If a \cos \theta– b \sin \theta= c, then a \cos \theta+ b \sin \theta=?

Correct Answer is : ±a2+b2−c2

RRB NTPC Mains or Stage 2 Model Paper 8 with solutions

Question : 20 of 120

Marks: +1, -0

a cos θ– b sin θ= c,

then

a cos θ+ b sin θ=?

Solution:

(b)

(a cos θ – b sin θ)^2 + (a cosθ + b sin θ)^2

= a^2 cos^2θ + b^2 sin^2 θ– 2ab sinθ cos θ + a^2cos^2 θ + b^2 sin^2 θ + 2ab cos θ. sin θ.

= a^2 (cos^2 θ + sin^2 θ) + b^2 (sin^2 θ + cos2 θ)

= a^2 × 1 + b^2 × 1

= a^2 + b^2.

∴ (a cos θ – b sin θ )^2 + (a cos θ + b sin θ)^2

= a^2 + b^2.

⇒ c^2 + (a cos θ + b sin θ)^2 = a^2 + b^2

⇒ a cos θ + b sin θ= ± a^2 + b^2 - c^2

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