If sin- - 12- then find the value of 1 - 2sin2-sin-

1 2sin^2θsinθ = 1 2(14)12 = 1 1212 = 1 ...

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If sinθ = 1...

Question

If $sin θ = \frac{1}{2}$ , then find the value of $\frac{1 - 2 {sin}^{2} θ}{sin θ}$

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The value of $\frac{2 (s i n 2 θ + 2 c o s^{2} θ - 1)}{c o s θ - s i n θ - c o s 3 θ + s i n 3 θ}$ is

\sqrt{(\frac{1 - sin θ}{1 + sin θ})} + \sqrt{(\frac{1 + sin θ}{1 - sin θ})}

If $θ$ lies in the second quadrant, then the value of $\sqrt{\frac{1 - s i n θ}{1 + s i n θ}}$ + $\sqrt{\frac{1 + s i n θ}{1 - s i n θ}}$ .

sin θ + 2 cos θ = 1

2 sin θ - cos θ

sin θ \geq 0

sin (\frac{1}{2} {cot}^{- 1} (- \frac{3}{4}))

sin θ / 2

sin θ / 2

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