(1+sec 2theta)(1+sec 4theta) = | vedclass

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Question

(B) Given expression: $(1+\sec 2	heta)(1+\sec 4	heta) = (1+\frac{1}{\cos 2	heta})(1+\frac{1}{\cos 4	heta})$$= (\frac{\cos 2	heta + 1}{\cos 2	het

Vedclass · Accepted Answer

$\cot 	heta 	an 4	heta$

Vedclass · Answer

(B) Given expression: $(1+\sec 2	heta)(1+\sec 4	heta) = (1+\frac{1}{\cos 2	heta})(1+\frac{1}{\cos 4	heta})$$= (\frac{\cos 2	heta + 1}{\cos 2	heta})(\frac{\cos 4	heta + 1}{\cos 4	heta})$Using the identity $1 + \cos 2A = 2\cos^2 A$,we get:$= (\frac{2\cos^2 	heta}{\cos 2	heta})(\frac{2\cos^2 2	heta}{\cos 4	heta})$$= \frac{2\cos^2 	heta}{\cos 2	heta} \cdot \frac{2\cos^2 2	heta}{\cos 4	heta}$$= \frac{2\cos^2 	heta}{\cos 2	heta} \cdot \frac{2\cos 2	heta \cdot \cos 2	heta}{\cos 4	heta}$$= \frac{2\cos^2 	heta \cdot 2\cos 2	heta}{\cos 4	heta} = \frac{2\cos 	heta \cdot (2\cos 	heta \cos 2	heta)}{\cos 4	heta}$$= \frac{2\cos 	heta \cdot (\sin 3	heta - \sin 	heta)}{\cos 4	heta}$ (This is not the simplest path).Let us simplify differently: $\frac{2\cos^2 	heta}{\cos 2	heta} \cdot \frac{2\cos^2 2	heta}{\cos 4	heta} = \frac{\sin 2	heta \cdot \cos 	heta}{\sin 	heta \cdot \cos 2	heta} \cdot \frac{2\cos^2 2	heta}{\cos 4	heta} = \cot 	heta \cdot 	an 2	heta \cdot \frac{2\cos^2 2	heta}{\cos 4	heta}$$= \cot 	heta \cdot \frac{\sin 2	heta}{\cos 2	heta} \cdot \frac{2\cos^2 2	heta}{\cos 4	heta} = \cot 	heta \cdot \frac{\sin 4	heta}{\cos 4	heta} = \cot 	heta 	an 4	heta$.

$(1+\sec 2\theta)(1+\sec 4\theta) = $

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