Question

Without using trigonometric tables, find the value of (sin^2 10° + sin^2 80°)/(sec^2 20°-cot^2 70°)-3tan80°tan 50° tan 45° tan 10° tan 40°.

Answer 1

We know,

$\sin^2 10^o+\sin^2 80^o=\sin^2 10^o+\cos^2 10^o=1\\ \sec^2 20^o-\cot^2 70^o=\sec^2 20^o-\tan^2 20^o=1\\ \tan 80^o=\cot 10^o=\frac{1}{\tan 10^o} \\ \tan 50^o=\cot 40^o=\frac{1}{\tan 40^o}$

Now the given expression is

$\frac{\sin^2 10^o+\sin^2 80^o}{\sec^2 20^o-\cot^2 70^o}-3\tan 80^o\tan 50^o\tan 45^o\tan 40^o\tan 10^o\\ =\frac{1}{1}-3\frac{1}{\tan 10^o} \frac{1}{\tan 40^o} \tan 45^o\tan 40^o\tan 10^o\\ =1-3 \tan 45^o\\ =1-3\\ =-2$