Rightarrow - x \tan \theta - \cot \theta \sin \theta + \sec \theta = 0 \]

\[ \Rightarrow - x \times \frac{\sin \theta}{\cos \theta} - \frac{\cos \theta}{\sin \theta} \times \sin \theta + \frac{1}{\cos \theta} = 0 \]

\[ \Rightarrow - x \times \frac{\sin \theta}{\cos \theta} - \cos\theta + \frac{1}{\cos \theta} = 0 \]

\[ \Rightarrow \frac{- x \sin \theta - \cos^2 \theta + 1}{\cos \theta} = 0 \]

`=>-xsintheta-cos^2theta+1=0`

`=>-xsintheta = - sin^2theta   ...[∵ 1 - cos^2theta = sin^2theta]`

`=>x=(-sin^2theta)/(-sintheta)`

`=>x=sintheta`