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Question: Prove the following trigonometric identity \(\left( {{\sec }^{2}}\theta -1 \right)\left( 1-{{\csc ...

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

Explanation

Solution

Hint: Convert into sines and cosines using the identities sec⁡θ=1cos⁡θ\sec \theta =\dfrac{1}{\cos \theta } and csc⁡θ=1sin⁡θ\csc \theta =\dfrac{1}{\sin \theta }. Use the identities sin⁡2θ=1−cos⁡2θ{{\sin }^{2}}\theta =1-{{\cos }^{2}}\theta and cos⁡2θ=1−sin⁡2θ{{\cos }^{2}}\theta =1-{{\sin }^{2}}\theta and hence prove that L.H.S. is equal to −sin⁡2θcos⁡2θcos⁡2θsin⁡2θ\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. =(sec⁡2θ−1)(1−csc⁡2θ)=\left( {{\sec }^{2}}\theta -1 \right)\left( 1-{{\csc }^{2}}\theta \right)
We know that sec⁡θ=1cos⁡θ\sec \theta =\dfrac{1}{\cos \theta } and csc⁡θ=1sin⁡θ\csc \theta =\dfrac{1}{\sin \theta }.
Using the above two identities, we get
L.H.S. =(1cos⁡2θ−1)(1−1sin⁡2θ)=\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⁡2θ{{\cos }^{2}}\theta as L.C.M. and in the right term of the product taking sin⁡2θ{{\sin }^{2}}\theta as L.C.M., we get
L.H.S. =1−cos⁡2θcos⁡2θ×sin⁡2θ−1sin⁡2θ=\dfrac{1-{{\cos }^{2}}\theta }{{{\cos }^{2}}\theta }\times \dfrac{{{\sin }^{2}}\theta -1}{{{\sin }^{2}}\theta }
We know that 1−cos⁡2θ=sin⁡2θ1-{{\cos }^{2}}\theta ={{\sin }^{2}}\theta and 1−sin⁡2θ=cos⁡2θ1-{{\sin }^{2}}\theta ={{\cos }^{2}}\theta .
Using the above two identities, we get
L.H.S. =sin⁡2θcos⁡2θ×−cos⁡2θsin⁡2θ=\dfrac{{{\sin }^{2}}\theta }{{{\cos }^{2}}\theta }\times \dfrac{-{{\cos }^{2}}\theta }{{{\sin }^{2}}\theta }
We know that ac×bd=ad×bc\dfrac{a}{c}\times \dfrac{b}{d}=\dfrac{a}{d}\times \dfrac{b}{c}
Hence, we have
L.H.S. =sin⁡2θsin⁡2θ×−cos⁡2θcos⁡2θ=1×(−1)=−1=\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⁡2θ−1=tan⁡2θ{{\sec }^{2}}\theta -1={{\tan }^{2}}\theta and csc⁡2θ−1=cot⁡2θ{{\csc }^{2}}\theta -1={{\cot }^{2}}\theta
Hence, we have
L.H.S. =tan⁡2θ(−cot⁡2θ)={{\tan }^{2}}\theta \left( -{{\cot }^{2}}\theta \right)
We know that cot⁡θ=1tan⁡θ\cot \theta =\dfrac{1}{\tan \theta }
Hence, we have
L.H.S. =tan⁡2θ(−1tan⁡2θ)=−1={{\tan }^{2}}\theta \left( \dfrac{-1}{{{\tan }^{2}}\theta } \right)=-1
Hence, we have
LHS = RHS