Question

CotACosA / cotA+cosA=cotA-cosA/cotAcosA

Answer 1

cotAcosA/cotA+cosA
=[(cosA/sinA)cosA]/[(cosA/sinA)+cosA]
=(cos²A/sinA)/[(cosA+sinAcosA)/sinA]
=cos²A/cosA(1+sinA)
=cosA/(1+sinA)
=[cosA(1-sinA)]/[(1+sinA)(1-sinA)]
 [multiplying the numerator and the denominator with (1-sinA)]
=[cosA(1-sinA)]/(1-sin²A)
=(cosA-cosAsinA)/cos²A
=[(cosA-cosAsinA)/sinA]/(cos²A/sinA)
[dividing the numerator and the denominator by sinA]
=[(cosA/sinA)-(cosAsinA/sinA)]/(cosA/sinA)cosA
=(cotA-cosA)/cotAcosA (Proved)

Answer 2

To prove:

$\frac {cot A \times cos A}{ cot A + cos A} = \frac {cot A - cos A}{cot A \times cos A}$

Answer:

Given that

$LHS = \frac {cot A \times cos A}{ cot A + cos A}$

$= \frac {\frac {cos A}{ sin A} \times cos A} { \frac {cos A}{sin A} + cos A}$

$= \frac {\frac {cos ^{2}A}{sin A}} {\frac {cos A + sin A \times cos A}{sin A}}$

$= \frac {cos ^{2} A}{ cos A \times (1 + sin A)}$

$= \frac {cos A }{1+ sin A}$

Multiply the both denominator and numerator with (1-SinA)

$= \frac {cos A}{ 1+ sin A} \times \frac {1- sin A}{ 1- sin A}$

$= \frac {cos A \times (1 - sin A)}{1- sin ^{2} A}$

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

Divide both denominator and numerator with Sin A

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

$= \frac { \frac {cos A}{sin A} - \frac {cos A \times sin A}{sin A}} {\frac {cos A} { sin A} \times {cos A}}$

$\frac {cot A \times cos A}{cot A + cos A} = \frac {cot A - cos A}{cot A \times cos A}$

Hence proved.