Question

if a cos theta + b sin theta =3 and a sin theta- b cos theta =4 then a^2+b^2has the value=

Answer 1

Solution :

Given that,

a cosθ + b sinθ = 3 --------(1)

a sinθ - b cosθ = 4 --------(2)

Squaring both sides in (1), We get

( a cosθ + b sinθ )² = 3²

a²cos²θ + b²sin²θ + 2ab cosθsinθ = 9
---------(3)

Squaring both sides in (2), We get

( a sinθ - b cosθ )² = 4²

a²sin²θ + b²cos²θ - 2ab sinθcosθ = 16
---------(4)

Adding (3) and (4), We get

a² ( cos²θ + sin²θ ) + b² ( sin²θ + cos²θ )
= 9 + 16

⇒ a² (1) + b² (1) = 25

⇒ a² + b² = 25

Answer 2

correct answer is 25