If tan A + sec A = K, then what is the value of sin A ?

tan A + sec A = K      ----(1)

sec A - tan A = \(\frac{1}{K}\)    ----(2)

Adding to 1 & 2

2 secA = K + \(\frac{1}{K}\) = \(\frac{ K² + 1}{K}\) 

secA = \(\frac{ K² + 1}{2K}\)

Subtracting 1 & 2

2 tanA = K - \(\frac{1}{K}\)  = \(\frac{ K² - 1}{K}\)

tanA = \(\frac{ K² - 1}{2K}\)

\(\frac{tanA}{secA}\) = \(\frac{ K² - 1}{2K}\)  × \(\frac{ 2K}{K² + 1}\)

sinA = \(\frac{ K² - 1}{K² + 1}\)