(Tan-1((SinA)/(CosA))^2?

Question

Tan-1 - inverse tan
Solve for A?

Answer 1

Is there more to the problem? it's not equal to anything currently so you can't really solve for A
Anyway suppose it's equal to some value B

(Tan-1((SinA)/(CosA))^2 = B
well Sin(A)/Cos(A) = tan(A)

[tan-1(tan(A))]^2 = B

If you take the tangent of something (assuming a is not 90 or 270) and then take the inverse you're just going to get that angle... or in this case A

A^2 = B

A = +/-sqrt(B)

Answer 2

y = {Tan-1(SinA)(CosA)]}^2

Tan-1 = inverse tan.

Solve for A.

A = √y.

Answer 3

(Tan-1((SinA)/(CosA))^2
=(tan-1(tanA))^2
=(A)^2
=A^2 ANS ^-^

x= A^2
A=+-sqrt(x)
A can be all real number except pi/2 and 3pi/2
because at pi/2 and 3pi/2 cosA is zero
A = R - {pi/2,3pi/2}

Answer 4

{arctan[(sinA)/(cosA)]}² = {arctan[tanA]}² = A²

The expression simplifies to A², but there is nothing to determine what A is.