Question

Tan^-1(√1-sinx÷√1+sinx)

Answer 1

tan^(-1)(|cos(x)| / (1 + sin(x)))

Explanation:

√(1 - sin(x)) / √(1 + sin(x))

(√(1 - sin(x)) / √(1 + sin(x))) * (√(1 + sin(x)) / √(1 + sin(x)))

√((1 - sin(x))(1 + sin(x))) / (1 + sin(x))

√(1 - sin^2(x)) / (1 + sin(x))

Then, using the trigonometric identity sin^2(x) + cos^2(x) = 1, we can substitute cos^2(x) for 1 - sin^2(x):

√(cos^2(x)) / (1 + sin(x))

|cos(x)| / (1 + sin(x))

tan^(-1)(|cos(x)| / (1 + sin(x)))

*Note : the absolute value |cos(x)| is used to ensure the argument of the inverse tangent is always positive.