Question

Check whether the following equation is true or false
\[\dfrac{{{\tan }^{3}}\theta }{1+{{\tan }^{2}}\theta }+\dfrac{{{\cot }^{3}}\theta }{1+{{\cot }^{2}}\theta }=\dfrac{1-2{{\sin }^{2}}\theta {{\cos }^{2}}\theta }{\sin \theta \cos \theta }\]

Answer 1

Hint: In this question, the left hand side contains terms involving ${tan}$ and ${cot}$ and the right hand side contains terms involving${sin}$ and ${cos}$. Therefore, the easiest way to solve this question will be to convert the ${tan}$ and ${cot}$ in the LHS into ${sin}$ and ${cos}$ using trigonometric identities and then use it to obtain the desired expression in the RHS.

Complete step-by-step solution -
From, trigonometry, the expressions for the tangent (tan) and cotangent (cot) in terms of sine (sin) and cosine (cos) is given by
$\tan \theta =\dfrac{\sin \theta }{\cos \theta }......................(1.1)$
and
$\cot \theta =\dfrac{\cos \theta }{\sin \theta }......................(1.2)$.
Therefore, we can expand the lhs by replacing tan and cot with sin and cos using (1.1) and (1.2) as
\[\begin{align}
  & \dfrac{{{\tan }^{3}}\theta }{1+{{\tan }^{2}}\theta }+\dfrac{{{\cot }^{3}}\theta }{1+{{\cot }^{2}}\theta }=\dfrac{\dfrac{{{\sin }^{3}}\theta }{{{\cos }^{3}}\theta }}{1+\dfrac{{{\sin }^{2}}\theta }{{{\cos }^{2}}\theta }}+\dfrac{\dfrac{{{\cos }^{3}}\theta }{{{\sin }^{3}}\theta }}{1+\dfrac{{{\cos }^{2}}\theta }{{{\sin }^{2}}\theta }} \\
 & =\dfrac{{{\cos }^{2}}\theta \times \dfrac{{{\sin }^{3}}\theta }{{{\cos }^{3}}\theta }}{{{\cos }^{2}}\theta +{{\sin }^{2}}\theta }+\dfrac{{{\sin }^{2}}\theta \times \dfrac{{{\cos }^{3}}\theta }{{{\sin }^{3}}\theta }}{{{\sin }^{2}}\theta +{{\cos }^{2}}\theta }=\dfrac{1}{{{\sin }^{2}}\theta +{{\cos }^{2}}\theta }\left( \dfrac{{{\sin }^{3}}\theta }{\cos \theta }+\dfrac{{{\cos }^{3}}\theta }{\sin \theta } \right).................(1.3) \\
\end{align}\]
Now, we can use the identity
${{\sin }^{2}}\theta +{{\cos }^{2}}\theta =1.............(1.4)$
In equation (1.3) to obtain
\[\dfrac{{{\tan }^{3}}\theta }{1+{{\tan }^{2}}\theta }+\dfrac{{{\cot }^{3}}\theta }{1+{{\cot }^{2}}\theta }=\dfrac{{{\cos }^{2}}\theta \times \dfrac{{{\sin }^{3}}\theta }{{{\cos }^{3}}\theta }}{{{\cos }^{2}}\theta +{{\sin }^{2}}\theta }+\dfrac{{{\sin }^{2}}\theta \times \dfrac{{{\cos }^{3}}\theta }{{{\sin }^{3}}\theta }}{{{\sin }^{2}}\theta +{{\cos }^{2}}\theta }=\dfrac{{{\sin }^{3}}\theta }{\cos \theta }+\dfrac{{{\cos }^{3}}\theta }{\sin \theta }=\dfrac{{{\sin }^{4}}\theta +{{\cos }^{4}}\theta }{\sin \theta \cos \theta }.................(1.5)\]

Now, we know that
${{\left( a+b \right)}^{2}}={{a}^{2}}+{{b}^{2}}+2ab\Rightarrow {{a}^{2}}+{{b}^{2}}={{\left( a+b \right)}^{2}}-2ab........(1.6)$
Taking $a={{\sin }^{2}}\theta $ and $b={{\cos }^{2}}\theta $ in equation (1.6), we get
${{\left( {{\sin }^{2}}\theta \right)}^{2}}+{{\left( {{\cos }^{2}}\theta \right)}^{2}}={{\left( {{\sin }^{2}}\theta +{{\cos }^{2}}\theta \right)}^{2}}-2{{\sin }^{2}}\theta {{\cos }^{2}}\theta ............(1.7)$
Now, we can use the formula in equation (1.4) to write equation (1.7) as
${{\sin }^{4}}\theta +{{\cos }^{4}}\theta ={{1}^{2}}-2{{\sin }^{2}}\theta {{\cos }^{2}}\theta =1-2{{\sin }^{2}}\theta {{\cos }^{2}}\theta ............(1.8)$
Finally, we can use equation (1.8) to write the numerator of equation (1.5) and thus we can rewrite equation (1.3) as
\[\dfrac{{{\tan }^{3}}\theta }{1+{{\tan }^{2}}\theta }+\dfrac{{{\cot }^{3}}\theta }{1+{{\cot }^{2}}\theta }=\dfrac{1-2{{\sin }^{2}}\theta {{\cos }^{2}}\theta }{\sin \theta \cos \theta }\]
Which is exactly the equation given in the question that we wanted to prove.

Note: We should be careful to use equation (1.6) to simplify the fourth powers of sin and cos and write it in terms of sin squares because in equation (1.6), the terms appearing in LHS are in terms of squares of a and b and those appearing in RHS are in terms of only a and b.