Question

Prove the trigonometric expression $\dfrac{{{\tan }^{3}}\theta }{1+{{\tan }^{2}}\theta }+\dfrac{{{\cot }^{3}}\theta }{1+{{\cot }^{2}}\theta }=\sec \theta \text{cosec}\theta -2\sin \theta \cos \theta $

Answer 1

Hint: To solve this question, you must know the following conversions:
$\begin{align}
  & \cot \theta =\dfrac{1}{\tan \theta } \\
 & \tan \theta =\dfrac{\sin \theta }{\cos \theta } \\
 & \sec \theta =\dfrac{1}{\cos \theta } \\
 & \text{cosec}\theta =\dfrac{1}{\sin \theta } \\
\end{align}$
There are also a few identities to keep in mind:
$\begin{align}
  & {{\sin }^{2}}\theta +{{\cos }^{2}}\theta =1 \\
 & {{\tan }^{2}}\theta +1={{\sec }^{2}}\theta \\
 & {{\cot }^{2}}\theta +1=\text{cosec}^{2}\theta \\
\end{align}$
Here, we will first simplify the left side of the equation (L.H.S) and equate it to the right side of the equation (R.H.S). This would prove our result.

Complete step-by-step solution:
Let us start with the L.H.S as
$\begin{align}
  & \dfrac{{{\tan }^{3}}\theta }{1+{{\tan }^{2}}\theta }+\dfrac{{{\cot }^{3}}\theta }{1+{{\cot }^{2}}\theta } \\
 & \\
\end{align}$
Using $\cot \theta =\dfrac{1}{\tan \theta }$, we get
$\begin{align}
  & \dfrac{{{\tan }^{3}}\theta }{1+{{\tan }^{2}}\theta }+\dfrac{\dfrac{1}{{{\tan }^{3}}\theta }}{1+\dfrac{1}{{{\tan }^{2}}\theta }} \\
 & =\dfrac{{{\tan }^{3}}\theta }{1+{{\tan }^{2}}\theta }+\dfrac{{{\tan }^{2}}\theta }{{{\tan }^{3}}\theta \left( 1+{{\tan }^{2}}\theta \right)} \\
 & =\dfrac{{{\tan }^{3}}\theta }{1+{{\tan }^{2}}\theta }+\dfrac{1}{\tan \theta \left( 1+{{\tan }^{2}}\theta \right)} \\
\end{align}$
Taking L.C.M of denominator, we get
$\dfrac{{{\tan }^{4}}\theta +1}{\tan \theta \left( 1+{{\tan }^{2}}\theta \right)}$
Replacing $\tan \theta =\dfrac{\sin \theta }{\cos \theta }$,
$\begin{align}
  & \dfrac{\dfrac{{{\sin }^{4}}\theta }{{{\cos }^{4}}\theta }+1}{\dfrac{\sin \theta }{\cos \theta }\left( 1+\dfrac{{{\sin }^{2}}\theta }{{{\cos }^{2}}\theta } \right)} \\
 & =\dfrac{{{\sin }^{4}}\theta +{{\cos }^{4}}\theta }{\dfrac{\sin \theta {{\cos }^{4}}\theta \left( {{\sin }^{2}}\theta +{{\cos }^{2}}\theta \right)}{{{\cos }^{3}}\theta }} \\
\end{align}$
Using the identity ${{\sin }^{2}}\theta +{{\cos }^{2}}\theta =1$,
$=\dfrac{{{\sin }^{4}}\theta +{{\cos }^{4}}\theta }{\sin \theta \cos \theta }$
Now, adding and subtracting $2{{\sin }^{2}}\theta {{\cos }^{2}}\theta $ from the numerator,
$\begin{align}
  & =\dfrac{{{\sin }^{4}}\theta +{{\cos }^{4}}\theta +2{{\sin }^{2}}\theta {{\cos }^{2}}\theta -2{{\sin }^{2}}\theta {{\cos }^{2}}\theta }{\sin \theta \cos \theta } \\
 & =\dfrac{{{\left( {{\sin }^{2}}\theta \right)}^{2}}+{{\left( {{\cos }^{2}}\theta \right)}^{2}}+2{{\sin }^{2}}\theta {{\cos }^{2}}\theta -2{{\sin }^{2}}\theta {{\cos }^{2}}\theta }{\sin \theta \cos \theta } \\
\end{align}$
Using the identity ${{\left( a+b \right)}^{2}}={{a}^{2}}+{{b}^{2}}+2ab$ , we get
$=\dfrac{{{\left( {{\sin }^{2}}\theta +{{\cos }^{2}}\theta \right)}^{2}}-2{{\sin }^{2}}\theta {{\cos }^{2}}\theta }{\sin \theta \cos \theta }$
As we know that ${{\sin }^{2}}\theta +{{\cos }^{2}}\theta =1$,
$\begin{align}
  & =\dfrac{1-2{{\sin }^{2}}\theta {{\cos }^{2}}\theta }{\sin \theta \cos \theta } \\
 & =\dfrac{1}{\sin \theta \cos \theta }-2\sin \theta \cos \theta \\
 & \\
\end{align}$
As we know that,
$\begin{align}
  & \sec \theta =\dfrac{1}{\cos \theta } \\
 & \text{cosec}\theta =\dfrac{1}{\sin \theta } \\
\end{align}$
Hence, we get
$=\sec \theta \text{cosec}\theta -2\sin \theta \cos \theta $
Comparing L.H.S with R.H.S, we get
$L.H.S=R.H.S$
Hence, we have proved the above result.

Note: Here is an alternative and much shorter way to prove the above expression:
L.H.S
$\dfrac{{{\tan }^{3}}\theta }{1+{{\tan }^{2}}\theta }+\dfrac{{{\cot }^{3}}\theta }{1+{{\cot }^{2}}\theta }\text{ }$
Using the identities
$\begin{align}
  & {{\tan }^{2}}\theta +1={{\sec }^{2}}\theta \\
 & {{\cot }^{2}}\theta +1=\text{cosec}^{2}\theta \\
\end{align}$
We get,
$=\dfrac{\tan \theta \left( {{\sec }^{2}}\theta -1 \right)}{{{\sec }^{2}}\theta }+\dfrac{\cot \theta \left( \text{cosec}^{2}\theta -1 \right)}{\text{cosec}^{2}\theta }\text{ }$
On breaking the numerator further to separate different terms, we can write it as
$=\dfrac{\tan \theta \left( {{\sec }^{2}}\theta \right)}{{{\sec }^{2}}\theta }-\dfrac{\tan \theta }{{{\sec }^{2}}\theta }+\dfrac{\cot \theta \left( \text{cosec}^{2}\theta \right)}{\text{cosec}^{2}\theta }-\dfrac{\cot \theta }{\text{cosec}^{2}\theta }$
Using the following conversions
$\begin{align}
  & \cot \theta =\dfrac{1}{\tan \theta } \\
 & \tan \theta =\dfrac{\sin \theta }{\cos \theta } \\
 & \sec \theta =\dfrac{1}{\cos \theta } \\
 & \text{cosec}\theta =\dfrac{1}{\sin \theta } \\
\end{align}$
We get,
$=\tan \theta -\sin \theta \cos \theta +\cot \theta -\sin \theta \cos \theta \text{ }$
$\begin{align}
  & =\dfrac{\sin \theta }{\cos \theta }+\dfrac{\cos \theta }{\sin \theta }-2\sin \theta \cos \theta \\
 & =\dfrac{{{\sin }^{2}}\theta +{{\cos }^{2}}\theta }{\cos \theta \sin \theta }-2\sin \theta \cos \theta \\
\end{align}$
Using the identity ${{\sin }^{2}}\theta +{{\cos }^{2}}\theta =1$, we get
$=\dfrac{1}{\sin \theta \cos \theta }-2\sin \theta \cos \theta $
Using,
$\begin{align}
  & \sec \theta =\dfrac{1}{\cos \theta } \\
 & \text{cosec}\theta =\dfrac{1}{\sin \theta } \\
\end{align}$
We get,
$=\sec \theta \text{cosec}\theta -2\sin \theta \cos \theta $
Since, $L.H.S=R.H.S$, hence we have proved the above relation.
While solving questions related to verifying an equation, we can proceed by simplifying either of the sides of the equation and reducing it further to get it equal to the other side of the equation. Just remember the formulas, identities and trigonometric conversions and you will obtain the right answer. Take care of the signs and the expression and avoid any calculation errors.