Question

Trig Question.

Answer 1

\[\frac{ cosec A }{\cot A + \tan A }=\cos A\]

Answer 2

@campbell_st

Answer 3

ok... so work on the denominator

cot = cos/sin and tan = sin/cos

so cot + tan = cos/sin + sin/cos

same as before... common denominator and multiply...

then cross multiply in the numerator

what would you get..?

Answer 4

cscAsecA

Answer 5

lets keep it as

\[\frac{1}{sonAcosA}\]

is that ok..?

Answer 6

oops sin not son

Answer 7

Okay.

Answer 8

so the numerator can be written as

cosecA = 1/sinA

is that ok..?

Answer 9

yes

Answer 10

1/sinA

Answer 11

careful flip and multiply

\[\frac{1}{sinA}\times \frac{sinAcosA}{1} = ?\]

Answer 12

oops I made a mistake earlier it should read

\[\frac{cosecA}{\frac1{sinAcosA}} = \frac{1}{sinA} \div \frac{1}{sinAcosA}\]

opps sorry about the typo

Answer 13

that will now allow you to get the required solution

Answer 14

I didn't get..

Answer 15

ok... so are you happy \[cosA + tanA = \frac{1}{sinAcosA}\]

Answer 16

I got that.

Answer 17

ok so the problem can be written as

\[\frac{cosecA}{costA + tanA} = \frac{cosecA}{\frac{1}{sinAcosA}}\]

is that ok..?

Answer 18

yes..

Answer 19

ok... I'll now change the numerator

\[\frac{cosecA}{\frac{1}{sinAcosA}} = \frac{\frac{1}{sinA}}{\frac{1}{sinAcosA}} ~~or~~ \frac{1}{sinA} \div \frac{1}{sinAcosA}\]

is that ok..?

Answer 20

yes..

Answer 21

ok... so flip the 2nd fraction and multiply... what do you get...?

Answer 22

\[\frac{1}{sinA} \times \frac{sinAcosA}{1} =?\]

Answer 23

cosA

Answer 24

great... so you can do this stuff...