Question

Prove that , 1 + sin 2a - cos 2a / sin a + cos a = 2 sina​

Answer 1

Answer:

Given,

Cos2A÷1-sin2A ------------ I

To prove,

Cos2A÷1-sin2A=cosA+sinA÷cosA-sinA

We can use 2 formulas,

cos(2A) = cos^{2}A - sin^{2}A

and,

sin(2A) = 2 sin(A)cos(A)

We, also substitute 1 = cos^{2}A + sin^{2}A

NOW try to solve the question on you own, with the following substitutions.

Make sure you have tried before actually seeing the solution.

Now, our I, becomes,

[cos^{2}(A) - sin^{2}(A)] / [sin^{2}A + cos^{2}(A) - 2 sin(A)cos(A)]

Now, denominator can be brought to the square form, and numerator simplified as,

[cos(A) + sin(A)] * [cos(A) - sin(A)] / [cos(A) - sin(A)]^{2}

Now, we get

cosA+sinA / cosA-sinA

Hence ,

Cos2A÷1-sin2A=cosA+sinA÷cosA-sinA

Hence, proved.

                                                          OR

LHS

=(1+ sin2A)/(cos2A)

=(sin2A+1)

/(cos^2A-sin^2A)

=(2cosA*sinA+cos^2A+sin^2A)

/(cosA+sinA)(cosA-sinA)

=(cosA+sinA)(cosA+sinA)

/(cosA+sinA)(cosA-sinA)

=(cosA+sinA)/(cosA-sinA)

RHS.

Hope it’s helpful......... ☺

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