Question

Prove the following trigonometric identities. $\frac{1+cosA}{sinA}=\frac{sinA}{1-cosA}$

Answer 1

Answer with Step-by-step explanation:

Given :  (1 + cosA)/sinA = sinA/(1 − cosA)

LHS = (1 + cosA)/sinA

= (1 + cosA) × (1 − cosA)/ sinA × (1 − cosA)

[By Multiplying both numerator and denominator with (1 – cos A)]

= (1 − cos²A) / sinA(1 − cosA)

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

= sin²A/ sinA(1 − cosA)

[By using  an identity, (1 - cos²θ) = sin²θ]

= sinA/(1 − cosA)

(1 + cosA)/sinA = sinA/(1 − cosA)

L.H.S = R.H.S  

Hence Proved..

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Answer 2

Step-by-step explanation:

L.H.S (1+cosA)/sinA

multiply and devide by (1 - cosA)

we get

(1^2 - cos^2A)/ sinA(1-cosA)

used formula, (1 - cos^2A = sin^2A )

= sin^2A/sinA(1- cosA)

= sinA/(1- cosA) R .H S : Proved