Question

Find $x$ from the following equation:
1.$coesc\left( {\dfrac{\pi }{2} + \theta } \right) + x\cos \theta \cot \left( {\dfrac{\pi }{2} + \theta } \right) = \sin \left( {\dfrac{\pi }{2} + \theta } \right)$
2.$x\cot \left( {\dfrac{\pi }{2} + \theta } \right) + \sin \theta \tan \left( {\dfrac{\pi }{2} + \theta } \right) + \cos ec\left( {\dfrac{\pi }{2} + \theta } \right) = 0$

Answer 1

First we will convert $\left( {\dfrac{\pi }{2} + \theta } \right)$ into $\theta$ after that we will convert all the trigonometric ratio by using $\sec \theta = \dfrac{1}{{\cos \theta }}$,$\tan \theta = \dfrac{{\sin \theta }}{{\cos \theta }}$ and $\cot \theta = \dfrac{{\cos \theta }}{{\sin \theta }}$ after simplifying that equate the linear equation with 0 and find the value of $x$.

Complete step-by-step answer:
1:- Given $coesc\left( {\dfrac{\pi }{2} + \theta } \right) + x\cos \theta \cot \left( {\dfrac{\pi }{2} + \theta } \right) = \sin \left( {\dfrac{\pi }{2} + \theta } \right)$
Convert $\left( {\dfrac{\pi }{2} + \theta } \right)$into $\theta$ we get
$\Rightarrow \sec \theta - x\cos \theta \tan \theta = \cos \theta$
We know that$\sec \theta = \dfrac{1}{{\cos \theta }}$ and $\tan \theta = \dfrac{{\sin \theta }}{{\cos \theta }}$ substituting these values, we get,
$\Rightarrow \dfrac{1}{{\cos \theta }} - x\sin \theta = \cos \theta$
Multiplying $\cos \theta$ on both side we get,
$\Rightarrow 1 - x\cos \theta \sin \theta = {\cos ^2}\theta$
We know that $1 - {\cos ^2}\theta = {\sin ^2}\theta$
$\Rightarrow {\sin ^2}\theta - x\sin \theta \cos \theta = 0$
Taking $\sin \theta$ common
$\Rightarrow \sin \theta \left( {\sin \theta - x\cos \theta } \right) = 0$
Now,
$\Rightarrow \sin \theta - x\cos \theta = 0 \\\ \therefore x = \tan \theta \\\$

2: - Given $x\cot \left( {\dfrac{\pi }{2} + \theta } \right) + \sin \theta \tan \left( {\dfrac{\pi }{2} + \theta } \right) + \cos ec\left( {\dfrac{\pi }{2} + \theta } \right) = 0$
Convert $\left( {\dfrac{\pi }{2} + \theta } \right)$ into $\theta$ we get
$\Rightarrow - x\tan \theta - sin\theta \cot \theta + \sec \theta = 0$
We know that $\sec \theta = \dfrac{1}{{\cos \theta }}$,$\tan \theta = \dfrac{{\sin \theta }}{{\cos \theta }}$ and $\cot \theta = \dfrac{{\cos \theta }}{{\sin \theta }}$ substituting these values, we get,
$\Rightarrow - x\dfrac{{\sin \theta }}{{\cos \theta }} - \cos \theta + \dfrac{1}{{\cos \theta }} = 0$
Multiplying $\cos \theta$on both side we get,
$\Rightarrow 1 - x\sin \theta = {\cos ^2}\theta$
We know that $1 - {\cos ^2}\theta = {\sin ^2}\theta$
$\Rightarrow {\sin ^2}\theta - x\sin \theta \cos \theta = 0$
Taking $\sin \theta$ common
$\Rightarrow \sin \theta \left( {\sin \theta - x\cos \theta } \right) = 0$
Now,
$\Rightarrow \sin \theta - x\cos \theta = 0 \\\ \therefore x = \tan \theta \\\$

Note: Trigonometric formulas used in this question $coesc\left( {\dfrac{\pi }{2} + \theta } \right) = \sec \theta$,$\cot \left( {\dfrac{\pi }{2} + \theta } \right) = - \tan \theta$,$\sin \left( {\dfrac{\pi }{2} + \theta } \right) = \cos \theta$and must be taken into consideration. We can also say as $\sin \theta = 0$then $\theta = n\pi$. But in this question, we cannot consider $\theta = n\pi$because when we consider it $x = 0$which contradicts the question.