SSC CGL Tier II 12 Apr 2015 Paper

Question : 41 of 100

Marks: +1, -0

If a sin θ + b cos θ = c, then a cos θ – b sin θ is equal to

Solution:

(3) a sinθ + b cosθ = c .... (i)
a cosθ – b sinθ = x (let) ... (ii)
On squaring equations (i) and (ii) and adding,

a^{2} \sin ^{2} \θ+b^{2} \cos ^{2} \θ+2 a b \sin\θ

\cos \θ+a^{2} \cos ^{2} \θ+b^{2} \sin ^{2} \θ-2 a b-

\sin \θ . \cos \θ=c^{2}+x^{2}

⇒ a^{2} (\sin ^{2}\θ+\cos ^{2} \θ)+b^{2} (\cos ^{2} \θ

+\sin ^{2} \θ )=c^{2}+x^{2}

⇒ a^{2}+b^{2}=c^{2}+x^{2}

⇒ x^{2}=a^{2}+b^{2}-c^{2}

⇒ x=± √{a^{2}+b^{2}-c^{2}}

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