tan A/ (1 + sec A) - tan A/ (1 - sec A) = 2 cosec A. Prove the following statement

Solution:

LHS = tan A/ (1 + sec A) - tan A/ (1 - sec A)

Taking the LCM

= [tan A (1 - sec A) - tan A (1 + sec A)]/ [(1 + sec A) (1 - sec A)]

Taking out tan A as common in the numerator

= [tan A (1 - sec A - 1 - sec A)]/ [(1 + sec A) (1 - sec A)]

Using the algebraic identity in the denominator

(a + b) (a - b) = a² - b²

= [tan A (-2 sec A)]/ (1 - sec² A)

Multiplying negative sign in numerator and denominator

= (2 sec A tan A)/ (sec² A - 1)

Here sec² A + tan² A = 1

= (2 sec A tan A)/ tan² A

We know that

sec A = 1/ cos A and tan A = sin A/cos A

= 2 sec A/ tan A

= 2/ sin A

= 2 cosec A

= RHS

Therefore, it is proved.

✦ Try This: Prove the following:

1/ (sec A - 1) + 1/ (sec A + 1) = 2 cosec A cot A

☛ Also Check: NCERT Solutions for Class 10 Maths Chapter 8

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NCERT Exemplar Class 10 Maths Exercise 8.3 Problem 2

tan A/ (1 + sec A) - tan A/ (1 - sec A) = 2 cosec A. Prove the following statement

Summary:

The secant function of a right triangle is its hypotenuse divided by its base. It is proved that tan A/ (1 + sec A) - tan A/ (1 - sec A) = 2 cosec A

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