Question

prove that Cos A by 1 minus tan a minus sin square A by Cos A minus sin a=sin a +cos a​

Answer 1

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$\mathtt{\huge{\underline{\red{Answer\: :}}}}$

$\: \large \green{ \bold{ \: step \: by \: step \: explaination}}$

$\quad\bullet\bold{\underline{Given:}}$

$\bullet\bf{\dfrac{CosA}{1-TanA}+\dfrac{sinA^2}{sinA-cosA}=SinA+CosA}$

$\Large\bf{\underline{LHS:\to}}$

$\: \large \implies\bf{ \frac{cosx}{1 - tanx} + \frac{ {sinx}^{2} }{sinx - cosx} }$

$\: \implies \displaystyle \sf{ \frac{cosx}{1 - \frac{sinx}{cosx} } + \frac{ {sinx}^{2} }{sinx - cosx} }$

$\: \implies \displaystyle \sf \: \frac{ {cosx}^{2} }{cosx - sinx} + \frac{ {sin}^{2}x }{sinx - cosx}$

$\: \implies \displaystyle \sf \: \frac{ - {cos}^{2}x }{sinx - cosx} + \frac{ {sin}^{2}x }{sinx - cosx}$

$\: \implies \displaystyle \sf \: \frac{ {sin}^{2} x - {cos}^{2}x }{sinx - cosx}$

$\: \large \ \pink {\bf{ \underline{by \: using \: property \: {a}^{2} - {b}^{2} = (a - b)(a + b)}}}$

$\: \implies \displaystyle \sf \: \frac{(sinx - cosx)(sinx + cosx)}{(sinx - cosx)}$

$\: \implies \displaystyle \sf \: \frac{ \cancel{(sinx - cosx)}(sinx + cosx)}{ \cancel{sinx - cosx)}}$

$\: \implies \sf \: sinx \: + cosx \: \: \: is \: the \: answer$

$\therefore\large\bf{\underline{LHS=RHS}}$

$\\ \therefore \boxed{ \bf{ \frac{cosx}{1 - tanx} + \frac{ {sin}^{2}x }{sinx - cosx} = sinx + cosx}}$

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