Question

If 3 cos²θ + 7 sin²θ = 4, show that cot θ = $\sqrt{3}$.
Or, if tan $\mathrm{\theta }=\frac{8}{15},\mathrm{evaluate}\frac{(2+2\mathrm{sin\theta })(1-\mathrm{sin\theta })}{(1+\mathrm{cos\theta })(2-2\mathrm{cos\theta })}.$

Answer 1

We have:
$3{\cos }^2\theta +7{\sin }^2\theta =4$
$\Rightarrow 3{\cos }^2\theta +3{\sin }^2\theta +4{\sin }^2\theta =4$
$\Rightarrow 3\left({\cos }^2\theta +{\sin }^2\theta \right)+4{\sin }^2\theta =4$ [ Since cos²θ + sin²θ = 1]
$\Rightarrow 3\times 1+4{\sin }^2\theta =4$
$\Rightarrow 4{\sin }^2\theta =1\Rightarrow {\sin }^2\theta =\frac{1}{4}$
∴ ${\cos }^2\theta =\left(1-{\sin }^2\theta \right)=\left(1-\frac{1}{4}\right)=\frac{3}{4}$
∴ ${\tan }^2\theta =\frac{{\sin }^2\theta }{{\cos }^2\theta }=\left(\frac{1}{4}\times \frac{4}{3}\right)=\frac{1}{3}$
$\Rightarrow \tan \theta =\frac{1}{\sqrt{3}}\Rightarrow cot\theta =\frac{1}{\tan \theta }=\sqrt{3}$

OR
We have:
$\frac{\left(2+2\sin \theta \right)\left(1-\sin \theta \right)}{\left(1+\cos \theta \right)\left(2-2\cos \theta \right)}=\frac{2\left(1+\sin \theta \right)\left(1-\sin \theta \right)}{2\left(1+\cos \theta \right)\left(1-\cos \theta \right)}=\frac{\left(1-{\sin }^2\theta \right)}{\left(1-{\cos }^2\theta \right)}=\frac{{\cos }^2\theta }{{\sin }^2\theta }=co{t}^2\theta =\left(cot\theta \right)$²
Given that,
$\tan \theta =\frac{8}{15}\Rightarrow cot\theta =\frac{15}{8}$
${\left(cot\theta \right)}^2={\left(\frac{15}{8}\right)}^2=\left(\frac{15}{8}\times \frac{15}{8}\right)=\frac{225}{64}$
Hence,
$\frac{\left(2+2\sin \theta \right)\left(1-\sin \theta \right)}{\left(1+\cos \theta \right)\left(2-2\cos \theta \right)}=\frac{225}{64}$