SSC CGL Tier II 21 Sep 2014 Paper

Question : 45 of 100

Marks: +1, -0

If x = a (sinθ + cosθ), y = b (sinθ – cosθ) then the value of

{x^{2}}/{a^{2}}+ {y^{2}}/{b^{2}}

Solution:

(3) x = a (sinθ + cosθ) and y = b (sinθ – cosθ)
⇒

x/a =

(3) x = a (sinθ + cosθ) and

y/b

= (sinθ – cosθ)

∴{x^{2}}/{a^{2}}+ {y^{2}}/{b^{2}}=(\sin \θ+\cos \θ)^{2}+

(\sin \θ-\cos \θ)^{2}

=\sin ^{2} \θ+\cos ^{2} \θ+2 \sin \θ . \cos \θ+

\sin ^{2} \θ+\cos ^{2} \θ-2 \sin \θ . \cos \θ

=2 (\sin ^{2} \θ+\cos ^{2} \θ)=2

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