Question

if cos A-sin A=m and cos A+sin A=n ,show that m2-n2/m2+n2=-2sin A .cos A=-2/tan A+cot A

Answer 1

i think the question is not clear can you attach the pic of that question

Ok as m=cos a-sin a and n= cos a+sin a

your sol is (cos a-sin a)^2 - (cos a + sin a)^2/ (cos a -sin a)^2 = [ cos^2 a + sin^2 a - 2 sin a cos a] -[ cos ^2 a + sin^2 a+ 2 sin a cos a]/

[cos^2 a+sin^2 a - 2 sin a cos a] + [cos^2 a+ sin^ 2 a + 2 sin a cos a]

= after all cancellation you will be remained with [ - 4 sin a cos a / 2]

as 4/2 is cancelled you remain with your first answer [ -2 sin a cos a]

Answer 2

Given $\cos A - \sin A=m, \cos A + \sin A=n$. Now,

$m^2-n^2=(\cos A - \sin A)^2-(\cos A + \sin A)^2\\ m^2-n^2=\cos^2 A + \sin^2 A-2\cos A \sin A-(\cos^2 A + \sin^2 A+2\cos A \sin A)\\ m^2-n^2=-4\cos A \sin A$

$m^2+n^2=(\cos A - \sin A)^2+(\cos A + \sin A)^2\\ m^2-n^2=\cos^2 A + \sin^2 A-2\cos A \sin A+(\cos^2 A + \sin^2 A+2\cos A \sin A)\\ m^2-n^2=1+1=2$

$LHS=\frac{m^2-n^2}{m^2+n^2} =\frac{-4\cos A \sin A}{2} =-2\cos A \sin A\\ =-2\frac{\cos A \sin A}{1} =-2\frac{\cos A \sin A}{\cos^2 A +\sin^2 A} \\ =-2\frac{1}{\frac{\cos^2 A +\sin^2 A}{\cos A \sin A}} =-\frac{2}{\tan A+\cot A} =RHS$

The proof is complete.