Question

can anyone please solve this
please​

Answer 1

Answer:

Step-by-step explanation:

Answer 2

Given that,

$\sqrt{3} \tan \theta = 3 \sin \theta \\\\\implies \sqrt{3} \cdot \dfrac{\sin \theta}{\cos \theta} = 3 \sin \theta\\\\\implies \dfrac{\sqrt{3}}{\cos \theta} = 3 \\\\\implies \dfrac{3}{\cos^2 \theta} = 9\\\\\implies 9\cos^2 \theta = 3\\\\\implies 3\cos^2 \theta = 1\\\\\implies \cos^2 \theta = \dfrac 13\\\\\text{Now,}\\\\\sin^2 \theta - \cos^2 \theta \\\\= 1-\cos^2 \theta - \cos^2 \theta \\\\ =1-2\cos^2 \theta \\\\=1- 2 \cdot \dfrac 13 = \dfrac 13$