Question

Prove the following trigonometric identity
$\left( {{\sec }^{2}}\theta -1 \right)\left( 1-{{\csc }^{2}}\theta \right)=-1$

Answer 1

Hint: Convert into sines and cosines using the identities $\sec \theta =\dfrac{1}{\cos \theta }$ and $\csc \theta =\dfrac{1}{\sin \theta }$. Use the identities ${{\sin }^{2}}\theta =1-{{\cos }^{2}}\theta $ and ${{\cos }^{2}}\theta =1-{{\sin }^{2}}\theta $ and hence prove that L.H.S. is equal to $\dfrac{-{{\sin }^{2}}\theta {{\cos }^{2}}\theta }{{{\cos }^{2}}\theta {{\sin }^{2}}\theta }$ and hence prove that the value of L.H.S. is equal to R.H.S.

Complete step-by-step answer:

We have
L.H.S. $=\left( {{\sec }^{2}}\theta -1 \right)\left( 1-{{\csc }^{2}}\theta \right)$
We know that $\sec \theta =\dfrac{1}{\cos \theta }$ and $\csc \theta =\dfrac{1}{\sin \theta }$.
Using the above two identities, we get
L.H.S. $=\left( \dfrac{1}{{{\cos }^{2}}\theta }-1 \right)\left( 1-\dfrac{1}{{{\sin }^{2}}\theta } \right)$
In the left term of the product taking ${{\cos }^{2}}\theta $ as L.C.M. and in the right term of the product taking ${{\sin }^{2}}\theta $ as L.C.M., we get
L.H.S. $=\dfrac{1-{{\cos }^{2}}\theta }{{{\cos }^{2}}\theta }\times \dfrac{{{\sin }^{2}}\theta -1}{{{\sin }^{2}}\theta }$
We know that $1-{{\cos }^{2}}\theta ={{\sin }^{2}}\theta $ and $1-{{\sin }^{2}}\theta ={{\cos }^{2}}\theta $.
Using the above two identities, we get
L.H.S. $=\dfrac{{{\sin }^{2}}\theta }{{{\cos }^{2}}\theta }\times \dfrac{-{{\cos }^{2}}\theta }{{{\sin }^{2}}\theta }$
We know that $\dfrac{a}{c}\times \dfrac{b}{d}=\dfrac{a}{d}\times \dfrac{b}{c}$
Hence, we have
L.H.S. $=\dfrac{{{\sin }^{2}}\theta }{{{\sin }^{2}}\theta }\times \dfrac{-{{\cos }^{2}}\theta }{{{\cos }^{2}}\theta }=1\times \left( -1 \right)=-1$
Hence, we have
L.H.S. – R.H.S.

Note: [1] Dealing with sines and cosines is easier than dealing with tangents, cotangents, secants and cosecants. This is why a student must always check whether converting to sines and cosines makes the solution of the problem easy or not as is done in the above case.
[2]Alternative solution:
We know that ${{\sec }^{2}}\theta -1={{\tan }^{2}}\theta $ and ${{\csc }^{2}}\theta -1={{\cot }^{2}}\theta $
Hence, we have
L.H.S. $={{\tan }^{2}}\theta \left( -{{\cot }^{2}}\theta \right)$
We know that $\cot \theta =\dfrac{1}{\tan \theta }$
Hence, we have
L.H.S. $={{\tan }^{2}}\theta \left( \dfrac{-1}{{{\tan }^{2}}\theta } \right)=-1$
Hence, we have
LHS = RHS