Let - -0-2 - be such that 2 cos-1-sin-sin 2-tan-2-2 cos-1- tan 2 -0 and -1-sin-3-2- Then - cannotsatisfy

The correct options are A 0 lt; ϕ lt; π/2 C 4 π/3 lt; ϕ lt;3 π/2 D 3 π/2 lt; ϕ lt; 2 π 2 cos θ (1 sin ϕ) =2 sin^2 θ/sin θcos ϕ 1 ∴ 2 cos θ 2 c ...

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Byju's Answer

Standard XI

Mathematics

Solving Simultaneous Trigonometric Equations

Let θ, ϕ∈[0,2...

Question

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

$0 < ϕ < \frac{π}{2}$

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$\frac{π}{2} < ϕ < \frac{4 π}{3}$

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$\frac{4 π}{3} < ϕ < \frac{3 π}{2}$

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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
$\frac{4 π}{3} < ϕ < \frac{3 π}{2}$

D
$\frac{3 π}{2} < ϕ < 2 π$

$2 c o s θ (1 - s i n ϕ) = \frac{2 s i n^{2} θ}{s i n θ} c o s ϕ - 1 ∴ 2 c o s θ - 2 c o s θ s i n ϕ = 2 s i n θ c o s ϕ - 1 ∴ 2 c o s θ + 1 = 2 s i n (θ + ϕ) t a n (2 π - θ) > 0 \Rightarrow t a n θ < 0$
and $- 1 < s i n θ < - \frac{\sqrt{3}}{2}$
$\Rightarrow θ \in (\frac{3 π}{2}, \frac{5 π}{3})$
$\Rightarrow 0 < c o s θ < \frac{1}{2}$
$\frac{1}{2} < s i n (θ + ϕ) < 1$
$\Rightarrow \frac{π}{6} + 2 π < s i n (θ + ϕ) < \frac{5 π}{6} + 2 π$
$2 π + \frac{π}{6} - θ_{m a x} < ϕ < 2 π + \frac{5 π}{6} - θ_{m i n}$
$\Rightarrow \frac{π}{2} < ϕ \frac{4 π}{3}$

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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]

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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

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- 1 < sin θ < \frac{\sqrt{3}}{2}

.
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and

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Q. Let

θ, φ ϵ 0, 2 π

be such that

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s i n^{2} θ (t a n \frac{θ}{2} + c o t \frac{θ}{2}) c o s φ - 1

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

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