if tanA+sinA=m and tanA-sinA=n then show (m2-n2)=16mn

On substituting the values of m and n in both L.H.S. as well as R.H.S. WE GET--->LHS--->m2-n2(tan@+sin@)^2 - (tan@-sin@)^2 now we can see is of the form a^2- b^2 .So it is = (a+b)(a-b)So we get--> (tan@+sin@+tan@-sin@)×(tan@+sin@-tan@+sin@) So tan@ and sin@ will be cancelled Then --->2tan@×2sin@=4sin^2@/cos^2@Now RHS 16 mn16 (tan@+sin@) (tan@-sin@)16(tan^2@-sin^2@)16(sin^2@/cos^2@-@-sin^2@)16(sin^2@-cos^2@@-sin^2@ divided by cos^2@)Now--->taking common sin^2@ we get16×sin^2@sin^2@/ cos^2@Finally [4 sin^2@/cos@]^2 PROVED.****But in the LHS part there must be be whole square of m2-n2 then only it is possible . THANK YOU. I THINK THIS SOLUTION WILL HELP YOU.