Prove that


Prove that

$\tan ^{2} \frac{\pi}{3}+2 \cos ^{2} \frac{\pi}{4}+3 \sec ^{2} \frac{\pi}{6}+4 \cos ^{2} \frac{\pi}{2}=8$



To prove: $\tan ^{2} \frac{\pi}{3}+2 \cos ^{2} \frac{\pi}{4}+3 \sec ^{2} \frac{\pi}{6}+4 \cos ^{2} \frac{\pi}{2}=8$

Taking LHS,

$=\tan ^{2} \frac{\pi}{3}+2 \cos ^{2} \frac{\pi}{4}+3 \sec ^{2} \frac{\pi}{6}+4 \cos ^{2} \frac{\pi}{2}$

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

$=\tan ^{2} \frac{180}{3}+2 \cos ^{2} \frac{180}{4}+3 \sec ^{2} \frac{180}{6}+4 \cos ^{2} \frac{180}{2}$

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

Now, we know that

$\tan 60^{\circ}=\sqrt{3}$

$\cos 45^{\circ}=\frac{1}{\sqrt{2}}$

$\sec 30^{\circ}=\frac{2}{\sqrt{3}}$

$\cos 90^{\circ}=0$

Putting the values, we get

$=(\sqrt{3})^{2}+2 \times\left(\frac{1}{\sqrt{2}}\right)^{2}+3 \times\left(\frac{2}{\sqrt{3}}\right)^{2}+4(0)^{2}$

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




$\therefore \mathrm{LHS}=\mathrm{RHS}$

Hence Proved


Leave a comment


Click here to get exam-ready with eSaral

For making your preparation journey smoother of JEE, NEET and Class 8 to 10, grab our app now.

Download Now