How do I prove this? Not solving for 0, but making a proof? (1+tanx)/(1+cotx) = (tanx)?

Answer 1

(1+tanx)/(1+cotx)
=(1+sinx/cosx)/(1+cosx/sinx)
=[(cosx+sinx)/cosx]/(sinx+cosx)/sinx]
=(cosx+sinx)(sinx)/(cosx+sinx)cosx)
=sinx/cosx
=tanx
hence proved

Answer 2

The key is to remember that tan x = sin x/cos x and that cot x = cos x/sin x. Then substitute this into the left hand side of the equation.

(1+sinx/cosx)/(1+cosx/sinx)

Convert the numerator and denominator into their own fractions, then reduce:

((cosx+sinx)/cosx) / ((sinx+cosx)/sinx) =
((cosx+sinx)/cosx) * (sinx/(sinx+cosx))

since sinx+cosx cancels out, you're left with

sinx/cosx = tanx

for all nonzero values of sinx, cosx, or sinx+cosx

Answer 3

1) Convert everything to sin and cos...
LHS = (1 + sinx/cosx) / (1+cosx/sinx) ...

2) Find a common denominator for everyting...
= ((cosx+sinx)/cosx) / ((sinx + cosx)/sinx) ...

3) Dividing by a/b is the same as multiplying by b/a, so...
= (cosx+sinx)*sinx / [cosx*(sinx + cosx)]

4) Cancel the common factors:
= sinx/cosx = tanx.

But I'd give placebo more than one thumb up if I could...

Answer 4

Here's another way to do this:
Use cot x = 1/tan x.
This gives
(1 + tan x) / (1 + 1/tan x).
Multiplying every term by tan x gives
(tan x + tan²x)/(1 + tan x).
Now factor tan x out of the numerator and cancel
to get your result. This is good everywhere
except when tan x = -1.

Answer 5

perchance i'll get you all started my solutions are many times for algebra its purely that perchance I see something (a million+tanx) / (a million+cotx) = (a million-tanx) / (cotx-a million) the right denominator a million/(cotx-a million)=a million/(-a million+cotx)= -a million/(a million-cotx) so (a million-tanx)/(cotx-a million)= -(a million-tanx)/(a million-cotx) delivers (a million+tanx) / (a million+cotx)= -(a million-tanx)/(a million-cotx) you are able to now make the denominators into 'massive difference of squares' left area (a million+tanx)/(a million+cotx)*(a million-cotx)/(a million-cotx)=(a million... top area -(a million-tanx)/(a million-cotx)*(a million+cotx)/(a million+cotx)= -(a million-tanx)(a million+cotx)/(a million-cotx^2) the (a million-cotx^2)'s cancel leaving (a million+tanx)(a million-cotx)= -(a million-tanx)(a million+cotx) perchance you are able to take it from there? sturdy success

Answer 6

Ans. (1+Tanx)/(1+Cotx)
= (1+Sinx/Cosx)/(1+Cosx/Sinx)
= (Cosx+Sinx/Cosx)(Sinx+Cosx/Sinx)
= 1/Cosx/Sinx -->Cutting (Cosx+Sinx)
= Sinx/Cosx
= Tanx.

Answer 7

1 + cot x = 1 + 1/ tan x = (tan x + 1) / tan x

or tan x = ( tan x + 1 ) /(1 +cot x)