8. Introduction to Trigonometry
easy

यदि $\sqrt{3} \tan \theta=1$ है, तो $\sin ^{2} \theta-\cos ^{2} \theta$ का मान ज्ञात कीजिए।

A

$\frac{3}{4}$

B

$\frac{1}{2}$

C

$-\frac{1}{2}$

D

$-\frac{3}{4}$

Solution

Given that, $\sqrt{3} \tan \theta=1$

$\tan \theta=\frac{1}{\sqrt{3}}=\tan 30^{\circ}$

$\theta=30^{\circ}$

Now, $\sin ^{2} \theta-\cos ^{2} \theta=\sin ^{2} 30^{\circ}-\cos ^{2} 30^{\circ}$

$=\left(\frac{1}{2}\right)^{2}-\left(\frac{\sqrt{3}}{2}\right)^{2}$

$=\frac{1}{4}-\frac{3}{4}=\frac{1-3}{4}=-\frac{2}{4}=-\frac{1}{2}$

Standard 10
Mathematics

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