Question

Which of the following is not an identity ?

• A. $(1-{\sin }^{2}A).{\sec }^{2}A=1$
• B. $({\sec }^2\theta -1)(1-\cos {\sec }^2\theta )=1$
• C. $\frac{\sin \theta }{1+\cos \theta }+\frac{1+\cos \theta }{\sin \theta }=\frac{2}{\sin \theta }$
• D. ${\sin }^4\theta -{\cos }^4\theta ={\sin }^2\theta -{\cos }^2\theta$

Answer 1

The correct option is B $({\sec }^2\theta -1)(1-\cos {\sec }^2\theta )=1$

$I]$Option A:

Taking L.H.S.

$(1-{\sin }^{2}A){\sec }^{2}A={\cos }^{2}A.{\sec }^{2}A$

$={\cos }^{2}A.\frac{1}{{\cos }^{2}A}=1=$ R.H.S.

$II]$Option B:

Taking L.H.S.

$({\sec }^2-1)(1-{\text{cosec}}^2\theta )=-{\tan }^2\theta \times {\cot }^2\theta =-1\ne$R.H.S. $(\begin{array}{ccc}\because {\sec }^2\theta & -{\tan }^2\theta & =1\\ cose{c}^2\theta & -{\cot }^2\theta & =1\\ & & \end{array})$
$III]$Option C:

Taking L.H.S.

$\frac{\sin \theta }{1+\cos \theta }+\frac{1+\cos \theta }{\sin \theta }=\frac{{\sin }^2\theta +{(1+\cos \theta )}^2}{\sin \theta (1+cos\theta )}$

$=\frac{{\sin }^2\theta +1+2\cos \theta +{\cos }^2\theta }{\sin \theta (1+\cos \theta )}$

$=\frac{2+2\cos \theta }{\sin \theta (1+\cos \theta )}$

$=\frac{2(1+\cos \theta )}{\sin \theta (1+\cos \theta )}$

$=\frac{2}{\sin \theta }=$ R.H.S.
$IV]$Option D:

Taking L.H.S.

${\sin }^4\theta -{\cos }^4\theta ={\left({\sin }^2\theta \right)}^2-{\left({\cos }^2\theta \right)}^2$

$=({\sin }^2\theta +{\cos }^2\theta )({\sin }^2\theta -{\cos }^2\theta )$

$={\sin }^2-{\cos }^2\theta (\because {\sin }^2\theta +{\cos }^2\theta =1)=$ R.H.S.
Hence, (b) is not an identity.