Question

Given that:-

sinθ 2cosθ = 1, then prove that 2sinθ – cosθ = 2.

Answer 1

HEY!!

___________________________

✴(sinθ + 2cosθ)^2 = 12

✴(sinθ + 2cosθ)^2 + (2sinθ – cosθ)^2 = 1 + (2sinθ – cosθ)^2

✴5sin^2θ + 5cos^2θ = 1 + (2sinθ – cosθ)^2

✴5 - 1 = (2sinθ – cosθ)^2

✴root4 = 2sinθ – cosθ

✔2sinθ – cosθ = 2.

Answer 2

Note: I am writing theta as A.

= > Given Equation is sinA + 2cosA = 1

= > sinA + cosA + cosA = 1

= > sinA + cosA = 1 - cosA

On Squaring both sides, we get

= > (sinA + cosA)^2 = (1 - cosA)^2

= > sin^2A + cos^2A + 2sinAcosA = 1 + cos^2A - 2cosA

= > 1 + 2sinAcosA = 1 + cos^2A - 2cosA

= > 2sinAcosA - cos^2A = -2cosA

= > cosA(2sinA - cosA) = -2cosA

= > 2sinA - cosA = -2cosA/cosA

= > 2sinA - cosA = -2



Hope this helps!