Question

Prove that 2 divided by cos square theta minus one by cos 4 theta minus 2 by sin square theta + 1 by sin 4 theta equals to cos 4 theta minus tan 4 theta.
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Answer 1

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$\frac{2}{Cos^{2}\: \theta}- \frac{1}{Cos^{4}\: \theta} - \frac{2}{Sin^{2}\: \theta}+\frac{1}{Sin^{4}\: \theta} = Cos^{4}\: \theta - tan^{4}\: \theta$

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$\large\underline{LHS=}$

=$\frac{2}{Cos^{2}\: \theta}- \frac{1}{Cos^{4}\: \theta} - \frac{2}{Sin^{2}\: \theta}+\frac{1}{Sin^{4}\: \theta}$

we have,

$Sec\: \theta = \frac{1}{Cos\: \theta}$ &
$Cosec\: \theta = \frac{1}{Sin\: \theta}$ &

So

=$2Sec^{2}\: \theta - Sec^{4}\: \theta - 2Cosec^{2}\: \theta+Cosec^{4}\: \theta$

=$2Sec^{2}\: \theta - (Sec^{2}\: \theta)^{2} - 2Cosec^{2}\: \theta+(Cosec^{2}\: \theta)^{2}$

We have,
$Sec^{2}\: \theta = 1 + tan^{2}\: \theta$
$Cosec^{2}\: \theta = 1 + Cot^{2}\: \theta$

=$2(1 + tan^{2}\: \theta) - (1 + tan^{2}\: \theta)^{2} - 2(1 + Cot^{2}\: \theta) + (1 + Cot^{2}\: \theta)^{2}$

=$2 + 2tan^{2}\: \theta - (1 + tan^{4}\: \theta + 2tan^{2}\: \theta) - 2 - 2Cot^{2}\: \theta + (1 + Cot^{4}\: \theta+2Cot^{2}\: \theta)$

= $2 + 2tan^{2}\: \theta - 1 - tan^{4}\: \theta - 2tan^{2}\: \theta - 2 - 2Cot^{2}\: \theta + 1 + Cot^{4}\: \theta+2Cot^{2}\: \theta$

=$Cot^{4}\: \theta - tan^{4}\: \theta$

$\large\underline{=RHS}$

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Answer 2

Step-by-step explanation:

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