Question

prove the identity
tan^2A - tan ^2B =sin^2A-sin^2B/cos^2Acos^2B=sec^2A-sec^2B​

Answer 1

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Answer 2

To prove:-

$\rm tan^2 \ A - tan^2 \ B = \dfrac{sin^2 \ A \sin^2 B}{\cos^2 A \cos^2 B} = \sec^2 A - \sec^2 B$

Solution:-

1.)

• tan A = $\dfrac{\sin A}{\cos A}$

$\rm \dashrightarrow tan^2 \ A - tan^2 \ B = \dfrac{\sin^2 A}{\cos^2 A}-\dfrac{\sin^2 B}{\cos^2 B}$

$\rm \dashrightarrow \dfrac{\sin^2 A \cos^2 B - \sin^2 B \cos^2 A}{\cos^2 A \cos^2 B}$

• cos² A = 1 - sin² A

$\rm \dashrightarrow \dfrac{\sin^2 A (1-\sin^2 B) - \sin^2 B (1-\sin^2 A)}{\cos^2 A \cos^2 B}$

$\rm \dashrightarrow \dfrac{\sin^2 A - \sin^2 A \sin^2 B - \sin^2 B + \sin^2 A \sin^2 B}{\cos^2 A \cos^2 B}$

$\rm \dashrightarrow \dfrac{\sin^2 A - \sin^2 B }{\cos^2 A \cos^2 B}$

= LHS

2.) tan² A - tan² B = sec² A - sec² B

• tan² A = sec² A -1

→ (sec² A - 1) - (sec² B - 1)

→ sec² A - 1 - sec² B + 1

→ sec² A - sec² B = LHS

$\therefore \rm\tan^2 A - \tan^2 B = \dfrac{\sin^2 A - \sin^2 B}{\cos^2 A \cos^2 B} = \sec^2 A - \sec^2 B$

Hence, proved.