Question

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

Answer 1

Hint: First of all, take LHS and convert the whole expression in terms of \[\sin \theta \] and \[\cos \theta \]. Now simplify the whole equation and use identity \[{{\sin }^{2}}x+{{\cos }^{2}}x=1\] to get the desired result.

Complete step- by-step-by-step solution -
In this question, we have to prove that
\[\left( \operatorname{cosec}\theta -\sin \theta \right)\left( \sec \theta -\cos \theta \right)\left( \tan \theta +\cot \theta \right)=1\]
First of all, let us take the LHS of the above equation.
\[E=\left( \operatorname{cosec}\theta -\sin \theta \right)\left( \sec \theta -\cos \theta \right)\left( \tan \theta +\cot \theta \right)\]
We know that \[\operatorname{cosec}x=\dfrac{1}{\sin x}\]. By using this in the above expression, we get,
\[E=\left( \dfrac{1}{\sin \theta }-\sin \theta \right)\left( \sec \theta -\operatorname{cos}\theta \right)\left( \tan \theta +\cot \theta \right)\]
Also, we know that \[\sec x=\dfrac{1}{\cos x}\]. By using this in the above expression, we get,
\[E=\left( \dfrac{1}{\sin \theta }-\sin \theta \right)\left( \dfrac{1}{\cos \theta }-\cos \theta \right)\left( \tan \theta +\cot \theta \right)\]
Now, we know that \[\tan x=\dfrac{\sin x}{\cos x}\] and \[\cot x=\dfrac{\cos x}{\sin x}\]. By using this in the above expression, we get,
\[E=\left( \dfrac{1}{\sin \theta }-\sin \theta \right)\left( \dfrac{1}{\cos \theta }-\cos \theta \right)\left( \dfrac{\sin \theta }{\cos \theta }+\dfrac{\cos \theta }{\sin \theta } \right)\]
By simplifying the above expression, we get,
\[E=\left( \dfrac{1-{{\sin }^{2}}\theta }{\sin \theta } \right)\left( \dfrac{1-{{\cos }^{2}}\theta }{\cos \theta } \right)\left( \dfrac{{{\sin }^{2}}\theta +{{\cos }^{2}}\theta }{\sin \theta \cos \theta } \right)\]
We know that \[{{\sin }^{2}}x+{{\cos }^{2}}x=1\]. By using this in the above expression, we get,
\[E=\left( \dfrac{1-{{\sin }^{2}}\theta }{\sin \theta } \right)\left( \dfrac{1-{{\cos }^{2}}\theta }{\cos \theta } \right)\left( \dfrac{1}{\sin \theta \cos \theta } \right)\]
Also, we know that \[1-{{\sin }^{2}}x={{\cos }^{2}}x\] and \[1-{{\cos }^{2}}x={{\sin }^{2}}x\]. By using this in the above expression, we get,
\[E=\left( \dfrac{{{\cos }^{2}}\theta }{\sin \theta } \right).\left( \dfrac{{{\sin }^{2}}\theta }{\cos \theta } \right)\left( \dfrac{1}{\sin \theta \cos \theta } \right)\]
We can also write the above equation as
\[E=\dfrac{\left( {{\cos }^{2}}\theta \right).\left( {{\sin }^{2}}\theta \right)}{\left( {{\cos }^{2}}\theta \right).\left( {{\sin }^{2}}\theta \right)}\]
By canceling the like terms from RHS of the above equation, we get,
E = 1
So, we get LHS = RHS
Hence, we have proved
\[\left( \operatorname{cosec}\theta -\sin \theta \right)\left( \sec \theta -\cos \theta \right)\left( \tan \theta +\cot \theta \right)=1\]

Note: In these types of questions which have multiple trigonometric ratios, it is always advisable to convert the whole expression in terms of \[\sin \theta \] and \[\cos \theta \] especially when RHS is a constant value like in the above question. Also, we not only need to remember \[{{\sin }^{2}}x+{{\cos }^{2}}x=1\] but its various forms like \[{{\sin }^{2}}x=1-{{\cos }^{2}}x\] or \[{{\cos }^{2}}x=1-{{\sin }^{2}}x\] also. We should not get confused with multiple formulas.