Question

proove that sin theta + cos theta÷ sin theta - cos theta÷ sin theta+cos theta= 2/2(sin²-1)
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Answer 1

Answer:

(sinA + 2cosA)^2+(2sinA - cos A)^2

=Sin^2(A)+2.sinA.2cosA+4cos^2(A)+4sin^A-2.2sinA.cosA+cos^2(A)

=5sin^2(A)+5cos^2(A)

=5*{sin^2(A)+cos^2(A)}

=5..

Now,

(sinA+2cosA)^2+(2cosA-sinA)^2=5

Or,1+(2cosA-sinA)^2=5

Or, (2cosA-sinA)^2=5–1=4.

Or, 2cosA-sinA=√4=±2

Or, 2cosA-sinA=±2 (proved).

Answer 2

Answer:

Step-by-step explanation:

sin ϴ + 2 cos ϴ = 1

Squaring both the sides

(sin ϴ + 2 cos ϴ) ² = (1) ²

sin² ϴ + 4 cos² ϴ + 4 sin ϴ cos ϴ = 1

because sin² ϴ = 1 - cos² ϴ & cos² ϴ=1- sin² ϴ

So replacing sin² ϴ by 1 - cos² ϴ and cos² ϴ by 1- sin² ϴ

we get

1 - cos² ϴ + 4 ( 1 - sin² ϴ ) + 4sin ϴ cos ϴ = 1

1 - cos² ϴ + 4 – 4sin² ϴ + 4 sin ϴ cos ϴ = 1

5 – 1 = cos² ϴ +4sin² ϴ - 4 sin ϴ cos ϴ

or

( cos ϴ – 2 sin ϴ ) ² = 4

cos ϴ -2sin ϴ = ± 2