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Given that sinθ + 2cosθ = 1, then prove that 2sinθ - cosθ = 2

Solution:

Given, sinθ + 2cosθ = 1

We have to prove that 2sinθ - cosθ = 2.

Squaring on both sides,

(sinθ + 2cosθ)² = 1

(a + b)² = a² + 2ab + b²

So, (sinθ + 2cosθ)² = sin²θ + 4sinθ cosθ + 4cos²θ

Now, sin²θ + 4sinθ cosθ + 4cos²θ = 1

cos² A = 1 - sin² A

sin² A = 1 - cos² A

So, (1 - cos² θ) + 4sinθ cosθ + 4(1 - sin² θ) = 1

1 - cos²θ + 4 - 4sin²θ + 4 sinθ cosθ = 1

5 - cos²θ - 4sin²θ + 4 sinθ cosθ = 1

On rearranging,

5 - 1 = cos²θ + 4sin²θ - 4 sinθ cosθ

cos²θ + 4sin²θ - 4sinθ cosθ = 4

4sin²θ + cos²θ - 4sinθ cosθ = 4 --------------- (1)

By using algebraic identity,

(a - b)² = a² - 2ab + b² --------------------- (2)

By comparing (1) and (2),

a² = 4sin²θ

a = 2sinθ

2ab = 4sinθ cosθ

ab = 2sinθ cosθ

b² = cos²θ

b = cosθ

So, 4sin²θ + cos²θ - 4sinθ cosθ = (2sinθ - cosθ)²

Now, (2sinθ - cosθ)² = 4

Taking square root,

2sinθ - cosθ = 2

Therefore, 2sinθ - cosθ = 2

✦ Try This: If 3cosθ - 4sinθ = 2cosθ + sinθ, find tanθ

☛ Also Check: NCERT Solutions for Class 10 Maths Chapter 8

NCERT Exemplar Class 10 Maths Exercise 8.4 Problem 5

Summary:

Given that sinθ + 2cosθ = 1, then it is proven that 2sinθ - cosθ = 2

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