Question

prove that 1+Cos2A/1-Cos2A=tan²A​

Answer 1

Proof:

We know that,

cos(A+B) =cosA.cosB-sinA.sinB

=>cos2A=cos(A+A)

=>cos2A=cosA.cosA - sinA.sinA

=>cos2A=cos²A-sin²A

=>cos2A=(cos²A-sin²A)/(cos²A+sin²A

Since {cos²A+sin²A=1}

Divide the numerator & the denominator by (cos²A) to get,

cos2A = {(cos²A-sin²A) ÷cos²A} / {(cos²A+sin²A) ÷cos²A}

cos2A ={(1-tan²A)/(1+tan²A)}

Then,

1-cos2A = 1-[{(1–tan²A)/(1+tan²A)}]

1-cos2A =(1+tan²A-1+tan²A)/(1+tan²A)

1-cos2A=(2tan²A)/(1+tan²A)

And now.......

1+cos2A=1+[{(1-tan²A)/(1+tan²A)}]

1+cos2A={1+tan²A+1-tan²A}/{1+tan²A}

1+cos2A=2/(1+tan²A)

So now,

(1-cos2A)/(1+cos2A)= {2tan²A/(1+tan²A)}÷{2/(1+tan²A)}

={(2tan²A)(1+tan²A)}÷{2(1+tan²A)}

={(2tan²A)(1+tan²A)}÷{2(1+tan²A)}

=tan²A

Answer 2

Step-by-step explanation:

here answer is tan square a