9. If sin theta + cos theta =root 2 sin theta, then show that sin theta-cos theta = root 2 sin theta.

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c o s Θ + s i n Θ = \sqrt{2} c o s Θ

c o s Θ - s i n Θ = \sqrt{2} s i n Θ

x^{3} - x^{2} (sin θ cos θ + sin θ + 1) + x ({sin}^{2} θ cos θ + sin θ cos θ + sin θ) - {sin}^{2} θ cos θ

x_{1}, x_{2} & x_{3}

\sec θ = \frac{13}{5}

\frac{2 \sin θ - 3 \cos θ}{4 \sin θ - 9 \cos θ} = 3

s i n θ + c o s θ = a, c o s θ - s i n θ = b

s i n θ (s i n θ - c o s θ) + s i n^{2} θ (s i n^{2} θ - c o s^{2} θ) + s i n^{3} θ (s i n^{3} θ - c o s^{3} θ) + . . . .

Cos theta+sin theta =√2cos theta show thatcos theta-sin theta =√2sin theta

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