Question

if tan Teeta+sin teeta= m and tan teeta-sin =n show that m square -n square = 4√mn

Answer 1

LHS = m^2−n^2

= (tantheta+sintheta)^2−(tantheta−sintheta)^2

= (tantheta+sintheta+tantheta−sintheta) (tantheta+sintheta−tantheta+sintheta)

= (2tantheta)(2sintheta)

= 4 * sintheta/costheta * sintheta

= 4 sin^2 theta/cos theta    --------------------- (1)

RHS = 4 root mn

= 4 root (tan theta+sintheta)(tantheta−sintheta)

= 4 root(tan^2 theta - sin^2 theta)

= 4 root (sin^2 theta/cos^2 theta - sin^2 theta)

= 4 root(sin ^2 theta - sin ^2 theta.cos^2 theta/cos^2 theta

= 4 root(sin ^2 theta(1-cos ^2 theta)/cos ^2 theta)

= 4 root (sin ^2 theta.sin^2 theta/cos ^2 theta)

= 4 sin ^2 theta/cos theta    -------------------- (2).


LHS = RHS.


hence proved.