Let - - - - -0- 2- - be such that 2cos- 1-sin- -sin 2- tan-2-2 cos- -1- tan 2- - -0 and -1-sin- -32- Then - cannot satisfy

2cosθ (1 sinϕ )=sin ^2θ ( 1sinθ2.cosθ2 )cosϕ 1 ⇒ 2cosθ (1 sinϕ )=2sinθ .cosϕ 1 ⇒ 2cosθ +1=2sin (θ +ϕ ) nbsp; nbsp; nbsp;...(i) Now, tanθ lt;0 and ...

You visited us 1 times! Enjoying our articles? Unlock Full Access!

Byju's Answer

Standard XII

Mathematics

Variable Separable Method

Let θ , φ ϵ [...

Question

Let $θ, φ ϵ [0, 2 π]$ 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

0 < φ < \frac{π}{2}

Right on! Give the BNAT exam to get a 100% scholarship for BYJUS courses

\frac{4 π}{3} < φ < \frac{3 π}{2}

Right on! Give the BNAT exam to get a 100% scholarship for BYJUS courses

\frac{3 π}{2} < φ < 2 π

Right on! Give the BNAT exam to get a 100% scholarship for BYJUS courses

\frac{π}{2} < φ < \frac{4 π}{3}

No worries! We‘ve got your back. Try BYJU‘S free classes today!

Open in App

Solution

The correct option is C $\frac{3 π}{2} < φ < 2 π$
$2 cos θ (1 - sin ϕ) = {sin}^{2} θ ⎛ ⎜ ⎜ ⎜ ⎝ \frac{1}{sin \frac{θ}{2} . cos \frac{θ}{2}} ⎞ ⎟ ⎟ ⎟ ⎠ cos ϕ - 1$

$\Rightarrow 2 cos θ (1 - sin ϕ) = 2 sin θ . cos ϕ - 1$

$\Rightarrow 2 cos θ + 1 = 2 sin (θ + ϕ)$ ...(i)

Now, $tan θ < 0 and - 1 < sin θ < \frac{- \sqrt{3}}{2}$ gives

$θ ϵ (\frac{3 π}{2}, \frac{5 π}{3})$

Using (i);

$sin (θ + ϕ) = cos θ + \frac{1}{2}$

$\Rightarrow \frac{1}{2} < sin (θ + ϕ) < 1$

i.e. $\frac{π}{6} < θ + ϕ < \frac{5 π}{6} or, \frac{13 π}{6} < θ + ϕ < \frac{17 π}{6}$

$\Rightarrow \frac{13 π}{6} - θ < ϕ < \frac{17 π}{6} - θ$

$\Rightarrow \frac{2 π}{3} < ϕ < \frac{7 π}{6}$

So, options (1), (3) and (4) are correct

Suggest Corrections

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

$\frac{1 + s i n θ}{1 - s i n θ} = {tan}^{2} (\frac{π}{4} + \frac{θ}{2})$

Q. There exists a value of

θ

between

0

and

2 π

that satisfies the equation

{sin}^{4} θ - 2 {sin}^{2} θ - 1 = 0

Q. Let

tan (2 π | sin θ |) = cot (2 π | cos θ |)

, where

θ \in R

and

f (x) = (| sin θ | + | cos θ |)^{x}, x \geq 1

. Then range of

f (x)

not include

Join BYJU'S Learning Program

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

Let $θ, φ ϵ [0, 2 π]$ 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