Question

IF TAN A + SINA = M AND TANA -SINA =N PROVE TAHT M2-N2 = 4 rootmn​

Answer 1

Given :-

$\sf tanA+sinA = m$

$\sf tanA-sinA=n$

To Prove :-

$\sf m^2-n^2 = 4 \sqrt{mn}$

Solution :-

$\sf L.H.S = m^2-n^2$

       $= \sf (tanA+sinA)^2 - (tanA-sinA)^2$

       $\sf= tan^2A+sin^2A+2sinAtanA-(tan^2A+sin^2A-2sinAtanA )$

       $\sf= tan^2A+sin^2A+2sinAtanA-tan^2A-sin^2A+2sinAtanA$

       $\sf= 2sinAtanA+2sinAtanA$

       $\sf= 4sinAtanA$

$\sf R.H.S = 4 \sqrt{mn}$

       $= \sf 4 \sqrt{(tanA+sinA)(tanA-sinA)}$

       $\sf= 4 \sqrt{tan^2A-sin^2A}$

       $\sf= 4 \sqrt{\dfrac{sin^2A}{cos^2A}-sin^2A}$

       $\sf= 4 \sqrt{sin^2A \left[ \dfrac{1}{cos^2A} -1 \right] }$

      $\sf= 4 \sqrt{sin^2A \left[ \dfrac{1-cos^2A}{cos^2A} \right]}$

      $\sf= 4 \sqrt{sin^2A \times \dfrac{sin^2A}{cos^2A}}$

      $\sf= 4 \sqrt{sin^2Atan^2A}$

      $\sf= 4 sinA tanA$

L.H.S = R.H.S

Hence proved.