Question

prove:
secA/secA-1 + secA/secA+1 = 2cosec^A

Answer 1

reply if can't understand

Answer 2

$\dfrac{\sec A}{\sec A-1} +\dfrac{\sec A}{\sec A+1} =2\csc^2 A$, proved.

Step-by-step explanation:

To prove the given identity: $\dfrac{\sec A}{\sec A-1} +\dfrac{\sec A}{\sec A+1} =2\csc^2 A$

L.H.S. = $\dfrac{\sec A}{\sec A-1} +\dfrac{\sec A}{\sec A+1}$

Taking the LCM of denominator part, we get

= $\dfrac{\sec A(\sec A+1)+\sec A(\sec A+1)}{(\sec A-1)(\sec A+1)}$

= $\dfrac{\sec^2 A+\sec A+\sec^2 A-\sec A}{(\sec A-1)(\sec A+1)}$

Using the algebraic identity,

$a^{2} -b^{2} =(a+b)(a-b)$

= $\dfrac{\sec^2 A+\sec^2 A}{\sec^2 A-1^2}$

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

Using the trigonometric identity,

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

⇒ $\tan^2 A=\sec^2 A-1$

= $\dfrac{2\sec^2 A}{\tan^2 A}$

Using the trigonometric identity,

$\sec A=\dfrac{1}{\cos A}$ and $\tan A=\dfrac{\sin A}{\cos A}$

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

= $\dfrac{2}{\sin^2 A}$

= $2\csc^2 A$

= L.H.S., proved.

Thus, $\dfrac{\sec A}{\sec A-1} +\dfrac{\sec A}{\sec A+1} =2\csc^2 A$, proved.