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Prove that: sinA/1+cosA +cosA/sinA=cosecA​

User Rcshon
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Answer:

We proceed to demonstrate the identity given on statement by algebraic and trigonometric means:

1)
(\sin A)/(1 + \cos A) + (\cos A)/(\sin A) Given

2)
(\sin^(2)A+\cos A\cdot (1+\cos A))/(\sin A\cdot (1 + \cos A))
(a)/(b) + (c)/(d) = (a\cdot d + b\cdot c)/(b\cdot d)/Definition of power

3)
(\sin^(2)A+\cos^(2)A + \cos A)/(\sin A\cdot (1 + \cos A)) Distributive property/Definition of power

4)
(1+\cos A)/(\sin A\cdot (1+\cos A)) Associative property/
\sin^(2)A + \cos^(2)A = 1

5)
(1)/(\sin A) Existence of multiplicative inverse/Modulative property

6)
\csc A
(1)/(\sin A) = \csc A/Result

Step-by-step explanation:

We proceed to demonstrate the identity given on statement by algebraic and trigonometric means:

1)
(\sin A)/(1 + \cos A) + (\cos A)/(\sin A) Given

2)
(\sin^(2)A+\cos A\cdot (1+\cos A))/(\sin A\cdot (1 + \cos A))
(a)/(b) + (c)/(d) = (a\cdot d + b\cdot c)/(b\cdot d)/Definition of power

3)
(\sin^(2)A+\cos^(2)A + \cos A)/(\sin A\cdot (1 + \cos A)) Distributive property/Definition of power

4)
(1+\cos A)/(\sin A\cdot (1+\cos A)) Associative property/
\sin^(2)A + \cos^(2)A = 1

5)
(1)/(\sin A) Existence of multiplicative inverse/Modulative property

6)
\csc A
(1)/(\sin A) = \csc A/Result

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