Question

7. If √3 tan theta = 3 sin theta, prove that (sin’o-cos²0) = 1/3​

Answer 1

$(sin^2 \theta-cos^2 \theta)=\dfrac{1}{3}$, proved.

Step-by-step explanation:

We have,

$\sqrt{3}\tan \theta=3 \sin \theta$

To prove that, $(sin^2 \theta-cos^2 \theta)=\dfrac{1}{3}$.

∴ $\sqrt{3}\tan \theta=3 \sin \theta$

⇒ $\sqrt{3}\dfrac{\sin \theta}{\cos \theta} =3 \sin \theta$

⇒ $\sqrt{3}\dfrac{1}{\cos \theta} =3$

⇒ $\cos \theta=\dfrac{1}{\sqrt{3}}$

∴ $\sin \theta=\sqrt{1-\cos^2 \theta}=\sqrt{1-(\dfrac{1}{\sqrt{3}})^2}$

$\sin \theta}=\sqrt{\dfrac{2}{3}}}$

L.H.S. = $sin^2 \theta-cos^2 \theta$

= $(\sqrt{\dfrac{2}{3}}})^2 -(\dfrac{1}{\sqrt{3}})^2$

= $\dfrac{2}{3}-\dfrac{1}{3}$

= $\dfrac{2-1}{3}$

= $\dfrac{1}{3}$

= R.H.S., proved.