Question

plz answer this question

Answer 1

$\csc( \alpha ) + \cot( \alpha ) = m \\ ( \frac{1}{ \sin( \alpha ) }) + ( \frac{ \cos(?) }{ \sin( \alpha ) } ) = m \\ (1 + \cos( \alpha ) ) = m \sin( \alpha ) \\ square \: both \: sides \\ 1 + {cos}^{2} \alpha + 2cos \alpha = m {}^{2} (1 - {cos}^{2} \alpha ) \\ {cos}^{2} \alpha (1 + m {}^{2} ) + 2 \cos( \alpha ) + (1 - {m}^{2} ) = 0 \\ apply \: quadrtic \: formula \\ \cos( \alpha ) = \frac{ - 2( + - ) \sqrt{4 - 4(1 + {m}^{2})(1 - {m}^{2} ) } }{2(1 + m {}^{2}) } \\ \cos( \alpha ) = \frac{ - 2( + - )2 {m}^{2} }{2(1 + {m}^{2} )}$
cancel 2 from numerator and denominator
then u will get ans

Answer 2

this is the answer of your question
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