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Find the value of the given expression.​​cosA1−tanA+sin2AsinA−cosA+2sinA−cosA\frac{cosA}{1-tanA} + \frac{sin^2A}{sinA - cosA} + 2sinA - cosA1−tanAcosA
Question

Find the value of the given expression.

​​cosA1tanA+sin2AsinAcosA+2sinAcosA\frac{cosA}{1-tanA} + \frac{sin^2A}{sinA - cosA} + 2sinA - cosA​​

A.

3sinA

B.

sinA

C.

3cosA

D.

cosA

Correct option is A

Given:

cosA1tanA+sin2AsinAcosA+2sinAcosA\frac{cosA}{1-tanA} + \frac{sin^2A}{sinA - cosA} + 2sinA - cosA  

Formula Used: 

a2b2=(a+b)(ab)a^2 - b^2 = (a+b)(a-b)

Solution:

cosA1sinAcosA+sin2AsinAcosA+2sinAcosA =cos2AcosAsinA+sin2AsinAcosA+2sinAcosA\frac{cosA}{1-\frac{sinA}{cosA}} + \frac{sin^2A}{sinA - cosA} + 2sinA - cosA \\\ \\ = \frac{cos^2A}{cosA - sin A} + \frac{sin^2A}{sinA - cosA} + 2sinA - cosA  

 cos2A1sinAcosA+sin2AsinAcosA+2sinAcosA =cos2Asin2AcosAsinA+2sinAcosA =(cosA+sinA)(cosAsinA)cosAsinA+2sinAcosA =(cosA+sinA)+2sinAcosA =3sinA\frac{cos^2A}{1-\frac{sinA}{cosA}} + \frac{sin^2A}{sinA - cosA} + 2sinA - cosA \\\ \\ = \frac{cos^2A- sin^2A}{cosA - sin A} + 2sinA - cosA \\\ \\= \frac{(cosA + sin A)(cosA - sin A)}{cosA - sin A}+2sinA - cosA \\\ \\= (cosA + sin A) + 2sinA - cosA \\\ \\= 3 sinA

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