Question

Question below.

Answer 1

\[\frac{ Cos(A) }{ 1 - \tan(A) } -\frac{ \sin^2(A) }{ \cos(A) - \sin(A) } = \cos(A) + \sin(A)\]

Answer 2

@IrishBoy123

Answer 3

multiply the very first term by cos A

Answer 4

top and bottom....of cour

Answer 5

of courSE

Answer 6

cos^2(A) over 1 - tan(A)

Answer 7

bottom as well....

Answer 8

\[\frac{ Cos(A) }{ 1 - \tan(A) } * \frac{ \cos(A) }{ \cos(A)}\]

Answer 9

cos^2(A) over (1-tan(A))*cos(A)

Answer 10

\[\frac{ Cos(A) }{ 1 - \tan(A) } * \frac{ \cos(A) }{ \cos(A)} = \frac{ Cos^2(A) }{ \cos A - \sin A }\]

then put that into your original and note that \(x^2 - y^2 = (x+y)(x-y)\)

that's pretty "hinty"...

Answer 11

what did you do at the buttom..??

Answer 12

hi
first convert tan A in terms of sinA n cos A

Answer 13

sin/cos

Answer 14

so in LHS
your denominator would be ?
for the first term

Answer 15

cos^2(A) over cos(A) - sin(A)

Answer 16

perfect!
now you see that the denominator is same in LHS

Answer 17

\[\rm~LHS= \frac{ \cos^2A }{cosA-sinA }-\frac{ \sin^2A }{cosA-sinA }\]

Answer 18

so now its cos^2A-sin^2A can be written as together in the numerator

Answer 19

thats 1

Answer 20

no,no sorry

Answer 21

nope!

Answer 22

(cosA + sinA)*(cosA - sinA)

Answer 23

got it !:)

Answer 24

perfect!

Answer 25

well @IrishBoy123 i hope u did not mind me helping him :)

Answer 26

@rvc
not at all mate! in fact, i appreciated it.