Question

Evaluate: $\frac{\sec \theta .cosec({90}^o-\theta )-\tan \theta .\cot ({90}^o-\theta )+{\sin }^2{55}^o+{\sin }^2{35}^o}{\tan {10}^o.\tan {20}^o.\tan {60}^o.\tan {70}^o.\tan {80}^o}$

Answer 1

Taking Numerator

$=\sec A\csc (90-A)-\tan A\cot (90-A)+{\sin }^{2}55+{\sin }^{2}35$

$=\sec A\times \sec A-\tan A\times \tan A+{\sin }^{2}55+{\cos }^2(90-35)$

$=1+1$

$=2$

As

${\sin }^{2}A+{\cos }^{2}A=1$

${\sec }^{2}A-{\tan }^{2}A=1$

$\csc (90-A)=\sec A$

$\cot (90-A)=\tan A$

We know that

$\tan A=\cot (90-A)=\frac{1}{\tan (90-A)}$

Taking denominator

$=\tan 10\tan 80\tan 60\tan 20\tan 70$

$=\tan 10\times \frac{1}{\tan 10}\times \tan 60\times \frac{1}{\tan 20}$

$=\tan 60$

$=\sqrt{3}$

$\therefore \frac{\sec A.\csc (90-A)-\tan A.\cot (90-A)+{\sin }^{2}55+{\sin }^{2}35}{\tan 10.\tan 20.\tan 60.\tan 70.\tan 80}=\frac{2}{\sqrt{3}}$