Question

Prove that cos A/1-tan A+Sin A/1-cotA=cos A+sin A​

Answer 1

$answer$

L.H.S. = cos A/1 – tanA + sin A/1 – cot A

= cos A/1 – sin A/cos A + sin A/1 – cos A/sin A

= cos A/(cos A – sin A)/cosA + sin A/(sin A – cos A)/sin A

= cos2 A/cos A – sinA + sin2 A/sinA – cos A

= cos2 A – sin2A/(cos A – sin A)

= (cos A + sin A)(cos A – sin A)/(cos A – sin A)

= cos A + sin A

= R.H.S.

Answer 2

$\displaystyle\huge\red{\underline{\underline{Solution}}}$

FORMULA TO BE IMPLEMENTED

1.

$\displaystyle \: {{cos}^{2} A- {sin}^{2} A} = \displaystyle \: {(cos A + sin A) \: (cos A - sin A) }$

2.

$\displaystyle \: tan A = \frac{sin A}{cos A }$

3.

$\displaystyle \: cot A = \frac{cos A}{sin A }$

CALCULATION

$\displaystyle \: \frac{cos A}{1 - tan A} + \frac{sin A}{1 - cot A}$

$\: = \displaystyle \: \frac{cos A}{1 - \frac{sinA}{cos A} } + \frac{sin A}{1 - \frac{cosA}{sin A} }$

$= \displaystyle \: \frac{{cos}^{2} A}{cos A - sin A} + \frac{{sin}^{2} A}{sin A - cos A}$

$= \displaystyle \: \frac{{cos}^{2} A}{cos A - sin A} - \frac{{sin}^{2} A}{cos A - sin A}$

$= \displaystyle \: \frac{{cos}^{2} A- {sin}^{2} A}{cos A - sin A}$

$= \displaystyle \: \frac{(cos A + sin A) \: (cos A - sin A) }{cos A - sin A}$

$= cos A + sin A$