Question

15. Prove the following identity, where the angles involved are acute
angles for which the expressions are defined.
1 - cos 0
(cosec 0 - cot 0)
1 + cos​

Answer 1

EXPLANATION.

$\sf \: \implies \: ( \csc( \theta) - \cot( \theta) ) {}^{2} = \dfrac{1 - \cos( \theta) }{1 + \cos( \theta) }$

$\sf \: \implies \: from \: LHS \: = ( \csc( \theta) - \cot( \theta)) {}^{2} \\ \\ \sf \: \implies \: it \: is \: in \: the \: form \: of \: (x - y) {}^{2} = {x}^{2} + {y}^{2} - 2xy \\ \\ \sf \: \implies \: ( \csc {}^{2} ( \theta) + \cot {}^{2} ( \theta) - 2 \csc( \theta) \cot( \theta) ) \\ \\ \sf \: \implies \: ( \frac{1}{ \sin {}^{2} ( \theta) } \: + \: \frac{ \cos {}^{2} ( \theta) }{ \sin {}^{2} ( \theta)} \: - \: 2 \times \frac{1}{ \sin( \theta) } \times \frac{ \cos( \theta) }{ \sin( \theta) } )$

$\sf \: \implies \: ( \dfrac{1 + \cos {}^{2} ( \theta) }{ \sin {}^{2} ( \theta) } \: - \: \dfrac{2 \cos( \theta) }{ \sin {}^{2} ( \theta) } ) \\ \\ \sf \: \implies \: ( \dfrac{1 + \cos {}^{2} ( \theta) - 2 \cos( \theta) }{ \sin {}^{2} ( \theta) } ) \\ \\ \sf \: \implies \: ( \frac{ \cos( \theta) - 1 }{ \sin( \theta) } ) {}^{2} = LHS \:$

$\sf \: \implies \: from \: RHS \: = \dfrac{1 - \cos( \theta) }{1 + \cos( \theta) } \\ \\\sf \: \implies \: rationalize \: the \: equation \: we \: get \\ \\ \sf \: \implies \: \frac{1 - \cos( \theta) }{1 + \cos( \theta) } \: \times \: \frac{1 - \cos( \theta) }{1 - \cos( \theta) } \\ \\ \sf \: \implies \: \frac{(1 - \cos( \theta)) {}^{2} }{1 - \cos {}^{2} ( \theta) }$

$\sf \: \implies \: \dfrac{1 - (\cos( \theta)) {}^{2} }{ \sin( \theta) {}^{2} } = ( \dfrac{(1 - \cos\theta) }{ \sin( \theta) } ) {}^{2}$

HENCE PROVED.