Question

If $\text{cosec}\theta -\cot \theta =p$ , then the value of $\text{cosec}\theta$ is
a. $\dfrac{1}{2}\left( p+\dfrac{1}{p} \right)$
b. $\dfrac{1}{2}\left( 1-\dfrac{1}{p} \right)$
c. $\dfrac{1}{2}\left( 1+\dfrac{1}{p} \right)$
d. $\left( 1-\dfrac{1}{p} \right)$

Answer 1

Hint: We need to find the value of $\text{cosec}\theta$ if $\text{cosec}\theta -\cot \theta =p$ . We start to solve the given question by expressing the given equation in terms of a trigonometric function $\sin \theta$ . Then, we find the value of $\text{cosec}\theta$ from the value of $\sin \theta$ .

Complete step by step answer:
We are given an equation $\text{cosec}\theta -\cot \theta =p$ and need to find the value of $\text{cosec}\theta$ in terms of $p$ . We will be solving the given question by expressing the given equation in terms of a trigonometric function $\sin \theta$ and then solving for $\text{cosec}\theta$
From trigonometry, we know that $\text{cosec}\theta$ is inverse of the trigonometric function $\sin \theta$ . It is expressed as follows,
$\Rightarrow \text{cosec}\theta =\dfrac{1}{\sin \theta }$
From trigonometry, we know that $\cot \theta$ is the ratio of the trigonometric functions $\cos \theta$ and $\sin \theta$ . It is expressed as follows,
$\Rightarrow \cot \theta =\dfrac{\cos \theta }{\sin \theta }$
The given equation is,
$\Rightarrow \text{cosec}\theta -\cot \theta =p$
Substituting the above values in the equation $\text{cosec}\theta -\cot \theta =p$ , we get,
$\Rightarrow \dfrac{1}{\sin \theta }-\dfrac{\cos \theta }{\sin \theta }=p$
Taking the denominator common from both the terms, we get,
$\Rightarrow \dfrac{\left( 1-\cos \theta \right)}{\sin \theta }=p$
Shifting $\sin \theta$ to other side of the equation, we get,
$\Rightarrow 1-\cos \theta =p\sin \theta$
$\Rightarrow 1-p\sin \theta =\cos \theta$
From the formulae of trigonometry, we know that ${{\sin }^{2}}\theta +{{\cos }^{2}}\theta =1$
We need to find the value of $\cos \theta$ from the above formula.
$\Rightarrow {{\sin }^{2}}\theta +{{\cos }^{2}}\theta =1$
$\Rightarrow {{\cos }^{2}}\theta =1-{{\sin }^{2}}\theta$
$\Rightarrow \cos \theta =\sqrt{1-{{\sin }^{2}}\theta }$
Substituting the value of $\cos \theta$ , we get,
$\Rightarrow 1-p\sin \theta =\sqrt{1-{{\sin }^{2}}\theta }$
Squaring the above equation on both sides, we get,
$\Rightarrow {{\left( 1-p\sin \theta \right)}^{2}}={{\left( \sqrt{1-{{\sin }^{2}}\theta } \right)}^{2}}$
Simplifying the above equation, we get,
$\Rightarrow 1+{{p}^{2}}{{\sin }^{2}}\theta -2p\sin \theta =1-{{\sin }^{2}}\theta$
Shifting all the terms to the left-hand side of the equation, we get,
$\Rightarrow {{p}^{2}}{{\sin }^{2}}\theta +{{\sin }^{2}}\theta -2p\sin \theta +1-1=0$
$\Rightarrow \left( {{p}^{2}}+1 \right){{\sin }^{2}}\theta -2p\sin \theta =0$
Taking the value of $\sin \theta$ common,
$\Rightarrow \sin \theta \left( \left( {{p}^{2}}+1 \right)\sin \theta -2p \right)=0$
$\Rightarrow \sin \theta =0\text{ or }\left( {{p}^{2}}+1 \right)\sin \theta -2p=0$
We need to express $\sin \theta$ in terms of $p$ .
Following the same,
$\Rightarrow \left( {{p}^{2}}+1 \right)\sin \theta -2p=0$
$\Rightarrow \left( {{p}^{2}}+1 \right)\sin \theta =2p$
$\Rightarrow \sin \theta =\dfrac{2p}{\left( {{p}^{2}}+1 \right)}$
We know that
$\Rightarrow$$\text{cosec}\theta =\dfrac{1}{\sin \theta }$
Substituting the value of $\sin \theta$,
$\Rightarrow \text{cosec}\theta =\dfrac{1}{\dfrac{2p}{\left( {{p}^{2}}+1 \right)}}$
$\Rightarrow \text{cosec}\theta =\dfrac{{{p}^{2}}+1}{2p}$
Dividing each term of numerator with $p$, we get,
$\Rightarrow \text{cosec}\theta =\dfrac{1}{2}\left( \dfrac{{{p}^{2}}}{p}+\dfrac{1}{p} \right)$
Simplifying the above equation, we get,
$\therefore \text{cosec}\theta =\dfrac{1}{2}\left( p+\dfrac{1}{p} \right)$

So, the correct answer is “Option a”.

Note: We must know that $\text{cosec}\theta$ is the inverse of the trigonometric function $\sin \theta$ and it is given by $\text{cosec}\theta =\dfrac{1}{\sin \theta }$ and not $\text{cosec}\theta ={{\sin }^{-1}}\theta$ . The given problem can be solved easily if we know the relationship between the various trigonometric functions such as $\cot \theta ,\sin \theta$ .