Question

Using the identity cosec²A = 1 + cot²A, prove that
(cosA-sinA+1)/(cosA+sinA-1) = cosecA + cotA

Answer 1

\[\frac{ \frac{cosa - sina +1 }{sina}}{\frac{cosa + sina -1}{sina}}\]

Answer 2

This Gives you \[\frac{cota -1 + coseca}{cota +1 - coseca }\]

Answer 3

Then \[ 1 = cot^2a - cosec^2a \]

Answer 4

after this...??

Answer 5

Sorry ...Typo

It's \[ 1 = cosec^2a - cot^2a \]

So the Whole Equation Becomes \[\frac{(cota + coseca ) ( 1 +cota - coseca)}{1 + cota - coseca }\]

Answer 6

Then You Get \[cota + coseca \]

Answer 7

What I did was to replace 1 in Numerator ... through the Identity

Answer 8

Harkirat Was it Helpful ?

Answer 9

next???

Answer 10

Well take \[cota + coseca \]Common and You will see Denominator cancels out ..
Giving you the answer

Answer 11

sorry, my internet went funny.....

thanks ishaan for yr help........
medal for u/.......

Answer 12

Thanks

Answer 13

Can u help me with one more...?

Answer 14

Yea Sure :D Post it

Answer 15

i have proved this by some manipulation but could not proceed from LHS to RHS.....

Prove that

1 1 1 1
---------- - ------ = ------ - -----------
secA + tanA cosA cosA secA - tanA

Answer 16

Ok

Answer 17

\[\frac{1}{seca - tana } + \frac {1}{tana + seca} = \frac{1}{cosa} + \frac{1}{cosa}\]

Answer 18

yes, i have used this manipulation to solve it..........

see if you can proceed from LHS and reach RHS without this manipulation......

Answer 19

So I should Solve it without manipulation ! agh No

Answer 20

Ok I'm ON it

Answer 21

I guess Your Internet went Funny Again :D lol

Answer 22

yeah, it is acting funny today.......

Answer 23

Hell No....can't do it without manipulation ! NNNoooo lol :D sorry it becomes something too complicated to operate

Answer 24

ok, thanks......☺

I'll try myself later, right now I am very tired and don't want to tax my brain.....lol !!!