Question

find the value of sec2 20 minus cot2 70 divided by sin2 70 plus sin2 20​

Answer 1

Step-by-step explanation:

quite quite easily done mark it as brilliance answer

Answer 2

$\dfrac{\sec^2 20-\cot^2 70}{\sin^2 70+\sin^2 20}$ = 1

Step-by-step explanation:

We have,

$\dfrac{\sec^2 20-\cot^2 70}{\sin^2 70+\sin^2 20}$

To find, the value of $\dfrac{\sec^2 20-\cot^2 70}{\sin^2 70+\sin^2 20}$ = ?

∴ $\dfrac{\sec^2 20-\cot^2 70}{\sin^2 70+\sin^2 20}$

= $\dfrac{\sec^2 20-\cot^2 (90-20)}{\sin^2 70+\sin^2 (90-70)}$

Using the trigonometric identity,

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

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

= $\dfrac{\sec^2 20-\tan^2 20}{\sin^2 70+\cos^2 70}$

Using the trigonometric identity,

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

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

= $\dfrac{1}{1}$

= 1

∴  $\dfrac{\sec^2 20-\cot^2 70}{\sin^2 70+\sin^2 20}$ = 1

Thus, the value of $\dfrac{\sec^2 20-\cot^2 70}{\sin^2 70+\sin^2 20}$ is equal to "one (1)".