Question

Prove the following trigonometric identities. $\frac{secA-tanA}{secA+tanA}=\frac{cos^{2}A }{(1+sinA)^{2} }$

Answer 1

Answer with Step-by-step explanation:

Given : (secA − tanA)/(secA + tanA) = cos²A/(1 + sinA)²

LHS = (secA − tanA)/(secA + tanA)

(secA − tanA) × (sec A + tan A)/(secA + tanA) × (sec A + tan A)

[By multiplying the denominator & numerator by (sec A + tan A)]  

= [sec²A − tan²A] /(secA + tanA)²

[By using identity , (a + b) (a - b) = a² - b²]

= 1/(secA + tanA)²

[By using  an identity, (sec²A − tan²A) = 1]

= 1/ (sec²A + tan²A + 2secAtanA)

[By using identity , (a + b)² = a² + 2ab + b²]

= 1/(1/cos²A + sin²A/cos²A + 2 × 1/cosA × sinA/cosA)

[By using the identity, secθ = 1/ cosθ & tanθ = sinθ/cosθ]

= 1/(1/cos²A + sin²A/cos²A + 2 sinA/cos²A)

= 1/ [(1 + sin²A + 2sinA)/cos²A]

= 1 ×  cos²A / [(1 + sin²A + 2sinA)]

= cos²A /(1 + sin²A)

[By using identity , a² + 2ab + b² = (a + b)² ]

(secA − tanA)/(secA + tanA) = cos²A/(1 + sinA)²

L.H.S = R.H.S  

Hence Proved..

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Answer 2

Answer:

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