RRB NTPC Mains or Stage 2 Model Paper 6 with solutions

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Question : 40 of 119

Marks: +1, -0

If x

sin^3 θ + y cos^3θ = sin θ cos θ ≠ 0 and x sin θ –y cos θ = 0,

then value of

(x^2 + y^2)

is

sin q – cos q
sin q + cos q
0
1

Solution:

(d) Given,

x sin^3 θ + y cos^3 θ = sin θ cos θ...(i)and x sin θ – y cos q = 0 ...(ii)

From Eq. (ii),

x sin θ = y cos θ

From Eq. (i)

y cos θ . sin^2θ + y cos^3 θ = sin θ cos θy cosθ (sin^2 θ + cos^2 θ) = sin θcos θy = sin θ

From Eq. (i)
x = cos θ

∴ x^2 + y^2 = sin^2 θ + cos^2θ = 1

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