Question

Without using table find the value of $\frac{Sin{70}^o}{\cos {20}^o}+\tan {25}^o\tan {65}^o+{\sec }^2{75}^o-{\cot }^{2}15-3\tan 5^o\tan 7^o\times \tan {83}^o\tan {85}^o$ ?

Answer 1

$\frac{\sin {70}^o}{\cos {20}^o}+\tan {25}^o\tan {65}^o+{\sec }^2{75}^o-{\cot }^2{15}^o-3\tan 5^o\tan 7^o\times \tan {85}^0\tan {83}^0$

$=\frac{\sin ({90}^o-{20}^o)}{\cos {20}^o}+\tan ({90}^0-{65}^0)\tan {65}^o+{\sec }^2{75}^o-\left[\cot \right({90}^o-{75}^o){]}^2-3\tan ({90}^o-{85}^o)\tan ({90}^o-{83}^o)\times \tan {85}^o\tan {83}^o$

$=\frac{\cos {20}^o}{\cos {20}^o}+\cot {65}^o\tan {65}^o+({\sec }^2{75}^o-{\tan }^2{75}^o)-3\cot {85}^o\cot {83}^0\times \tan {85}^0\tan {83}^o\left(\sin \right(90-x)=\cos x)\left(tan\right(90-x)=\cot x)\left(cot\right(90-x)=\tan x)$

$=1+1+1-3\times 1){(\tan }^{2}x+1={\sec }^{2}x)=0$