Question

Which is true for all values of $\theta$?

• A. ${\sin }^2\theta -{\cos }^2\theta =1$
• B. ${\text{cosec}}^2\theta -{\cot }^2\theta =1$
• C. ${\text{cosec}}^2\theta -{\cos }^2\theta =1$
• D. ${\sec }^2\theta -{\sin }^2\theta =1$

Answer 1

Answer: (B)

${\text{cosec}}^2\theta -{\cot }^2\theta =1$

We need to verify every option

$\left(A\right)$

Consider, ${\sin }^2\theta -{\cos }^2\theta$

$=1-{\cos }^2\theta -{\cos }^2\theta$

$=1-2{\cos }^2\theta$

$=-(2{\cos }^2\theta -1)=-\cos 2\theta$

Now, $-\cos 2\theta =1$ if $\cos 2\theta =-1⟹2\theta ={\cos }^{-1}(-1)=(2n+1)\pi$

$⟹\theta =\frac{(2n+1)\pi }{2},n\in Z$

Thus, it is not true for every $\theta$

$\left(B\right)$

Consider, ${\text{cosec}}^2\theta -{\cot }^2\theta$

$=\frac{1}{{\sin }^2\theta }-\frac{{\cos }^2\theta }{{\sin }^2\theta }$

$=\frac{1-{\cos }^2\theta }{{\sin }^2\theta }=\frac{{\sin }^2\theta }{{\sin }^2\theta }=1$

$\therefore {\text{cosec}}^2\theta -{\cot }^2\theta =1,$ For every value of $\theta$

$\left(C\right)$

${\text{cosec}}^2\theta -{\cos }^2\theta =1⟹1-{\sin }^2\theta {\cos }^2\theta ={\sin }^2\theta$

$⟹1-{\sin }^2\theta ={\sin }^2\theta {\cos }^2\theta$

$⟹{\sin }^2\theta =1⟹\theta =\frac{(2n+1)\pi }{2}$

Thus, ${\text{cosec}}^2\theta -{\cos }^2\theta =1$ is true for $\theta =\frac{(2n+1)\pi }{2}$ but not for all values of $\theta$.

$\left(D\right)$

${\sec }^2\theta -{\sin }^2\theta =1⟹\frac{1}{{\cos }^2\theta }-{\sin }^2\theta =1$

$⟹1-{\sin }^2\theta {\cos }^2\theta ={\cos }^2\theta$

$⟹{\sin }^2\theta ={\sin }^2\theta {\cos }^2\theta ⟹{\cos }^2\theta =1$

$⟹\cos \theta =\pm 1⟹\theta =n\pi ,n\in Z$

${\sec }^2\theta -{\sin }^2\theta =1$ for $\theta \in Z$ but not for all values of $\theta$.

Hence, option B is correct.