If a sinθ + b cosθ = c, then prove that a cosθ - b sinθ = √(a2 + b2 - c2)
Solution:
Given, a sinθ + b cosθ = c
We have to prove that a cosθ - b sinθ = √(a² + b² - c²)
On squaring both sides,
(a sinθ + b cosθ)² = c²
By using algebraic identity,
(x + y)² = x² + 2xy + y²
Now, (a sinθ + b cosθ)² = a²sin²θ + 2absinθ cosθ + b²cos²θ
So, a²sin²θ + 2absinθ cosθ + b²cos²θ = c²
By using trigonometric identities,
cos² A + sin² A = 1
sin² A = 1 - cos² A
cos² A = 1 - sin² A
Now, a²(1 - cos² θ) + 2absinθ cosθ + b²(1 - sin² θ) = c²
a² - a²cos²θ + 2absinθ cosθ + b² - b²sin²θ = c²
a² + b² - c² = a²cos²θ - 2absinθ cosθ + b²sin²θ ---------- (1)
By using algebraic identity,
(x - y)² = x² - 2xy + y² ------------------------------------------- (2)
By comparing (1) and (2),
x² = a²cos²θ
x = acosθ
y² = b²sin²θ
y = bsinθ
So, a²cos²θ - 2absinθ cosθ + b²sin²θ = (acosθ - bsinθ)²
Now, (acosθ - bsinθ)² = a² + b² - c²
Taking square root,
acosθ - bsinθ = √(a² + b² - c²)
✦ Try This: If sinθ = 3/5, find the value of (cosθ - 1/tanθ)/2cotθ
☛ Also Check: NCERT Solutions for Class 10 Maths Chapter 8
NCERT Exemplar Class 10 Maths Exercise 8.4 Problem 11
If a sinθ + b cosθ = c, then prove that a cosθ - b sinθ = √(a2 + b2 - c2)
Summary:
Trigonometric ratios are the ratios of the length of sides of a triangle. It is proved that a cosθ - b sinθ = √(a²+b²-c²) if a sinθ + b cosθ = c
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