Solve this Identity of inverse trignometry.?

Question

every cot has inverse 1 power on it.its an identity of inverse trignometry....plzz prove it....i really need help.


cot x +cot y =cot (xy-1/x+y)

Answer 1

cotx+coty = cosx/sinx + cosy/siny
=( cosxsiny +sinxcosy) / (sinxsiny)
= sin(x+y) / (sinxsiny)
= tan(x+y) cos(x+y)/ (sinxsiny)
= tan(x+y) ( cosxcosy -sinx siny) / (sinxsiny)
=tan(x+y) (cotx coty - 1)
= (cotxcoty -1)/cot(x+y)

Answer 2

Let's start with the trig identity
tan (A+B) = (tan A + tan B) / (1 - tan A tan B)
Now tan X = 1/cot X, so we get
cot (A+B) = (1 - tan A tan B) / (tan A + tan B)
and dividing top and bottom by (tan A tan B) gives
cot (A+B) = (cot A cot B - 1) / (cot B + cot A)
and so
A + B = cot^-1((cot A cot B - 1) / (cot A + cot B))

Now let A = cot^-1 x, B = cot^-1 y; then we get
cot^-1 x + cot^-1 y = cot^-1 ((xy - 1) / (x + y))
as required.

Answer 3

cot ^-1 x +cot ^-1 y =cot ^ -1 (xy-1/x+y)

let cot ^-1 x = a
and cot ^-1 y = b
then x = cot a
y = cot b
we know that the normal trignomteric identity
cot(a+b) = (cot a cot b - 1)/(cot a + cot b)
so
cot(a+b) = xy -1 /x+y
or a+b = cot ^ -1 (xy -1 /x+y)
or cot ^-1 x +cot ^-1 y =cot ^ -1 (xy-1/x+y)

Answer 4

cot ^-1 x +cot ^-1 y =cot ^ -1 (xy-1/x+y)

let cot ^-1 x =A
and cot ^-1 y =B
then x = cotA
and y = cotB

Lets start from R.H.S:
======================
Cot^-1[xy -1 /x+y]
=Cot^-1[(CotA*CotB -1)/CotA+CotB]
=Cot^-1Cot(A+B) [standard trigonometric identity Cot(A+B)=(CotA*CotB -1)/CotA+CotB]
=A+B
=cot ^-1 x +cot ^-1 y
=L.H.S(proved)