Question

Prove that $\dfrac{{{\tan }^{3}}\theta }{1+{{\tan }^{2}}\theta }+\dfrac{{{\cot }^{3}}\theta }{1+{{\cot }^{2}}\theta }=\dfrac{1-2{{\sin }^{2}}\theta {{\cos }^{2}}\theta }{\sin \theta \cos \theta }$

Answer 1

Hint: Convert L.H.S. into sines and cosines using the identities $\tan \theta =\dfrac{\sin \theta }{\cos \theta }$ and $\cot \theta =\dfrac{\cos \theta }{\sin \theta }$. Use the identity ${{\sin }^{2}}\theta +{{\cos }^{2}}\theta =1$ to simplify the expression. Hence prove L.H.S $=\dfrac{{{\sin }^{3}}\theta }{\cos \theta }+\dfrac{{{\cos }^{3}}\theta }{\sin \theta }$.

Complete step-by-step answer:
Take $\sin \theta \cos \theta$ as L.C.M. and use the identity ${{a}^{4}}+{{b}^{4}}={{\left( {{a}^{2}}+{{b}^{2}} \right)}^{2}}-2{{a}^{2}}{{b}^{2}}$ to simplify the resulting expression. Hence prove that L.H.S. is equal to R.H.S.

We have L.H.S. $=\dfrac{{{\tan }^{3}}\theta }{1+{{\tan }^{2}}\theta }+\dfrac{{{\cot }^{3}}\theta }{1+{{\cot }^{2}}\theta }$
Using $\tan \theta =\dfrac{\sin \theta }{\cos \theta }$ and $\cot \theta =\dfrac{\cos \theta }{\sin \theta }$, we get
L.H.S. $=\dfrac{\dfrac{{{\sin }^{3}}\theta }{{{\cos }^{3}}\theta }}{1+\dfrac{{{\sin }^{2}}\theta }{{{\cos }^{2}}\theta }}+\dfrac{\dfrac{{{\cos }^{3}}\theta }{{{\sin }^{3}}\theta }}{1+\dfrac{{{\cos }^{2}}\theta }{{{\sin }^{2}}\theta }}$
Multiplying the numerator and the denominator of the first expression by ${{\cos }^{3}}\theta$ and that of the second expression by ${{\sin }^{3}}\theta$, we get
L.H.S. $=\dfrac{{{\sin }^{3}}\theta }{\cos \theta \left( {{\sin }^{2}}\theta +{{\cos }^{2}}\theta \right)}+\dfrac{{{\cos }^{3}}\theta }{\sin \theta \left( {{\cos }^{2}}\theta +{{\sin }^{2}}\theta \right)}$
We know that ${{\cos }^{2}}\theta +{{\sin }^{2}}\theta =1$
Using the above identity, we get
L.H.S. $=\dfrac{{{\sin }^{3}}\theta }{\cos \theta }+\dfrac{{{\cos }^{3}}\theta }{\sin \theta }$
Taking $\sin \theta \cos \theta$ as L.C.M., we get
L.H.S. $=\dfrac{{{\cos }^{4}}\theta +{{\sin }^{4}}\theta }{\cos \theta \sin \theta }$
Adding and subtracting $2{{\sin }^{2}}\theta {{\cos }^{2}}\theta$ in the numerator of the above expression, we get
L.H.S. $=\dfrac{{{\cos }^{4}}\theta +2{{\sin }^{2}}\theta {{\cos }^{2}}\theta +{{\sin }^{4}}\theta -2{{\sin }^{2}}\theta {{\cos }^{2}}\theta }{\cos \theta \sin \theta }$
We know that ${{a}^{2}}+2ab+{{b}^{2}}={{\left( a+b \right)}^{2}}$
Using the above identity, we get
L.H.S. $=\dfrac{\left( {{\cos }^{2}}\theta +{{\sin }^{2}}\theta \right)-2{{\sin }^{2}}\theta {{\cos }^{2}}\theta }{\sin \theta \cos \theta }$
We know that ${{\cos }^{2}}\theta +{{\sin }^{2}}\theta =1$
Using the above identity, we get
L.H.S. $=\dfrac{1-2{{\sin }^{2}}\theta {{\cos }^{2}}\theta }{\sin \theta \cos \theta }$
Hence, we have
L.H.S = R.H.S

Note: [1] Dealing with sines and cosines is usually easier than dealing with tangents, cotangents, secants and cosecants. So while proving a trigonometric identity, we usually convert tangents, cotangents, secants and cosecants into sines and cosines.
[2] Alternative solution.
We know that $1+{{\tan }^{2}}\theta ={{\sec }^{2}}\theta$ and $1+{{\cot }^{2}}\theta ={{\csc }^{2}}\theta$
Hence, we have
L.H.S $={{\tan }^{3}}\theta {{\cos }^{2}}\theta +{{\cot }^{3}}\theta {{\csc }^{2}}\theta$
We know that $\tan \theta \cos \theta =\sin \theta$ and $\cot \theta \csc \theta =\cos \theta$
Hence, we have
L.H.S $={{\sin }^{2}}\theta \tan \theta +{{\cos }^{2}}\theta \cot \theta$
Hence, we have L.H.S $=\dfrac{{{\sin }^{3}}\theta }{\cos \theta }+\dfrac{{{\cos }^{3}}\theta }{\sin \theta }$, which is the same as obtained above. Proceeding similarly, we can prove L.H.S = R.H.S.