Question

Evaluate :$\frac{\sec \theta \csc (90-\theta )\tan \theta \cot (90-\theta )+({sin}^2{35}^{\circ }+{sin}^2{55}^{\circ })}{\tan 10\tan 20\tan 45\tan 70\tan 80}$

Answer 1

We have, $\frac{\sec \theta cosec(90-\theta )-\tan \theta \cot (90-\theta )+(si{n}^{2}35+si{n}^{2}55)}{\tan 10\tan 20\tan 45\tan 70\tan 80}$

$=\frac{\sec \theta \sec \theta -\tan \theta \tan \theta +({\sin }^2-35)-{\sin }^2(90-35)}{\tan 10\tan 20\tan (90-20)\tan (90-10)\tan 45}$

$=\frac{{\sec }^2\theta -{\tan }^2\theta +{\sin }^{2}35+{\cos }^{2}35}{\tan 10\cot 10\tan 20\cot 20\tan 45}$

$=\frac{1+1}{\tan 45}=2$