Question

sinA/1-cotA+cosA/1-tanA=sinA+cotA​

Answer 1

Given :

To prove that :

$\\ \bullet \bf \frac{sinx}{1 - cotx} + \frac{cosx}{1 - tanx} = sinx + cosx$

To Find :

$\\ \bullet \bf \: \frac{sinx}{1 - cotx} + \frac{cosx}{1 - tanx} = sinx + cosx$

Formulas :

$\\ \bullet \bf \: tanx = \frac{sinx}{cosx}$

$\\ \bullet \bf \: cotx = \frac{1}{tanx} = \frac{cosx}{sinx}$

$\\ \bullet \bf \: secx = \frac{1}{cosx}$

Solution :

$\\ \bullet \bf \: \frac{cosx}{1 - tanx} + \frac{sinx}{1 - cotx} = sinx + cosx$

Taking L.H.S :

$\\ \implies \sf \: \frac{cosx}{1 - tanx} + \frac{sinx}{1 - cotx}$

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

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

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

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

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

RHS,

$\\ \implies \boxed{ \sf{ \frac{cosx}{1 - tanx} + \frac{sinx}{1 - cotx} = sinx + cosx}}$

Answer 2

$\frac{cosA}{1 - tanA} + \frac{sinA}{1 - cotA} \\ \\ = \frac{ {cos}^{2}A }{cosA - sinA} + \frac{ {sin}^{2}A }{sinA - cosA} \\ \\ = \frac{ {cos}^{2} A}{cosA - sinA} - \frac{ {sin}^{2}A }{cosA- sinA} \\ \\ = \frac{ {cos}^{2}A - {sin}^{2} A}{cosA - sinA} \\ \\ = cosA + sinA$