13. Which of the following is not a correct relation?

A. sin ⁡ 2 = 1 − cos ⁡ 2 sin 2 θ=1−cos 2 θ

B. sec ⁡ 2 = 1 + tan ⁡ 2 sec 2 θ=1+tan 2 θ

C. cosec ⁡ 2 = 1 + cot ⁡ 2 cosec 2 θ=1+cot 2 θ

D. sec ⁡ 2 = 1 − tan ⁡ 2 sec 2 θ=1−tan 2 θ​

13 Which of the following is not a correct relation A sin 2 1 cos 2 sin 2 θ1cos 2 θ B sec 2 1 tan 2 sec 2 θ1tan 2 θ C cosec 2 1 cot 2 cosec 2 θ1cot 2 θD sec 2 1 class=


Answer :

answer of this question is

option (B)

Answer:

[tex]\textsf{D)}\quad \sec^2\theta=1-\tan^2\theta[/tex]

Step-by-step explanation:

To determine which of the given equations is not a correction relation, we can use the Pythagorean Identity, sin²θ + cos²θ = 1.

[tex]\dotfill[/tex]

Option A

If we subtract cos²θ from both sides of the Pythagorean Identity, we get:

[tex]\sin^2\theta +\cos^2\theta-\cos^2\theta= 1- \cos^2\theta \\\\ \sin^2\theta = 1- \cos^2\theta[/tex]

Therefore, sin²θ = 1 - cos²θ is a correct relation.

[tex]\dotfill[/tex]

Option B

Divide both sides of the Pythagorean Identity by cos²θ:

[tex]\dfrac{\sin^2\theta}{\cos^2\theta} +\dfrac{\cos^2\theta}{\cos^2\theta} =\dfrac{1}{\cos^2\theta} \\\\\\ \tan^2\theta+1=\sec^2\theta[/tex]

Switch sides:

[tex]\sec^2\theta=1 + \tan^2\theta[/tex]

Therefore, sec²θ = 1 + tan²θ is a correct relation.

[tex]\dotfill[/tex]

Option C

Divide both sides of the Pythagorean Identity by sin²θ:

[tex]\dfrac{\sin^2\theta}{\sin^2\theta} +\dfrac{\cos^2\theta}{\sin^2\theta} =\dfrac{1}{\sin^2\theta} \\\\\\ 1+\cot^2\theta=\text{cosec}^2\theta[/tex]

Switch sides:

[tex]\text{cosec}^2\theta=1 + \cot^2\theta[/tex]

Therefore, cosec²θ = 1 + cot²θ is a correct relation.

[tex]\dotfill[/tex]

Option D

Divide both sides of the Pythagorean Identity by cos²θ:

[tex]\dfrac{\sin^2\theta}{\cos^2\theta} +\dfrac{\cos^2\theta}{\cos^2\theta} =\dfrac{1}{\cos^2\theta} \\\\\\ \tan^2\theta+1=\sec^2\theta[/tex]

Switch sides:

[tex]\sec^2\theta=1 + \tan^2\theta[/tex]

Therefore, sec²θ = 1 - tan²θ is NOT a correct relation.