Question

Given $a={\sec }^2\alpha ,b={cosec}^2\alpha ,and,c=\frac{1}{1-{\sin }^2\alpha {\cos }^2\alpha }.$ If $\frac{\tan 3\alpha }{\tan \alpha }=\frac{\lambda a}{b},then\lambda =$

• A. $\frac{3b-4}{4-3a}$
• B. $\frac{4-3a}{3b-4}$
• C. $\frac{3b+4}{4+3a}$
• D. $\frac{4+3a}{3b+4}$

Answer 1

The correct option is A $\frac{3b-4}{4-3a}$

$\tan 3\alpha =\frac{\sin 3\alpha }{\cos 3\alpha }=\frac{3\sin \alpha -4{\sin }^3\alpha }{4{\cos }^3\alpha -3\cos \alpha }=\tan \alpha \left(\frac{3-4{\sin }^2\alpha }{4{\cos }^2\alpha -3}\right)$
$=\tan \alpha \frac{(3-\frac{4}{b})}{(\frac{4}{a}-3)}=\tan \alpha \left(\frac{3b-4}{4-3a}\right)\frac{a}{b}$
Hence $\lambda =\frac{3b-4}{4-3a}$