Question

Solve $\frac{1-\sec 8\theta }{1-\sec 4\theta }$

• A. $\sin 8\theta .\cos 2\theta$
• B. $\tan 8\theta .\cot 2\theta$
• C. $\sec 8\theta .\cot 2\theta$
• D. $\tan 8\theta .\tan 2\theta$

Answer 1

The correct option is B $\tan 8\theta .\cot 2\theta$

LHS

$\frac{(\sec 8\theta -1)}{(\sec 4\theta -1)}$

$=\frac{\frac{1}{\cos 8\theta }-1}{(\frac{1}{\cos 4\theta }-1)}=\frac{\frac{1-\cos 8\theta }{\cos 8\theta }}{\frac{1-\cos 4\theta }{\cos 4\theta }}$

$=\frac{2{\sin }^{2}4\theta }{\cos 8\theta }\times \frac{\cos 4\theta }{2{\sin }^2\theta }[\because 1-\cos 2\theta =2{\sin }^2\theta ]$

$=\frac{{2\left(2\sin 2\theta \cos 2\theta \right)}^2}{\cos 8\theta }\times \frac{\cos 4\theta }{2{\sin }^{2}2\theta }$

$=\frac{4{\cos }^{2}2\theta \cos 4\theta }{\cos 8\theta }$

RHS

$\tan 8\theta \tan 2\theta$

$=\frac{\sin 8\theta }{\cos 8\theta }\times \frac{\cos 2\theta }{\sin 2\theta }$

$=\frac{2\sin 4\theta \cos 4\theta }{\cos 8\theta }\times \frac{\cos 2\theta }{\sin 2\theta }$

$=\frac{2.2\sin 2\theta \cos 2\theta .\cos 4\theta }{\cos 8\theta }\times \frac{\cos 2\theta }{\sin 2\theta }$

$=\frac{{4\cos }^{2}2\theta \cos 4\theta }{\cos 8\theta }=LHS$