Question

Which one of the following identities (wherever defined) is not correct?

• A. $\frac{{\sin }^{4}x-{\cos }^{4}x}{{\sin }^{2}x-{\cos }^{2}x}=1$
• B. $\frac{\cot x}{1+\tan x}=\frac{\cot x-1}{2-{\sec }^{2}x}$
• C. ${cosec}^{2}x+{\sec }^{2}x={cosec}^{2}x.{\sec }^{2}x$
• D. $\left[(1+\cot x-cosecx)(1+\tan x+\sec x)\right]=1$

Answer 1

The correct option is D $\left[(1+\cot x-cosecx)(1+\tan x+\sec x)\right]=1$

(1) $\frac{{\sin }^{4}x-{\cos }^{4}x}{{\sin }^{2}x-{\cos }^{2}x}=1$
L.H.S: $\frac{{\sin }^{4}x-{\cos }^{4}x}{{\sin }^{2}x-{\cos }^{2}x}$
= $\frac{({\sin }^{2}x-{\cos }^{2}x)({\sin }^{2}x+{\cos }^{2}x)}{{\sin }^{2}x-{\cos }^{2}x}=1$
= $1$
Thus, $L.H.S=R.H.S.$

(2) $\frac{\cot x}{1+\tan x}=\frac{\cot x-1}{2-{\sec }^{2}x}$
R.H.S: $\frac{\cot x-1}{2-{\sec }^{2}x}$
= $\frac{\cot x-1}{1+(1-{\sec }^{2}x)}$
= $\frac{\frac{1}{\tan x}-1}{1-{\tan }^{2}x}$
= $\frac{1-\tan x}{\tan x(1-\tan x)(1+\tan x)}$
= $\frac{\cot x}{1+\tan x}$
thus, $L.H.S=R.H.S$

(3) ${cosec}^{2}x+{\sec }^{2}x={cosec}^{2}x.{\sec }^{2}x$
L.H.S: $cose{c}^{2}x+{\sec }^{2}x=\frac{1}{{\sin }^{2}x}+\frac{1}{{\cos }^{2}x}$
= $\frac{{\sin }^{2}x+{\cos }^{2}x}{{\sin }^{2}x{\cos }^{2}x}$
= ${cosec}^{2}x{\sec }^{2}x$
Thus, $L.H.S=R.H.S$

All the three are identities, hence the fourth is not an identity.