Question

Plese solve the question

Answer 1

Solution

$l.h.s. = \frac{1 - \cos(\theta) + \sin(\theta) }{ \cos(\theta) + \sin(\theta) - 1} \\ = \frac{ \sin(\theta) + [(1 - \cos(\theta) \: ] }{ \sin(\theta) - [1 - \cos(\theta)]} \\ now \: multipying \: by \: [\sin(\theta) + (1 - \cos(\theta) ] \: both \: \\ numaerator \: and \: denomintor \: we \: get...... \\ = \frac{ [\sin(\theta) + (1 - \cos(\theta) )] {}^{2} }{[\sin(\theta) - (1 - \cos(\theta) )][(( \sin(\theta) + (1 - \cos(\theta) )] \: } \\ = \frac{ [\sin {}^{2} (\theta) + 2 \times \sin(\theta) (1 - \cos(\theta) + (1 - \cos(\theta) ) {}^{2} ] }{ [\sin {}^{2} (\theta) - (1 - \cos(\theta) ) {}^{2} ] } \\ now \: putting \: \sin {}^{2} (\theta) = [1 - \cos {}^{2} (\theta) ]\: \\ = \frac{(1 - \cos {}^{2} (\theta) ) + 2 \sin(\theta) (1 - \cos(\theta)) + (1 - \cos(\theta) ) {}^{2} }{(1 - \cos {}^{2} (\theta)) - (1 - \cos(\theta)) {}^{2} } \\ = \frac{(1 - \cos(\theta) )(1 + \cos(\theta)) +2 \sin(\theta) (1 - \cos(\theta)) + (1 - \cos(\theta) ) {}^{2} \: }{(1 - \cos(\theta) )(1 + \cos(\theta)) - (1 - \cos(\theta)) {}^{2} } \\ now \: taking \: common \: of \: (1 - \cos(\theta) ) \\ = \frac{(1 - \cos(\theta))[(1 + \cos(\theta) + 2 \sin(\theta) + 1 - \cos(\theta)] }{(1 - \cos(\theta) )[(1 + \cos(\theta) - 1 + \cos(\theta)] } \\ = \frac{2 + 2 \sin(\theta) }{2 \cos(\theta) } \\ = \frac{2[1 + \sin(\theta) ]}{2 \cos(\theta)} \\ = \frac{1 + \sin(\theta) }{ \cos(\theta) } \: \: = r.h.s.(proved)$

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Answer 2

Answer:

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Solution in the attachment ♡ ♡ ♡

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