Question

6. Prove (1 + tan A)2 + (1 - tan A)? = 2 sec A.
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Answer 1

Question:

prove that

[1 + tan(a)]² + [1 - tan(a)]² = 2 sec²(a)

Used formulas:

→ sec²θ - tan²θ = 1

sec²θ = 1 + tan²θ

tan²(a) = sec²(a) - 1

→ (a + b)² = a² + 2ab + b²

(a - b)² = a² - 2ab + b²

Answer:

Taking LHS

= 1 + 2tan(a) + tan²(a) + 1 - 2tan(a) + tan²(a)

= 2 + 2tan²(a)

= 2 + 2 [sec²(a) - 1]

= 2 + 2sec²(a) - 2

= 2sec²(a)

LHS = RHS

Hence proved

Concepts used:

⟩⟩ trigonometric identities

⟩⟩ trigonometric ratios