Question

solve the above question​

Answer 1

Answer:

$tan^{2} \frac{\pi }{3} + 2cos^{2} \frac{\pi }{4} + 3sec^{2}\frac{\pi }{6} + 4cos^{2}\frac{\pi }{2} = 6\\\\LHS\\\\ tan^{2} \frac{\pi }{3} + 2cos^{2} \frac{\pi }{4} + 3sec^{2}\frac{\pi }{6} + 4cos^{2}\frac{\pi }{2}\\\\In trigo, \pi = 180\\substituting\\\\tan^{2} \frac{180 }{3} + 2cos^{2} \frac{180 }{4} + 3sec^{2}\frac{180 }{6} + 4cos^{2}\frac{180 }{2}\\\\tan^{2} 60 + 2cos^{2}45 + 3sec^{2}30 + 4cos^{2}90$

$(\sqrt{3})^{2} + 2(\frac{1}{\sqrt{2}})^{2} + 3(\frac{2}{\sqrt{3}})^{2} + 4(0)^{2} \\\\\\3 +1 + 2 =6$

LHS=RHS hence, proved

There is an error in the question. The RHS will be 6 instead of 8