`secx = 2=secpi/3` or `sec(2pi-pi/3)` `rArr x=pi/3` or `(2pi-pi/3)` `rArr x=pi/3`or `(5pi)/(3)` Therefore, principal solution of equation `secx=2` is `x=pi/3` or `(5pi)/(3)`. `secx=2` `1/(cosx)=2` `cosx=1/2=cos60^(@)` `cosx=cospi/3` and general solution is `x=2npi+-pi/3, n in Z` Ans.
(i) sinx=(sqrt(3))/(2) (ii) cosx=1 (iii) secx=sqrt(2)
If secx=sqrt2 and (3pi)/(2)lt x lt 2pi find the value of (1-tanx-cosecx)/(1-cotx-cosecx)
∫dx/(sinx + secx) = 1/(2√3) log |(√3 + s)/(√3 - s)| + tan^(-1)t then, find the value of s and t?
The integral int (sec^2x)/(secx+tanx)^(9/2)dx equals to (for some arbitrary constant K ) (A) -1/(secx+tanx)^(11/2){1/11-1/7(secx+tanx)^2}+K (B) 1/(secx+tanx)^(11/2){1/11-1/7(secx+tanx)^2}+K (C) -1/(secx+tanx)^(11/2){1/11+1/7(secx+tanx)^2}+K (D) 1/(secx+tanx)^(11/2){1/11+1/7(secx+tanx)^2}+K
int(secx+tanx)^(2)dx=
(secx.secy + tanx.tany)^(2)-(secx.tany + tanx.secy)^(2) in its simplest form, is
