Find x from the following equations:

$90^{\circ}=\frac{\pi}{2}$

(i) We have,

$\operatorname{cosec}\left(90^{\circ}+\theta\right)+x \cos \theta \cot \left(90^{\circ}+\theta\right)=\sin \left(90^{\circ}+\theta\right)$

$\Rightarrow \sec \theta+x \cos \theta[-\tan \theta]=\cos \theta$

$\Rightarrow \sec \theta-x \cos \theta \tan \theta=\cos \theta$

$\Rightarrow \sec \theta-x \cos \theta \times \frac{\sin \theta}{\cos \theta}=\cos \theta$

$\Rightarrow \sec \theta-x \sin \theta=\cos \theta$

$\Rightarrow \sec \theta-\cos \theta=x \sin \theta$

$\Rightarrow \frac{1}{\cos \theta}-\cos \theta=x \sin \theta$

$\Rightarrow \frac{1-\cos ^{2} \theta}{0}=x \sin \theta$

$\Rightarrow \frac{\sin ^{2} \theta}{\cos \theta}=x \sin \theta$

$\Rightarrow \frac{\sin ^{2} \theta}{\cos \theta \sin \theta}=x$

$\Rightarrow \frac{\sin \theta}{\cos \theta}=x$

$\Rightarrow \tan \theta=x$

$\therefore \quad x=\tan \theta$

(ii) We have,

$x \cot \left(90^{\circ}+\theta\right)+\tan \left(90^{\circ}+\theta\right) \sin \theta+\operatorname{cosec}\left(90^{\circ}+\theta\right)=0$

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

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

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

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

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

$\Rightarrow-x \sin \theta-\cos ^{2} \theta+1=0$

$\Rightarrow-x \sin \theta+\sin ^{2} \theta=0$

$\Rightarrow x \sin \theta=\sin ^{2} \theta$

$\Rightarrow x=\frac{\sin ^{2} \theta}{\sin \theta}$

$\Rightarrow x=\sin \theta$