The expression
(1+sec2A)(1+sec4A)(1+sec8A) is equal to -

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Solution

$We have (1 + \sec 2 A) (1 + \sec 4 A) (1 + \sec 8 A) = (1 + \frac{1}{\cos 2 A}) (1 + \frac{1}{\cos 4 A}) (1 + \frac{1}{\cos 8 A}) = \frac{(\cos 2 A + 1)}{\cos 2 A} \times \frac{(\cos 4 A + 1)}{\cos 4 A} \times \frac{(\cos 8 A + 1)}{\cos 8 A} = \frac{2 (\cos 2 A + 1)}{2 \cos 2 A} \times \frac{2 (\cos 4 A + 1)}{2 \cos 4 A} \times \frac{2 (\cos 8 A + 1)}{2 \cos 8 A} = \frac{2 (2 \cos^{2} A)}{2 \cos 2 A} \times \frac{2 (2 \cos^{2} 2 A)}{2 \cos 4 A} \times \frac{2 (2 \cos^{2} 4 A)}{2 \cos 8 A} = \frac{8 \cos^{2} A \cos^{2} 2 A \cos^{2} 4 A}{\cos 2 A \cos 4 A \cos 8 A} = \frac{8 \cos^{2} A \cos 2 A \cos 4 A}{\cos 8 A} = \frac{8 sinA \cos^{2} A \cos 2 A \cos 4 A}{sinA \cos 8 A} = \frac{4 \times 2 sinA cosA cosA \cos 2 A \cos 4 A}{sinA \cos 8 A} = \frac{2 \times 2 \sin 2 A \cos 2 A \cos 4 A cosA}{sinA \cos 8 A} = \frac{2 \sin 4 A \cos 4 A cosA}{sinA \cos 8 A} = \frac{\sin 8 A cosA}{sinA \cos 8 A} = \tan 8 A cotA Note : \cos 2 x = 2 \cos^{2} x - 1 \sin 2 x = 2 sinx cosx$

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