can anyone solvethe questions above
Answers
aCosθ -bSinθ = √a² + b² - c² if aSinθ + bCosθ = c
Step-by-step explanation:
aSinθ + bCosθ = c
squaring both sides
=>( aSinθ + bCosθ)² = c²
=> a²Sin²θ + b²Cos²θ +2aSinθbCosθ = c²
using Sin²θ + Cos²θ = 1
=> a²(1 - Cos²θ ) + b²(1 - Sin²θ ) + 2abSinθCosθ = c²
=> a² - a²Cos²θ + b² - b²Sin²θ + 2abSinθCosθ = c²
=> a² + b² - c² = a²Cos²θ + b²Sin²θ - 2abSinθCosθ
=> a² + b² - c² = (aCosθ)² + (-bSinθ)² + 2aCosθ(-bSinθ)
=> a² + b² - c² = (aCosθ -bSinθ)²
Taking square root both sides
=> √a² + b² - c² = aCosθ -bSinθ
=> aCosθ -bSinθ = √a² + b² - c²
QED
Proved
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