3.Trigonometrical Ratios, Functions and Identities
easy

સાબિત કરો કે : $\sin ^{2} \frac{\pi}{6}+\cos ^{2} \frac{\pi}{3}-\tan ^{2} \frac{\pi}{4}=-\frac{1}{2}$

Option A
Option B
Option C
Option D

Solution

$L.H.S$ $=\sin ^{2} \frac{\pi}{6}+\cos ^{2} \frac{\pi}{3}-\tan ^{2} \frac{\pi}{4}$

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

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

$= R . H.S$

Standard 11
Mathematics

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