Question

Show that sec h^-1(sin θ) =log cot (`theta/2` ).

Answer 1

LHS = sec h^-1(sin θ)
Let y = sec h^-1(sin θ)
sec hy = sin θ
`1/sintheta= 1/(secℎy)`
cos hy = cosec θ
y = cos h^-1(cosec θ)
but cos h^-1x = log |𝑥+ `sqrt(x2−1)`|
∴ y = log |cosec θ+ `sqrt(cosec2 θ−1)`|
∴ y = log |𝑐𝑜𝑠𝑒𝑐 𝜃+cot𝜃|
= log |`1/sintheta + costheta/sintheta` |
= log |`(1+costheta)/sintheta`|
= log |`(2 cos^2  theta/2)/(2cos  theta/2 sin  theta/2)`|
= log |`(cos  theta/2)/(sin  theta/2)`|
= log cot (`theta/2`).
= RHS
∴ LHS = RHS
Hence proved