If tan theta + sin theta =m and tan theta- sin theta = n prove that m square – n square = 4(mn)1/2
Theta is angle
Tan +theta
Plzzz solve this ques
If (tan theta+sin theta)=m and (tan theta -sin theta )=n.prove that (m^2-n^2)^2=16mn
If (tan theta+sin theta)=m and (tan theta -sin theta )=n.prove that (m^2-n^2)^2=16mn
Given, {tex}\\tan \\theta + \\sin \\theta = m{/tex} and {tex}\\tan \\theta – \\sin \\theta = n{/tex}L.H.S =\xa0{tex}m^{2}-n^{2}{/tex}{tex}=(tan\\theta+sin\\theta)^{2}-(tan\\theta-sin\\theta)^{2}{/tex}{tex}=tan^{2}\\theta+sin^{2}\\theta+2tan\\theta sin\\theta-[tan^{2}\\theta+sin^{2}\\theta-2tan\\theta sin\\theta]{/tex}{tex}=tan^{2}\\theta+sin^{2}\\theta+2tan\\theta sin\\theta-tan^{2}\\theta-sin^{2}\\theta+2tan\\theta sin\\theta{/tex}{tex}=4tan\\theta sin\\theta{/tex}R.H.S = 4{tex}\\sqrt{mn}{/tex}{tex}=4\\sqrt{(tan\\theta+sin\\theta)(tan\\theta-sin\\theta)}{/tex}{tex}=4\\sqrt{tan^{2}\\theta-sin^{2}\\theta}{/tex}{tex}=4\\sqrt{\\frac{sin^{2}\\theta}{cos^{2}\\theta}-sin^{2}\\theta}{/tex}{tex}=4\\sqrt{\\frac{sin^{2}\\theta-sin^{2}\\theta cos^{2}\\theta}{cos^{2}\\theta}}{/tex}{tex}=\\frac{4}{cos\\theta}\\sqrt{sin^{2}\\theta-sin^{2}\\theta cos^{2}\\theta}{/tex}{tex}=\\frac{4}{cos\\theta}\\sqrt{sin^{2}\\theta (1-cos^{2}\\theta)}{/tex}{tex}=\\frac{4}{cos\\theta} \\times sin\\theta \\times \\sqrt{sin^{2}\\theta}{/tex}{tex}=4tan\\theta sin\\theta{/tex}Hence, L.H.S = R.H.S