Question

Find number of real roots of the equation sec(theta )+cosec(theta)=sqr(15) lying between 0 to pi
by the way answer 3
plz answer the question step by step

Answer 1

Given,

$\small\longrightarrow\sec\theta+\csc\theta=\sqrt{15},\quad\!\theta\in(0,\ \pi)$

$\small\longrightarrow\dfrac{1}{\cos\theta}+\dfrac{1}{\sin\theta}=\sqrt{15}$

$\small\longrightarrow\sin\theta+\cos\theta=\sqrt{15}\,\sin\theta\cos\theta\quad\dots(1)$

Here both sides should be either positive or negative.

Assume,

$\small\longrightarrow\sin\theta\cos\theta>0$

$\small\Longrightarrow\sin\theta>0,\quad\cos\theta>0$

Then θ should lie in first quadrant.

$\small\longrightarrow\theta\in\left(0,\ \dfrac{\pi}{2}\right)\quad\dots(i)$

Or,

$\small\Longrightarrow\sin\theta<0,\quad\cos\theta<0$

Then θ should lie in third quadrant but $\theta\in(0,\ \pi)$ so this is not possible.

And,

$\small\longrightarrow\sin\theta+\cos\theta>0$

$\small\longrightarrow\sin\theta\cdot\dfrac{1}{\sqrt2}+\cos\theta\cdot\dfrac{1}{\sqrt2}>0\cdot\dfrac{1}{\sqrt2}$

$\small\longrightarrow\sin\left(\theta+\dfrac{\pi}{4}\right)>0$

$\small\Longrightarrow\theta+\dfrac{\pi}{4}\in(2n\pi,\ 2n\pi+\pi),\quad\!n\in\mathbb{Z}$

$\small\longrightarrow\theta\in\left(2n\pi-\dfrac{\pi}{4},\ 2n\pi+\dfrac{3\pi}{4}\right)\cap(0,\ \pi)$

$\small\longrightarrow\theta\in\left(0,\ \dfrac{3\pi}{4}\right)\quad\dots(ii)$

Taking (i) ∧ (ii),

$\small\longrightarrow\theta\in\left(0,\ \dfrac{\pi}{2}\right)\quad\dots(a)$

Assume,

$\small\longrightarrow\sin\theta\cos\theta<0$

$\small\Longrightarrow\sin\theta>0,\quad\cos\theta<0$

Then θ should lie in second quadrant.

$\small\longrightarrow\theta\in\left(\dfrac{\pi}{2},\ \pi\right)\quad\dots(iii)$

Or,

$\small\Longrightarrow\sin\theta<0,\quad\cos\theta>0$

Then θ should lie in fourth quadrant but $\theta\in(0,\ \pi)$ so this is not possible.

And,

$\small\longrightarrow\sin\theta+\cos\theta<0$

$\small\longrightarrow\sin\left(\theta+\dfrac{\pi}{4}\right)<0$

$\small\Longrightarrow\theta+\dfrac{\pi}{4}\in(2n\pi-\pi,\ 2n\pi),\quad\!n\in\mathbb{Z}$

$\small\longrightarrow\theta\in\left(2n\pi-\dfrac{5\pi}{4},\ 2n\pi-\dfrac{\pi}{4}\right)\cap(0,\ \pi)$

Since $\theta\in(0,\ \pi),$

$\small\longrightarrow\theta\in\left(\dfrac{3\pi}{4},\ \pi\right)\quad\dots(iv)$

Taking (iii) ∧ (iv),

$\small\longrightarrow\theta\in\left(\dfrac{3\pi}{4},\ \pi\right)\quad\dots(b)$

Taking (a) ∨ (b),

$\small\longrightarrow\theta\in\left(0,\ \dfrac{\pi}{2}\right)\cup\left(\dfrac{3\pi}{4},\ \pi\right)$

This is domain of (1).

Squaring (1) we get,

$\small\longrightarrow1+2\sin\theta\cos\theta=15\sin^2\theta\cos^2\theta$

Taking $x=\sin\theta\cos\theta=\dfrac{\sin(2\theta)}{2},$

$\small\longrightarrow15x^2-2x-1=0$

$\small\longrightarrow(3x-1)(5x+1)=0$

$\small\text{ \Longrightarrow x\in\left\{\dfrac{1}{3},\ -\dfrac{1}{5}\right\} }$

$\small\text{ \longrightarrow\dfrac{\sin(2\theta)}{2}\in\left\{\dfrac{1}{3},\ -\dfrac{1}{5}\right\} }$

$\small\text{ \longrightarrow\sin(2\theta)\in\left\{\dfrac{2}{3},\ -\dfrac{2}{5}\right\} }$

Note that $2\theta\in(0,\ 2\pi).$ If $\sin(2\theta)=\dfrac{2}{3}>0$ then 2θ has two values, one in first quadrant and other in second quadrant. Then,

$\small\longrightarrow2\theta_1=\sin^{-1}\left(\dfrac{2}{3}\right),\quad2\theta_2=\pi-\sin^{-1}\left(\dfrac{2}{3}\right)$

$\small\longrightarrow\theta_1=\dfrac{1}{2}\sin^{-1}\left(\dfrac{2}{3}\right),\quad\theta_2=\dfrac{\pi}{2}-\dfrac{1}{2}\sin^{-1}\left(\dfrac{2}{3}\right)$

Here $\sin^{-1}\left(\dfrac{2}{3}\right)\in\left(0,\ \dfrac{\pi}{2}\right)$ followed by,

• $\small\theta_1\in\left(0,\ \dfrac{\pi}{4}\right)$

• $\small\theta_2\in\left(\dfrac{\pi}{4},\ \dfrac{\pi}{2}\right)$

Both are roots since they lie in the domain.

If $\sin(2\theta)=-\dfrac{2}{5}<0$ then 2θ has two values, one in third quadrant and other in fourth quadrant. Then,

$\small\longrightarrow2\theta_3=\pi+\sin^{-1}\left(\dfrac{2}{5}\right),\quad2\theta_4=2\pi-\sin^{-1}\left(\dfrac{2}{5}\right)$

$\small\longrightarrow\theta_3=\dfrac{\pi}{2}+\dfrac{1}{2}\sin^{-1}\left(\dfrac{2}{5}\right),\quad\theta_4=\pi-\dfrac{1}{2}\sin^{-1}\left(\dfrac{2}{5}\right)$

Here $\sin^{-1}\left(\dfrac{2}{5}\right)\in\left(0,\ \dfrac{\pi}{2}\right)$ followed by,

• $\small\theta_3\in\left(\dfrac{\pi}{2},\ \dfrac{3\pi}{4}\right)$

• $\small\theta_4\in\left(\dfrac{3\pi}{4},\ \pi\right)$

Here θ₃ does not lie in domain so it's not a root and θ₄ lies in domain so it's a root.

Hence there are 3 roots.

Answer 2

Answer:

There are 3 real roots of the equation sec(theta )+cosec(theta)=sqr(15) lying between 0 to π

Hope it helps!!!

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