Question

Cos^2(45+A)+cos^2(45-A)/tan(60+A)*tan(30-A) = 1

Answer 1

Cos^2(45+A)+cos^2(45-A)/tan(60+A)*tan(30-A) = 1



cos2(45+a) + cos2(45-a)
= cos2(45+a) + sin2(90-45+a)
= cos2(45+a) + sin2(45+a)
=1


tan(60+a) *tan(30-a)
= tan(60+a) * cot(90-60+a)
= tan(60+a) * cot(60+a)
= 1

Cos^2(45+A)+cos^2(45-A)/tan(60+A)*tan(30-A) = 1
1 / 1 = 1
1 = 1

RHS=LHS

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THANKS ☺️

Answer 2

Solution:

To Show:

\dfrac{cos^2(45+\theta)+cos^2(45-\theta)}{tan(60+\theta).tan(30-\theta)}=1

Formulas Used:

cosA=sin(90-A)\\\\tanA=cot(90-A)\\\\cos^2A+sin^2A=1\\\\tanA.cotA=1

Proof:

Taking Left Hand Side,\dfrac{cos^2(45+\theta)+cos^2(45-\theta)}{tan(60+\theta).tan(30-\theta)}\\\\\\=\dfrac{cos^(45+\theta)+sin^2(90-45+\theta)}{tan(60+\theta).cot(90-30+\theta)}\\\\\\=\dfrac{cos^2(45+\theta)+sin^2(45+\theta)}{tan(60+\theta).cot(60+\theta)}\\\\\\=\dfrac{1}{1}\\\\\\=1

Right Hand Side = Left Hand Side.

Hence Proved.