Let - - 0-2- be such that 2cos- 1-sin- -sin 2 - tan- 2 - 2 cos- -1 tan 2- - -0 and -1-sin- - 3 2- Then - can not satisfy

The correct options are A 0 C 4π 2 D 3π 2 ...

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Byju's Answer

Standard X

Mathematics

Standard Values of Trigonometric Ratios

Let θ ,ϕ∈[ ...

Question

Let $θ, ϕ \in [0, 2 π]$ be such that $2 cos θ (1 - sin ϕ) = {sin}^{2} θ (tan \frac{θ}{2} + cot \frac{θ}{2}) cos ϕ - 1$
$tan (2 π - θ) > 0$ and $- 1 < sin θ < \frac{\sqrt{3}}{2}$ .
Then $ϕ$ can not satisfy

0 < ϕ < \frac{π}{2}

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\frac{π}{2} < ϕ < \frac{4 π}{3}

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\frac{4 π}{2} < ϕ < \frac{3 π}{2}

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\frac{3 π}{2} < ϕ < 2 π

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Solution

The correct options are
A $0 < ϕ < \frac{π}{2}$
C $\frac{4 π}{2} < ϕ < \frac{3 π}{2}$
D $\frac{3 π}{2} < ϕ < 2 π$

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Similar questions

Let $θ, ϕ \in [0, 2 π]$ be such that $2 c o s θ (1 - s i n ϕ) = s i n^{2} θ (t a n \frac{θ}{2} + c o t \frac{θ}{2}) c o s θ - 1, t a n (2 π - θ) > 0$ and -1 < $s i n θ < - \frac{\sqrt{3}}{2}$ . Then $ϕ$ cannot satisfy

Q. Let

θ, φ \in [0, 2 π]

be such that

2 cos θ (1 - sin φ) = {sin}^{2} θ (tan \frac{θ}{2} + cot \frac{θ}{2}) cos φ - 1

tan (2 π - θ) > 0

and

- 1 < sin θ < - \frac{\sqrt{3}}{2}

then

φ

cannot satisfy

Q. Show that

\frac{cos θ + cos ϕ}{sin θ + sin ϕ} = cot (\frac{θ + ϕ}{2})

Then find the value of sin

Q. If both

θ

and

ϕ

are acute angles such that

sin θ = \frac{1}{2}

and

cos ϕ = \frac{1}{3}

, then

(θ + ϕ)

belongs to

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Let θ,ϕ∈[0,2π] be such that 2cosθ(1−sinϕ)=sin2θ(tanθ2+cotθ2)cosϕ−1tan(2π−θ)>0 and −1<sinθ<√32. Then ϕ can not satisfy

The correct options are A 0<ϕ<π2 C 4π2<ϕ<3π2 D 3π2<ϕ<2π

Let $θ, ϕ \in [0, 2 π]$ be such that $2 cos θ (1 - sin ϕ) = {sin}^{2} θ (tan \frac{θ}{2} + cot \frac{θ}{2}) cos ϕ - 1$
$tan (2 π - θ) > 0$ and $- 1 < sin θ < \frac{\sqrt{3}}{2}$ .
Then $ϕ$ can not satisfy

The correct options are
A $0 < ϕ < \frac{π}{2}$
C $\frac{4 π}{2} < ϕ < \frac{3 π}{2}$
D $\frac{3 π}{2} < ϕ < 2 π$