could you please prove this trigonometric equation:(secA+tanA)/(cosecA+cotA)=(cosecA-cotA)/(secA-tanA).
I have to prove LHS=RHS.

Generally the best approach to proving trig identities is to convert all the tangent, cotangent, secant, and cosecant functions to their sine and cosine equivalents, then simplify the resulting fractions while looking for obvious identities like sin^2x + cos^2x = 1, and indeed that approach works here. But for this one it's a bit less work to start by cross-multiplying; if the identity is correct then the numerator of the LHS times the denominator of the RHS equals the denominator of the LHS times the numerator of the RHS. Do that first, then convert to sine and cosine functions, and it works right out.

try the following to show the LS = RS
(secA + tanA) * (secA - tanA)
----------------------------------------
(cscA + cotA) * (secA - tanA)

sec^2 A - tan^2 A
--------------------------
(cscA + cotA) * (secA - tanA)

1
---------------------------
(cscA + cotA) * (secA - tanA)

1 1
-------------------- ----------------
(cscA + cotA) * (secA - tanA)

csc^2 A - cot^2 A 1
------------------------- ----------------
(cscA + cotA) * (secA - tanA)


(cscA + cotA) (cscA - cotA) 1
------------------------------------- ----------------
(cscA + cotA) * (secA - tanA)


(cscA - cotA) 1
-------------------- ----------------
* (secA - tanA)


(cscA - cotA) (cscA - cotA)
-------------------- = --------------------
(secA - tanA) (secA - tanA)

hope this is what you were looking for