Question

Which of the following is not an identity? A. Sin2θ + Cos2θ = 1 B. Sin θ= tanθ.cosθ C. 1 + cot2θ= cos2θ D. 1 – sec2θ = tan2θ

Answer 1

Incorrect identities: $1+cot^{2}\theta=cos^{2}\theta$ and $1-sec^{2}\theta=tan^{2}\theta$

We take an angle $\theta=45^{\circ}$ to find which of the given relations is not an identity.

A. $sin^{2}\theta+cos^{2}\theta=1$

When $\theta=45^{\circ}$,

$\quad sin^{2}45^{\circ}+cos^{2}45^{\circ}-1$

$=(\dfrac{1}{\sqrt{2}})^{2}+(\dfrac{1}{\sqrt{2}})^{2}-1$

$=\dfrac{1}{2}+\dfrac{1}{2}-1$

$=1-1$

$=0$

$\Rightarrow sin^{2}\theta+cos^{2}\theta=1$ is correct.

B. $sin\theta=tan\theta\:cos\theta$

When $\theta=45^{\circ}$,

$\quad sin45^{\circ}-tan45^{\circ}\:cos45^{\circ}$

$=\dfrac{1}{\sqrt{2}}-1.\dfrac{1}{\sqrt{2}}$

$=\dfrac{1}{\sqrt{2}}-\dfrac{1}{\sqrt{2}}$

$=0$

$\Rightarrow sin\theta=tan\theta\:cos\theta$ is correct.

C. $1+cot^{2}\theta=cos^{2}\theta$

When $\theta=45^{\circ}$,

$\quad 1+cot^{2}45^{\circ}-cos^{2}45^{\circ}$

$=1+(1)^{2}-(\dfrac{1}{\sqrt{2}})^{2}$

$=1+1-\dfrac{1}{2}$

$=\dfrac{3}{2}$

$\neq 0$

$\Rightarrow 1+cot^{2}\theta=cos^{2}\theta$ is incorrect.

D. $1-sec^{2}\theta=tan^{2}\theta$

When $\theta=45^{\circ}$,

$\quad 1-sec^{2}45^{\circ}-tan^{2}45^{\circ}$

$=1-(\sqrt{2})^{2}-(1)^{2}$

$=1-2-1$

$=-2$

$\neq 0$

$\Rightarrow 1-sec^{2}\theta=tan^{2}\theta$ is incorrect.

Answer 2

Answer:

Step-by-step explanation: