Question

Prove the following expression
$\operatorname{cosec}\theta \left( \sec \theta -1 \right)-\cot \theta \left( 1-\cos \theta \right)=\tan \theta -\sin \theta$

Answer 1

Hint: First of all, convert the whole expression in LHS in terms of $\sin \theta$ and $\cos \theta$ by using $\operatorname{cosec}x=\dfrac{1}{\sin x}$, $\sec x=\dfrac{1}{\cos x}$ and $\cot x=\dfrac{\cos x}{\sin x}$. Now simplify the expression to get the desired result.

Complete step-by-step solution -
In this question, we have to prove that
$\operatorname{cosec}\left( \sec \theta -1 \right)-\cot \theta \left( 1-\cos \theta \right)=\tan \theta -\sin \theta$
Let us take the LHS of the expression given in the question, we get,
$E=\operatorname{cosec}\theta \left( \sec \theta -1 \right)-\cot \theta \left( 1-\cos \theta \right)$
We know that $\operatorname{cosec}x=\dfrac{1}{\sin x}$. By using this in the above expression, we get,
$E=\dfrac{1}{\sin \theta }\left( \sec \theta -1 \right)-\cot \theta \left( 1-\cos \theta \right)$
We know that $\sec x=\dfrac{1}{\cos x}$. By using this in the above expression, we get,
$E=\dfrac{1}{\sin \theta }\left( \dfrac{1}{\cos \theta }-1 \right)-\cot \theta \left( 1-\cos \theta \right)$
We also know that $\cot x=\dfrac{\cos x}{\sin x}$. By using this in the above expression, we get,
$E=\dfrac{1}{\sin \theta }\left( \dfrac{1}{\cos \theta }-1 \right)-\dfrac{\cos \theta }{\sin \theta }\left( 1-\cos \theta \right)$
By simplifying the above expression, we get,
$E=\dfrac{1}{\sin \theta \cos \theta }-\dfrac{1}{\sin \theta }-\dfrac{\cos \theta }{\sin \theta }+\dfrac{{{\cos }^{2}}\theta }{\sin \theta }$
By taking $\sin \theta \cos \theta$ as LCM and further simplifying the above expression, we get,
$E=\dfrac{1-\cos \theta -{{\cos }^{2}}\theta +{{\cos }^{3}}\theta }{\sin \theta \cos \theta }$
We can also write the above expression as
$E=\dfrac{1\left( 1-\cos \theta \right)-{{\cos }^{2}}\theta \left( 1-\cos \theta \right)}{\sin \theta \cos \theta }$
By taking out $\left( 1-\cos \theta \right)$ common from the above expression, we get,
$E=\dfrac{\left( 1-\cos \theta \right)\left( 1-{{\cos }^{2}}\theta \right)}{\sin \theta \cos \theta }$
We know that $1-{{\cos }^{2}}x={{\sin }^{2}}x$. By using this in the above expression, we get,
$E=\dfrac{\left( 1-\cos \theta \right)\left( {{\sin }^{2}}\theta \right)}{\sin \theta \cos \theta }$
By canceling the like term from the above expression, we get,
$E=\dfrac{\left( 1-\cos \theta \right)\sin \theta }{\cos \theta }$
By taking $\sin \theta$ inside the bracket, we get,
$E=\dfrac{\left( \sin \theta -\sin \theta \cos \theta \right)}{\cos \theta }$
$E=\dfrac{\sin \theta }{\cos \theta }-\dfrac{\sin \theta \cos \theta }{\cos \theta }$
By canceling the like terms from the above expression, we get,
$E=\dfrac{\sin \theta }{\cos \theta }-\sin \theta$
We know that $\dfrac{\sin x}{\cos x}=\tan x$. By using this in the above expression, we get,
$E=\tan \theta -\sin \theta$
So, we get, LHS = RHS
Hence, we have proved that
$\operatorname{cosec}\theta \left( \sec \theta -1 \right)-\cot \theta \left( 1-\cos \theta \right)=\tan \theta -\sin \theta$

Note: In these types of questions, when you could not think of any other identity by looking at the expression, it is advisable to convert the whole expression in terms of $\sin \theta$ and $\cos \theta$. In case, we don’t get the expression which is given in the RHS, then we can take RHS also and convert it in terms of $\sin \theta$ and $\cos \theta$ and show it to be equal to LHS. So, basically in these questions of proof, we can take each side individually, solve them, and prove them equal.