Question

Prove:-
This one is a tuff proof

Answer 1

\[\frac{ tanx+sinx }{ sinx-tanx }=secx*\frac{ 1+cosx }{ 1-secx }\]
(secx is not equal to 1)

Answer 2

It is probably a good thing to first replace tanx by sinx/cosx.

Answer 3

plz help me with starting steps i'll do the rest

Answer 4

OK, so if I do what I suggested:\[LHS=\frac{ \frac{ \sin x }{ \cos x } +\sin x}{ \sin x-\frac{ \sin x }{ \cos x } }\].
Now try to write it as one fraction, or multiply numerator and denominator with cosx.

Answer 5

Then I would leave the LHS alone and take up the RHS:\[\sec x \cdot \frac{1+\cos x}{1-\sec x}=\frac{1}{\cos x}\cdot \frac{1+\cos x}{1-\frac{1}{\cos x}}\]
Now write the denominator of the second fraction as one and then simplify.

Answer 6

I am stuck here:-
\[\frac{ 1+cosx }{ cosx-1 }\]

Answer 7

That is OK! It's fine. That is why I said: leave it and move on to the right-hand side.
After a few steps you will get the same.

Answer 8

but I want to prove it using a single side

Answer 9

now divide numerator and deno by cos(x)

Answer 10

You can do it by using a single side afterwards: once you have proven it, you can trace the steps back.

Answer 11

I Multiply and divided by conjugate but got the same result

Answer 12

Conjugate not necessary!

Answer 13

using conjugates will give you cos^2(x)
dont do that

Answer 14

in your simplification where you got cos(x) both top and bottom, divide num and deno by cos(x)

Answer 15

ok wait

Answer 16

I got the same result again

Answer 17

It is really much simpler to work from both sides and arrive at the same result.
Afterwards, if you wish, you can reverse the steps done at the RHS, to see how it would be done only from left to right.

It is simpler, because you don't have to get brilliant ideas :)
Once you worked both sides, you arrive at the same result, infact the result you had with the LHS.

Answer 18

i Multiply and divided by cosx i got me back to 1+cosx/1-cosx

Answer 19

ok @ZeHanz i'll try

Answer 20

Here is a start: \[RHS=\frac{1}{\cos x}\cdot \frac{1+\cos x}{1-\frac{1}{\cos x}}=\frac{1}{\cos x}\cdot \frac{1+\cos x}{\frac{\cos x -1 }{\cos x}}\]

Answer 21

Now just one more step and you're done!

Answer 22

might i make a small suggestion? replace \(\cos(x)\) by \(a\) and \(\sin(x)\) by \(b\) and do the algebra that way. it is often more clear, because most of it is basic algebra

Answer 23

@satellite73: Good plan, although I think @koli123able can do it either way!

Answer 24

in fact in this case there is no sine, so you will just have two rational expressions in \(a\)

Answer 25

i proved it i reversed the step i did in R.H.S

Answer 26

Nice work!
You can use this trick a lot.
After tidying up afterwards, it seems you had some brilliant ideas to get from left to right, but all you did was work from both ways. There is nothing wrong with that!