Question

Value of 6736 $co{s}^2{18}^o+421ta{n}^2{36}^o$ is

Answer 1

Consider ,$6736{\cos }^2{18}^o+421{\tan }^2{36}^o$ is
$=421(16{\cos }^2{18}^o+{\tan }^2{36}^o)$
$=421[10+2\sqrt{5}+5-2\sqrt{5}]$ ($\because \cos {18}^0=\frac{1}{4}\sqrt{10+2\sqrt{5}},\tan {36}^0=\frac{1}{4}\sqrt{5-2\sqrt{5}})$
$=421\times 15$
$=6315$.