Question

$\huge\bigstar\huge\tt\underline\blue{QUESTION}\bigstar$


$\frac{1 + sec ( \alpha )}{sec (\alpha) } = \frac{sin {}^{2} ( \alpha )}{1 - cos( \alpha )}$
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Answer 1

Answer:

$</p><p></p><p>\frac{1}{sec^{2} \alpha } = cos {}^{2} \alphasec2α1=cos2α</p><p></p><p>cos^{4} \alpha + 2cos^{2} (1 - \cos {}^{2} \alpha )cos4α+2cos2(1−cos2α)</p><p></p><p>$

Hope it helps you

Answer 2

Answer:

$\huge\bigstar\huge\tt\underline\red{ANsWER}\bigstar$

$given -$

$\: L.H.S = \frac{1 + sec ( \alpha )}{sec (\alpha) } = \frac{sin {}^{2} ( \alpha )}{1 - cos( \alpha )} \\ \\ \frac{1 + sec ( \alpha )}{sec (\alpha) } = \frac{1 + \frac{1}{cos( \alpha )} }{ \frac{1}{cos( \alpha )} } \\ = \frac{ \frac{cos( \alpha ) + 1}{cos( \alpha )} }{ \frac{1}{cos( \alpha )} } = cos ( \alpha ) + 1 \\ = \frac{(1 - cos( \alpha ))(1 + cos( \alpha ))}{1 - cos( \alpha )} \\ = \frac{1 - cos {}^{2} ( \alpha )}{1 - cos( \alpha )} = \frac{sin {}^{2} ( \alpha )}{1 - cos( \alpha )} \\ = R.H.S$