Prove that

Solution:

To prove: $4 \sin \frac{\pi}{6} \sin ^{2} \frac{\pi}{3}+3 \cos \frac{\pi}{3} \tan \frac{\pi}{4}+\operatorname{cosec}^{2} \frac{\pi}{2}=4$

Taking LHS,

$=4 \sin \frac{\pi}{6} \sin ^{2} \frac{\pi}{3}+3 \cos \frac{\pi}{3} \tan \frac{\pi}{4}+\operatorname{cosec}^{2} \frac{\pi}{2}$

Putting $\pi=180^{\circ}$

$=4 \sin \frac{180}{6} \sin ^{2} \frac{180}{3}+3 \cos \frac{180}{3} \tan \frac{180}{4}+\operatorname{cosec}^{2} \frac{180}{2}$

$=4 \sin 30^{\circ} \sin ^{2} 60^{\circ}+3 \cos 60^{\circ} \tan 45^{\circ}+\operatorname{cosec}^{2} 90^{\circ}$

Now, we know that,

$\sin 30^{\circ}=\frac{1}{2}$

$\sin 60^{\circ}=\frac{\sqrt{3}}{2}$

$\cos 60^{\circ}=\frac{1}{2}$

$\tan 45^{\circ}=1$

$\operatorname{cosec} 90^{\circ}=1$

Putting the values, we get

$=4 \times \frac{1}{2} \times\left(\frac{\sqrt{3}}{2}\right)^{2}+3 \times \frac{1}{2} \times 1+(1)^{2}$

$=2 \times \frac{3}{4}+\frac{3}{2}+1$

$=\frac{3}{2}+\frac{3}{2}+1$

$=\frac{3+3+2}{2}$

= 4

= RHS

∴ LHS = RHS

Hence Proved