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.