Prove that : (1 + tan A. tan B) 2 + (tan A - tan B) 2 = sec 2 A sec 2 B

By formula, sec 2 θ = 1 + tan 2 θ Solving L.H.S. of the equation : ⇒ (1 + tan A. tan B) 2 + (tan A - tan B) 2 ⇒ 1 + tan 2 A tan 2 B + 2 tan A tan B + tan 2 A + tan 2 B - 2 tan A tan B ⇒ 1 + tan 2 A tan 2 B + tan 2 A + tan 2 B ⇒ 1 + tan 2 A + tan 2 A tan 2 B + tan 2 B ⇒ sec 2 A + tan 2 B(tan 2 A + 1) ⇒ sec 2 A + tan 2 B sec 2 A ⇒ sec 2 A(1 + tan 2 B) ⇒ sec 2 A sec 2 B. Since, L.H.S. = R.H.S. Hence, proved that (1 + tan A. tan B) 2 + (tan A - tan B) 2 = sec 2 A sec 2 B.

Mathematics

Prove that :
(1 + tan A. tan B)² + (tan A - tan B)² = sec² A sec² B

Trigonometric Identities

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Answer

By formula,

sec² θ = 1 + tan² θ

Solving L.H.S. of the equation :

⇒ (1 + tan A. tan B)² + (tan A - tan B)²

⇒ 1 + tan² A tan² B + 2 tan A tan B + tan² A + tan² B - 2 tan A tan B

⇒ 1 + tan² A tan² B + tan² A + tan² B

⇒ 1 + tan² A + tan² A tan² B + tan² B

⇒ sec² A + tan² B(tan² A + 1)

⇒ sec² A + tan² B sec² A

⇒ sec² A(1 + tan² B)

⇒ sec² A sec² B.

Since, L.H.S. = R.H.S.

Hence, proved that (1 + tan A. tan B)² + (tan A - tan B)² = sec² A sec² B.

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Mathematics

Prove that :
(1 + tan A. tan B)² + (tan A - tan B)² = sec² A sec² B

Trigonometric Identities

Answer

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