if `tantheta+sectheta=l ` then `tantheta=`

Options

A. `(l^2+2)/2`
B. `(l^2+1)/l`
C. `(l^2+l)/2`
D. None

Answer

B

Explanation

Given, `tantheta+sectheta=l`
[multiply by `(sectheta-tantheta)` on numerator and denominator LHS]
`((tantheta+sectheta)(sectheta-tantheta))/((sectheta-tantheta))=l rArr (sec^(2)theta-tan^(2)theta)/(sectheta-tantheta)=l`
`1/(sectheta-tantheta)=l` `[therefore sec^(2)theta - tan^(2)theta=1]`
`rArr sectheta-tantheta=1/l` ...............(ii)
On adding Eqs. (i) and (ii), we get
`2sectheta= l + 1/l`
`sectheta=(l^(2)+1)/(2l)` Hence Proved.