Question

Prove that `sin^(-1)(cosec  theta)=pi/2+i.log(cot  theta/2)`

Answer 1

we have to prove this `sin^(-1)(cosec  theta) = pi/2+i.log (cot  theta/2)`

`("cosec"  theta) = sin[pi/2+i.log (cot  theta/2)]`

R.H.S = `sin[pi/2+i.log(cot  theta/2)]`

`cos[i.log(cot  theta/2)]`  ..........`{sin(pi/2+x)=cosx}`

`=cos hlog(cot  theta/2)` ……….{ cos ix = cos hx }

`=1/2[e^(log(cot  theta/2)+e6(-log(cot  theta/2)]` ………..`{ cos hx = 1/2[e^x+e^(-x)] }`

`1/2[cot  theta/2+1/(cot  theta/2)]`

`1/2tan  theta/2[1+cot^2  theta/2]`

`1/2tan  theta/2[(sin^2  theta/2+cos^2  theta/2)/(sin^2  theta/2)]`  ..............`{tantheta=sintheta/costheta=1/cottheta}`

`1/2xx(sin  theta/2)/(cos  theta/2)xx1/(sin^2  theta/2)`

`1/sintheta`

= cosecθ = L.H.S

`therefore sin^(-1)(cosec  theta)=pi/2+i.log(cot  theta/2)`

Hence Proved.