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) $
$\Rightarrow \left(\frac{\alpha}{3}\right)^{4 / 3}=\alpha^{2}\left(1-\left(\frac{\alpha}{3}\right)^{2 / 3}\right)$
$\left[\right.$ From Eq. (iii) $\left.\sin \theta=\left(\frac{\alpha}{3}\right)^{1 / 3}\right]$
$\Rightarrow \frac{\alpha^{4}}{81}=\alpha^{6}\left[1-\left(\frac{\alpha}{3}\right)^{2}-3\left(\frac{\alpha}{3}\right)^{2 / 3}\left(1-\left(\frac{\alpha}{3}\right)^{2 / 3}\right)\right]$
$\Rightarrow \frac{1}{81}=\alpha^{2}\left[1-\frac{\alpha^{2}}{9}-3\left(\frac{\alpha}{3}\right)^{2 / 3} \frac{\left(\frac{\alpha}{3}\right)^{4 / 3}}{\alpha^{2}}\right]$
$\left[\because \alpha^{2}\left(1-\left(\frac{\alpha}{3}\right)^{2 / 3}=\left(\frac{\alpha}{3}\right)^{4 / 3}\right]\right.$
$\Rightarrow \frac{1}{81}=\alpha^{2}\left[1-\frac{\alpha^{2}}{9}-\frac{1}{3}\right]$
$\Rightarrow \frac{1}{81}=\alpha^{2}\left(\frac{2}{3}-\frac{\alpha^{2}}{9}\right)$
$ \Rightarrow 9 \alpha^{2}\left(6-\alpha^{2}\right)=1$