Question

Solve: $\frac{1-sec8\theta }{1-sec4\theta }$

• A. $sin8\theta \cos 2\theta$
• B. $tan8\theta \cot 2\theta$
• C. $sec8\theta \cot 2\theta$
• D. None of these

Answer 1

Answer: (B)

$tan8\theta \cot 2\theta$

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

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

$=\frac{-2{\sin }^{2}4\theta \times \cos 4\theta }{-2{\sin }^{2}2\theta \times \cos 8\theta }$..............$1-cos2\theta =2si{n}^2\theta$

$=\frac{(2\sin 4\theta .\cos 4\theta )\sin 4\theta }{\cos 8\theta .2{\sin }^{2}2\theta }$

$=\frac{\sin 8\theta }{\cos 8\theta }.\frac{2\sin 2\theta .\cos 2\theta }{2\sin 2\theta .\sin 2\theta }$.............$sin2\theta =2sin\theta .cos\theta$

$=\tan 8\theta .\cot 2\theta$

Hence, the answer is $\tan 8\theta .\cot 2\theta .$