Question

If tan + sin = m; tan - sin = n, then show that m2
-n
2= 4√mn

Answer 1

GIVEN :-

$\bullet\sf \ tan\theta+sin\theta= m \\\\\bullet \ tan\theta-sin\theta= n$

TO PROVE :-

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

SOLUTION :-

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

$\longrightarrow\sf m^2-n^2= (tan\theta+sin\theta)^2-(tan\theta-sin\theta)^2\\\\\longrightarrow m^2-n^2 = tan^2\theta+sin^2\theta+2tan\theta sin\theta -(tan^2\theta+sin^2\theta-2tan\theta sin\theta ) \\\\\longrightarrow m^2-n^2= tan^2\theta+sin^2\theta+2tan\theta sin\theta-tan^2\theta-sin^2\theta+2tan\theta sin\theta\\\\\longrightarrow m^2-n^2= 4tan\theta sin\theta$

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

$\longrightarrow \sf 4\sqrt{mn} = 4 \sqrt{(tan\theta+sin\theta)(tan\theta-sin\theta) } \\\\\longrightarrow 4 \sqrt{mn} = 4 \sqrt{tan^2\theta-sin^2\theta} \\\\\longrightarrow 4\sqrt{mn} = 4\sqrt{\dfrac{sin^2\theta}{cos^2\theta}-sin^2\theta}\\\\\longrightarrow 4\sqrt{mn} = 4 \sqrt{\dfrac{sin^2\theta-sin^2\theta cos^2\theta}{cos^2\theta} } \\\\\longrightarrow 4\sqrt{mn} = 4\sqrt{\dfrac{sin^2\theta(1-cos^2\theta)}{cos^2\theta}}$

$\bullet\bf \ 1-cos^2\theta = sin^2 \theta$

$\longrightarrow \sf 4\sqrt{mn} = 4 \sqrt{sin^2 \theta \times \dfrac{sin^2\theta}{cos^2\theta}} \\\\\longrightarrow 4\sqrt{mn} = 4\sqrt{sin^2\theta tan^2\theta} \\\\\longrightarrow 4\sqrt{mn} = 4 tan\theta sin\theta$

∴ L.H.S = R.H.S

Hence proved.