Question

If $\cot \theta +\tan \theta =3$ and $1-{\cos }^2\theta -\alpha \cos \theta =0$ then

• A. $6{\alpha }^2\left(9-{\alpha }^2\right)=1$
• B. $6{\alpha }^2\left({\alpha }^2-9\right)=1$
• C. $9{\alpha }^2\left(6-{\alpha }^2\right)=1$
• D. $9{\alpha }^2\left({\alpha }^2-6\right)=1$

Answer 1

Answer: (C)

If $\cot \theta +\tan \theta =3$
$\frac{1}{\sin \theta \cos \theta }=3$
$3\sin \theta \cos \theta =1...(i)$
and $1-{\cos }^2\theta -\alpha \cos \theta =0$
$\Rightarrow {\sin }^2\theta =\alpha \cos \theta ...(ii)$
From Eqs. (i) and (ii), we get
${\sin }^3\theta =\frac{\alpha }{3}...(iii)$
Now, from Eq. (ii), we get
${\sin }^4\theta ={\alpha }^2{\cos }^2\theta$
$\Rightarrow {\sin }^4\theta ={\alpha }^2\left(1-{\sin }^2\theta \right)$
$\Rightarrow {\left(\frac{\alpha }{3}\right)}^{4/3}={\alpha }^2\left(1-{\left(\frac{\alpha }{3}\right)}^{2/3}\right)$
$[$ From Eq. (iii) $\sin \theta ={\left(\frac{\alpha }{3}\right)}^{1/3}]$
$\Rightarrow \frac{{\alpha }^4}{81}={\alpha }^6\left[1-{\left(\frac{\alpha }{3}\right)}^2-3{\left(\frac{\alpha }{3}\right)}^{2/3}\left(1-{\left(\frac{\alpha }{3}\right)}^{2/3}\right)\right]$
$\Rightarrow \frac{1}{81}={\alpha }^2\left[1-\frac{{\alpha }^2}{9}-3{\left(\frac{\alpha }{3}\right)}^{2/3}\frac{{\left(\frac{\alpha }{3}\right)}^{4/3}}{{\alpha }^2}\right]$
$[\because {\alpha }^2\left(1-{\left(\frac{\alpha }{3}\right)}^{2/3}={\left(\frac{\alpha }{3}\right)}^{4/3}\right]$
$\Rightarrow \frac{1}{81}={\alpha }^2\left[1-\frac{{\alpha }^2}{9}-\frac{1}{3}\right]$
$\Rightarrow \frac{1}{81}={\alpha }^2\left(\frac{2}{3}-\frac{{\alpha }^2}{9}\right)$
$\Rightarrow 9{\alpha }^2\left(6-{\alpha }^2\right)=1$