Question

(1+ sinA)^2 + (1-sinA)^2/2cos^2A=1+sin^2A/1-sin^2A​

Answer 1

Step-by-step explanation:

How can we prove that (1+sin2A+cos2A) ^2=4cos^2A (1+sin2A)?

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(1+sin2A+cos2A)2=4cos22A(1+sin2A)

(a+b+c)2=a2+b2+c2+2ab+2bc+2ac

Left hand side

(1+sin2A+cos2A)2

=1+sin22A+cos22A+2sin2A+2sin2Acos2A+2cos2A

=2+2sin2A+2sin2Acos2A+2cos2A∵(cos2θ+sin2θ=1)

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

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

=2(1+sin2A)(1+cos2A)∵(2cos2θ=1+cos2θ)

=2(1+sin2A)(2cos2A)

=4cos2A(1+sin2A)

=Right hand side

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