Question

Prove that \[\left( 1+\dfrac{1}{{{\tan }^{2}}\theta } \right)\left( 1+\dfrac{1}{{{\cot }^{2}}\theta } \right)=\dfrac{1}{{{\sin }^{2}}\theta -{{\sin }^{4}}\theta }\].

Answer 1

Hint: We start solving the problem by considering the L.H.S (Left hand side) of the given proof and substituting $\dfrac{1}{\tan \theta }=\cot \theta $ and $\dfrac{1}{\cot \theta }=\tan \theta $ in it. We then make use of the trigonometric identities $\left( 1+{{\cot }^{2}}\theta \right)={{\operatorname{cosec}}^{2}}\theta $ and $\left( 1+{{\tan }^{2}}\theta \right)={{\sec }^{2}}\theta $ to proceed through the problem. We then make use of the results $\sec \theta =\dfrac{1}{\cos \theta }$, $\operatorname{cosec}\theta =\dfrac{1}{\sin \theta }$ and ${{\cos }^{2}}\theta =1-{{\sin }^{2}}\theta $ to proceed further into the problem. We then make the necessary calculations to complete the required proof.

Complete step by step answer:
According to the problem, we need to prove the result \[\left( 1+\dfrac{1}{{{\tan }^{2}}\theta } \right)\left( 1+\dfrac{1}{{{\cot }^{2}}\theta } \right)=\dfrac{1}{{{\sin }^{2}}\theta -{{\sin }^{4}}\theta }\].
Let us consider the L.H.S (Left Hand Side) and try to prove it equal to the R.H.S (Right Hand Side).
So, we have \[\left( 1+\dfrac{1}{{{\tan }^{2}}\theta } \right)\left( 1+\dfrac{1}{{{\cot }^{2}}\theta } \right)\].
We know that $\dfrac{1}{\tan \theta }=\cot \theta $ and $\dfrac{1}{\cot \theta }=\tan \theta $. Let us substitute these results.
\[\Rightarrow \left( 1+\dfrac{1}{{{\tan }^{2}}\theta } \right)\left( 1+\dfrac{1}{{{\cot }^{2}}\theta } \right)=\left( 1+{{\cot }^{2}}\theta \right)\left( 1+{{\tan }^{2}}\theta \right)\] ---(1).
We know that $\left( 1+{{\cot }^{2}}\theta \right)={{\operatorname{cosec}}^{2}}\theta $ and $\left( 1+{{\tan }^{2}}\theta \right)={{\sec }^{2}}\theta $. Let us substitute these results in equation (1).
\[\Rightarrow \left( 1+\dfrac{1}{{{\tan }^{2}}\theta } \right)\left( 1+\dfrac{1}{{{\cot }^{2}}\theta } \right)=\left( {{\operatorname{cosec}}^{2}}\theta \right)\left( {{\sec }^{2}}\theta \right)\] ---(2).
We know that $\sec \theta =\dfrac{1}{\cos \theta }$ and $\operatorname{cosec}\theta =\dfrac{1}{\sin \theta }$. Let us substitute these results in equation (2).
\[\Rightarrow \left( 1+\dfrac{1}{{{\tan }^{2}}\theta } \right)\left( 1+\dfrac{1}{{{\cot }^{2}}\theta } \right)=\left( \dfrac{1}{{{\sin }^{2}}\theta } \right)\left( \dfrac{1}{{{\cos }^{2}}\theta } \right)\].
\[\Rightarrow \left( 1+\dfrac{1}{{{\tan }^{2}}\theta } \right)\left( 1+\dfrac{1}{{{\cot }^{2}}\theta } \right)=\dfrac{1}{{{\sin }^{2}}\theta {{\cos }^{2}}\theta }\] ---(3).
We know that ${{\sin }^{2}}\theta +{{\cos }^{2}}\theta =1\Leftrightarrow {{\cos }^{2}}\theta =1-{{\sin }^{2}}\theta $. Let us substitute this result in equation (3).
\[\Rightarrow \left( 1+\dfrac{1}{{{\tan }^{2}}\theta } \right)\left( 1+\dfrac{1}{{{\cot }^{2}}\theta } \right)=\dfrac{1}{{{\sin }^{2}}\theta \left( 1-{{\sin }^{2}}\theta \right)}\].
\[\Rightarrow \left( 1+\dfrac{1}{{{\tan }^{2}}\theta } \right)\left( 1+\dfrac{1}{{{\cot }^{2}}\theta } \right)=\dfrac{1}{{{\sin }^{2}}\theta -{{\sin }^{4}}\theta }\].
So, we can see that L.H.S (Left hand side) is equal to R.H.S (Right Hand Side).

∴ We have proved the trigonometric result \[\left( 1+\dfrac{1}{{{\tan }^{2}}\theta } \right)\left( 1+\dfrac{1}{{{\cot }^{2}}\theta } \right)=\dfrac{1}{{{\sin }^{2}}\theta -{{\sin }^{4}}\theta }\].

Note: We should perform each step carefully in order to complete the proof properly. We can also solve this problem as shown below:
We have \[\left( 1+\dfrac{1}{{{\tan }^{2}}\theta } \right)\left( 1+\dfrac{1}{{{\cot }^{2}}\theta } \right)\] ---(4).
We know that $\tan \theta =\dfrac{\sin \theta }{\cos \theta }$ and $\cot \theta =\dfrac{\cos \theta }{\sin \theta }$. Let us substitute this results in equation (4).
$\Rightarrow \left( 1+\dfrac{1}{{{\tan }^{2}}\theta } \right)\left( 1+\dfrac{1}{{{\cot }^{2}}\theta } \right)=\left( 1+\dfrac{1}{\dfrac{{{\sin }^{2}}\theta }{{{\cos }^{2}}\theta }} \right)\left( 1+\dfrac{1}{\dfrac{{{\cos }^{2}}\theta }{{{\sin }^{2}}\theta }} \right)$.
$\Rightarrow \left( 1+\dfrac{1}{{{\tan }^{2}}\theta } \right)\left( 1+\dfrac{1}{{{\cot }^{2}}\theta } \right)=\left( 1+\dfrac{{{\cos }^{2}}\theta }{{{\sin }^{2}}\theta } \right)\left( 1+\dfrac{{{\sin }^{2}}\theta }{{{\cos }^{2}}\theta } \right)$.
$\Rightarrow \left( 1+\dfrac{1}{{{\tan }^{2}}\theta } \right)\left( 1+\dfrac{1}{{{\cot }^{2}}\theta } \right)=\left( \dfrac{{{\sin }^{2}}\theta +{{\cos }^{2}}\theta }{{{\sin }^{2}}\theta } \right)\left( \dfrac{{{\cos }^{2}}\theta +{{\sin }^{2}}\theta }{{{\cos }^{2}}\theta } \right)$ ---(5).
We know that ${{\sin }^{2}}\theta +{{\cos }^{2}}\theta =1$. Let us substitute this in equation (5).
$\Rightarrow \left( 1+\dfrac{1}{{{\tan }^{2}}\theta } \right)\left( 1+\dfrac{1}{{{\cot }^{2}}\theta } \right)=\left( \dfrac{1}{{{\sin }^{2}}\theta } \right)\left( \dfrac{1}{{{\cos }^{2}}\theta } \right)$.
$\Rightarrow \left( 1+\dfrac{1}{{{\tan }^{2}}\theta } \right)\left( 1+\dfrac{1}{{{\cot }^{2}}\theta } \right)=\dfrac{1}{{{\sin }^{2}}\theta {{\cos }^{2}}\theta }$ ---(6).
We know that ${{\sin }^{2}}\theta +{{\cos }^{2}}\theta =1\Leftrightarrow {{\cos }^{2}}\theta =1-{{\sin }^{2}}\theta $. Let us substitute this result in equation (6).
\[\Rightarrow \left( 1+\dfrac{1}{{{\tan }^{2}}\theta } \right)\left( 1+\dfrac{1}{{{\cot }^{2}}\theta } \right)=\dfrac{1}{{{\sin }^{2}}\theta \left( 1-{{\sin }^{2}}\theta \right)}\].
\[\Rightarrow \left( 1+\dfrac{1}{{{\tan }^{2}}\theta } \right)\left( 1+\dfrac{1}{{{\cot }^{2}}\theta } \right)=\dfrac{1}{{{\sin }^{2}}\theta -{{\sin }^{4}}\theta }\].