Question

IF TAN THETA +SIN THETA =M AND TAN THETA -SINTHETA =N THEN PROVE M^2-N^2=4ROOTMN

Answer 1

Let A = theta

Given as m=tan A+sinA and n= tanA-sinA now squaring them and subtratcing it is given also,m2-n2= (tan A+sinA)2-( tanA-sinA)2 =4tanA.SinA1 mn= (tan A+sinA) (tanA-sinA) mn=tan2A-Sin2A mn= Sin2A(1/Cos2A-1) mn= Sin2A(1-Cos2A/Cos2A) mn= Sin2A(Sin2A/Cos2A) mn=Sin2A. tan2A sqrt(mn)=SinA.tanA2 from 1, m2-n2=4tanA.SinA from 2, m2-n2=4sqrt(mn) 

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Answer 2

adding both eqn. we get 2tantheta=M+N and 2sintheta=M-N
multiplying both
M^2-N^2=4tantheta×sintheta
MN=tan^2theta-sin^theta
MN=(sin^theta-cos^theta×sin^2theta)/cos^theta
MN=[sin^theta(1-cos^theta)]/cos^theta
MN=[sin^theta×sin^theta]/cos^theta
MN=tan^theta×sin^theta
root MN =tantheta×sintheta
from above we have
M^2-N^2=4tantheta×sintheta and
as root MN =tantheta×sintheta
M^2-N^2=4root MN