Question

prove that (Cosec theta-Cot theta)² = 1-Cos theta/1+Cos theta

Answer 1

Hii friend,

LHS = (Cosec@-Cot@)²

= (1/Sin@-Cos@/Sin@)² = (1-Cos@/Sin@)²

= (1-Cos@)²/Sin²@ = (1-Cos@)(1-Cos@)/(1-Cos@) [ Sin²@ = 1-Cos²@]

= (1-Cos@)(1-Cos@)/(1+Cos@)(1+Cos@) = (1-Cos@)/(1+Cos@) = RHS.


HOPE IT WILL HELP YOU.... :-)

Answer 2

Hi...☺

Here is your answer...✌
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★LHS

$=(cosec (\theta) - cot(\theta) )^{2} \\ \\ = ( \: \frac{1}{ \sin(\theta)} - \frac{ \cos(\theta) }{ \sin(\theta) } \: ) {}^{2} \\ \\ = ( \: \frac{1 - \cos(\theta) }{ \sin(\theta) } \: ) {}^{2} \\ \\ = \frac{( { 1- \cos(\theta)) }^{2} }{ \sin {}^{2} (\theta) } \\ \\ = \frac{( { 1- \cos(\theta)) }^{2} }{1 - { \cos^{2}(\theta) } } \\ \\ = \frac{(1 - \cos (\theta) ) {}^{2} }{(1 + \cos(\theta) )( 1- \cos(\theta) )} \\ \\ = \frac{1 - \cos(\theta) }{1 + \cos(\theta) }$

= RHS★

Hence Proved ⭐