- Home
- Standard 10
- Mathematics
8. Introduction to Trigonometry
easy
જો $\sqrt{3} \tan \theta=1$ હોય, તો $\sin ^{2} \theta-\cos ^{2} \theta$નું મૂલ્ય શોધો.
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
