Question

Without using trigonometric table, evaluate- cos^20°+ cos^70°/sec^50°- cot^40°×2 sec^60°

Answer 1

Hope this will help you

Answer 2

$\dfrac{\cos^2 20+\cos^2 70}{\sec^2 50-\cot^2 40}\times 2 sec^2 60=8$

Step-by-step explanation:

We have,

$\dfrac{\cos^2 20+\cos^2 70}{\sec^2 50-\cot^2 40}\times 2 sec^2 60$

To find, $\dfrac{\cos^2 20+\cos^2 70}{\sec^2 50-\cot^2 40}\times 2 sec^2 60=?$

∴ $\dfrac{\cos^2 20+\cos^2 70}{\sec^2 50-\cot^2 40}\times 2 sec^2 60$

=$\dfrac{\cos^2 20+\cos^2 (90-20)}{\sec^2 50-\cot^2 (90-50)}\times 2 sec^2 60$

$=\dfrac{\cos^2 20+\sin^2 20}{\sec^2 50-\tan^2 50}\times 2 sec^2 60$

Using trigonometric identity,

$\cos (90-A)=\sin A$ and

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

$=\dfrac{1}{1}\times 2 sec^2 60$

Using trigonometric identity,

$\cos^2 A+\sin^2 A=1$ and

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

$= 2\times (2)^2$

= 8

Hence, $\dfrac{\cos^2 20+\cos^2 70}{\sec^2 50-\cot^2 40}\times 2 sec^2 60=8$