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Question: If \[{\text{tan}}\theta {\text{ + sin}}\theta {\text{ = }}m\], \[{\text{tan}}\theta - {\text{sin}}\t...

If tanθ + sinθ = m{\text{tan}}\theta {\text{ + sin}}\theta {\text{ = }}m, tanθ−sinθ = n{\text{tan}}\theta - {\text{sin}}\theta {\text{ = }}n and m≠nm \ne n, then show that m2−n2=4mn{m^2} - {n^2} = 4\sqrt {mn} .

Explanation

Solution

Hint: -Here, we have the value of m and n so, we go through by simply putting the value of mm and nn to proceed further.

Given,  m=(tanθ+sinθ)\;m = \left( {{\text{tan}}\theta + {\text{sin}}\theta } \right) n=tanθ−sinθn = {\text{tan}}\theta - {\text{sin}}\theta
We need to show m2−n2=4mn{m^2} - {n^2} = 4\sqrt {mn}
TakingL.H.S{\text{L}}{\text{.H}}{\text{.S}}.
m2−n2=(tanθ+sinθ)2−(tanθ−sinθ)2{m^2} - {n^2} = {\left( {{\text{tan}}\theta + {\text{sin}}\theta } \right)^2} - {\left( {{\text{tan}}\theta - {\text{sin}}\theta } \right)^2}
Here, we know that (a+b)2=a2+2ab+b2{\left( {a + b} \right)^2} = {a^2} + 2ab + {b^2}and (a−b)2=a2−2ab+b2{\left( {a - b} \right)^2} = {a^2} - 2ab + {b^2}
By applying these formula,

=tan2θ+sin⁡2θ+2tan⁡θsin⁡θ−tan⁡2θ−sin⁡2θ+2tan⁡θsin⁡θ =4tanθ⋅sinθ  = {\text{ta}}{{\text{n}}^2}\theta + {\sin ^2}\theta + 2\tan \theta \sin \theta - {\tan ^2}\theta - {\sin ^2}\theta + 2\tan \theta \sin \theta \\\ = 4{\text{tan}}\theta \cdot {\text{sin}}\theta \\\

Now, taking R.H.S{\text{R}}{\text{.H}}{\text{.S}}.
4mn=4(tanθ+sinθ)(tanθ−sinθ)4\sqrt {mn} = 4\sqrt {\left( {{\text{tan}}\theta + {\text{sin}}\theta } \right)\left( {{\text{tan}}\theta - {\text{sin}}\theta } \right)}
And here we know that (a+b)(a−b)=a2−b2\left( {a + b} \right)\left( {a - b} \right) = {a^2} - {b^2}
=4tan⁡2θ−sin⁡2θ= 4\sqrt {{{\tan }^2}\theta - {{\sin }^2}\theta } (∵tan⁡2θ=sin⁡2θcos⁡2θ)\left( {\because {{\tan }^2}\theta = \frac{{{{\sin }^2}\theta }}{{{{\cos }^2}\theta }}} \right)

=4sin⁡2θcos⁡2θ−sin⁡2θ =4sin⁡2θ(1−cos⁡2θ)cos⁡2θ  = 4\sqrt {\frac{{{{\sin }^2}\theta }}{{{{\cos }^2}\theta }} - {{\sin }^2}\theta } \\\ = 4\sqrt {\frac{{{{\sin }^2}\theta (1 - {{\cos }^2}\theta )}}{{{{\cos }^2}\theta }}} \\\

=4tan2θ⋅sin2θ= 4\sqrt {{\text{ta}}{{\text{n}}^2}\theta \cdot {\text{si}}{{\text{n}}^2}\theta } (∵(1−cos⁡2θ)=sin⁡2θ)\left( {\because (1 - {{\cos }^2}\theta ) = {{\sin }^2}\theta } \right)
=4tan⁡θ⋅sin⁡θ= 4\tan \theta \cdot \sin \theta
Therefore, L.H.S = R.H.S.{\text{L}}{\text{.H}}{\text{.S}}{\text{ = R}}{\text{.H}}{\text{.S}}{\text{.}}
Hence proved.
Note:-Whenever we face such type of question try it solving by taking L.H.S{\text{L}}{\text{.H}}{\text{.S}}and R.H.S{\text{R}}{\text{.H}}{\text{.S}} and then equate it to proof the question.