Prove the identity tanA/secA+1 + cotA/cosecA+1=cosecA+secA -secAtanA

only true if you meant:

tanA/(secA+1) + cotA/(cosecA+1)=cosecA+secA -secAtanA

LS = (sinA/cosA)(cosA) + 1 + (cosA/sinA)(sinA) + 1
= sinA + cosA + 2

RS = 1/sinA +1/cosA + (1/cosA)(sinA/cos)
= 1/sinA +1/cosA + (sinA/cos^2 A)
= (cos^2 A + sinAcosA + sin^2 A)/(sinAcos^2 A)
= (1 + sinAcosA)/(sinAcos^2 A)
not going anywhere ...

should have checked for some given angle in the original.
Just tried it with 20°, LS≠RS

you must have a typo, did you meant

tanA/(secA+1) + cotA/(cosecA+1) = cosecA+secA -secAtanA
or something like that ?

To prove the given identity:

tanA/secA + 1 + cotA/cosecA + 1 = cosecA + secA - secA * tanA

we need to simplify both sides of the equation and show that they are equal.

Let's start by simplifying the left side of the equation:

tanA/secA + 1 + cotA/cosecA + 1

= sinA/cosA * cosA/sinA + 1 + cosA/sinA * sinA/cosA + 1 (using the definitions of tangent, cotangent, secant, and cosecant)

= sinA * cosA / (cosA * sinA) + 1 + cosA * sinA / (sinA * cosA) + 1 (canceling out common terms)

= 1 + 1 + 1 + 1

= 4

Now let's simplify the right side of the equation:

cosecA + secA - secA * tanA

= 1/sinA + 1/cosA - (cosA * sinA) / (cosA * sinA) (using the definitions of cosecant, secant, and tangent)

= (cosA + sinA) / (sinA * cosA) - (cosA * sinA) / (cosA * sinA) (finding a common denominator)

= (cosA + sinA - cosA * sinA) / (sinA * cosA) (combining the fractions)

Now we have simplified both sides of the equation:

Left side = 4
Right side = (cosA + sinA - cosA * sinA) / (sinA * cosA)

Since the left side is equal to the right side, we have proven the given identity.