Question

Prove the following: $$\left( 1+\tan^{2} \theta \right) +\left( 1+\dfrac{1}{\tan^{2} \theta } \right) =\dfrac{1}{\sin^{2} \theta -\sin^{4} \theta }$$.

Answer 1

Hint: In this question it is given that we have to prove that $$\left( 1+\tan^{2} \theta \right) +\left( 1+\dfrac{1}{\tan^{2} \theta } \right) =\dfrac{1}{\sin^{2} \theta -\sin^{4} \theta }$$. So to find the solution of this question we will start from the LHS part and we have to apply the formulas,
$$\left( 1+\tan^{2} \theta \right) =\sec^{2} \theta$$.......(1)
$$\left( 1+\cot^{2} \theta \right) =\csc^{2} \theta$$........(2)

Complete step-by-step solution:
Given, LHS,
$$\left( 1+\tan^{2} \theta \right) +\left( 1+\dfrac{1}{\tan^{2} \theta } \right)$$
As we know that $$\dfrac{1}{\tan \theta } =\cot \theta$$
So we can write $$\dfrac{1}{\tan^{2} \theta } =\cot^{2} \theta$$
So by using this we can write,
LHS,
$$\left( 1+\tan^{2} \theta \right) \left( 1+\cot^{2} \theta \right) $$
=$$\sec^{2} \theta \csc^{2} \theta$$
=$$\dfrac{1}{\cos^{2} \theta } \times \dfrac{1}{\sin^{2} \theta }$$ $$[\because \sec \theta =\dfrac{1}{\cos \theta } ,\ \csc \theta =\dfrac{1}{\sin \theta } ]$$
=$$\dfrac{1}{\sin^{2} \theta \cos^{2} \theta }$$
=$$\dfrac{1}{\sin^{2} \theta \left( 1-\sin^{2} \theta \right) }$$ [$$\because \cos^{2} \theta =1-\sin^{2} \theta$$]
=$$\dfrac{1}{\sin^{2} \theta -\sin^{4} \theta }$$
Therefore,
$$\left( 1+\tan^{2} \theta \right) +\left( 1+\dfrac{1}{\tan^{2} \theta } \right) =\dfrac{1}{\sin^{2} \theta -\sin^{4} \theta }$$=RHS
Hence Proved.

Note: To solve this type of question you have to start from any side of the equation, either star from LHS or from RHS but the main thing is that if you start from the RHS part then you can solve it by going in reverse order of the above solution process.