# If \[{\text{tan}}\theta {\text{ + sin}}\theta {\text{ = }}m\], \[{\text{tan}}\theta - {\text{sin}}\theta {\text{ = }}n\] and \[m \ne n\], then show that \[{m^2} - {n^2} = 4\sqrt {mn} \].

Last updated date: 25th Mar 2023

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Answer

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

Given,$\;m = \left( {{\text{tan}}\theta + {\text{sin}}\theta } \right)$ $n = {\text{tan}}\theta - {\text{sin}}\theta $

We need to show ${m^2} - {n^2} = 4\sqrt {mn} $

Taking${\text{L}}{\text{.H}}{\text{.S}}$.

${m^2} - {n^2} = {\left( {{\text{tan}}\theta + {\text{sin}}\theta } \right)^2} - {\left( {{\text{tan}}\theta - {\text{sin}}\theta } \right)^2}$

Here, we know that ${\left( {a + b} \right)^2} = {a^2} + 2ab + {b^2}$and ${\left( {a - b} \right)^2} = {a^2} - 2ab + {b^2}$

By applying these formula,

\[

= {\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 \[{\text{R}}{\text{.H}}{\text{.S}}\].

\[4\sqrt {mn} = 4\sqrt {\left( {{\text{tan}}\theta + {\text{sin}}\theta } \right)\left( {{\text{tan}}\theta - {\text{sin}}\theta } \right)} \]

And here we know that $\left( {a + b} \right)\left( {a - b} \right) = {a^2} - {b^2}$

\[ = 4\sqrt {{{\tan }^2}\theta - {{\sin }^2}\theta } \] $\left( {\because {{\tan }^2}\theta = \frac{{{{\sin }^2}\theta }}{{{{\cos }^2}\theta }}} \right)$

\[

= 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 }}} \\

\]

\[ = 4\sqrt {{\text{ta}}{{\text{n}}^2}\theta \cdot {\text{si}}{{\text{n}}^2}\theta } \] \[\left( {\because (1 - {{\cos }^2}\theta ) = {{\sin }^2}\theta } \right)\]

\[ = 4\tan \theta \cdot \sin \theta \]

Therefore, \[{\text{L}}{\text{.H}}{\text{.S}}{\text{ = R}}{\text{.H}}{\text{.S}}{\text{.}}\]

Hence proved.

Note:-Whenever we face such type of question try it solving by taking ${\text{L}}{\text{.H}}{\text{.S}}$and \[{\text{R}}{\text{.H}}{\text{.S}}\] and then equate it to proof the question.

Given,$\;m = \left( {{\text{tan}}\theta + {\text{sin}}\theta } \right)$ $n = {\text{tan}}\theta - {\text{sin}}\theta $

We need to show ${m^2} - {n^2} = 4\sqrt {mn} $

Taking${\text{L}}{\text{.H}}{\text{.S}}$.

${m^2} - {n^2} = {\left( {{\text{tan}}\theta + {\text{sin}}\theta } \right)^2} - {\left( {{\text{tan}}\theta - {\text{sin}}\theta } \right)^2}$

Here, we know that ${\left( {a + b} \right)^2} = {a^2} + 2ab + {b^2}$and ${\left( {a - b} \right)^2} = {a^2} - 2ab + {b^2}$

By applying these formula,

\[

= {\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 \[{\text{R}}{\text{.H}}{\text{.S}}\].

\[4\sqrt {mn} = 4\sqrt {\left( {{\text{tan}}\theta + {\text{sin}}\theta } \right)\left( {{\text{tan}}\theta - {\text{sin}}\theta } \right)} \]

And here we know that $\left( {a + b} \right)\left( {a - b} \right) = {a^2} - {b^2}$

\[ = 4\sqrt {{{\tan }^2}\theta - {{\sin }^2}\theta } \] $\left( {\because {{\tan }^2}\theta = \frac{{{{\sin }^2}\theta }}{{{{\cos }^2}\theta }}} \right)$

\[

= 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 }}} \\

\]

\[ = 4\sqrt {{\text{ta}}{{\text{n}}^2}\theta \cdot {\text{si}}{{\text{n}}^2}\theta } \] \[\left( {\because (1 - {{\cos }^2}\theta ) = {{\sin }^2}\theta } \right)\]

\[ = 4\tan \theta \cdot \sin \theta \]

Therefore, \[{\text{L}}{\text{.H}}{\text{.S}}{\text{ = R}}{\text{.H}}{\text{.S}}{\text{.}}\]

Hence proved.

Note:-Whenever we face such type of question try it solving by taking ${\text{L}}{\text{.H}}{\text{.S}}$and \[{\text{R}}{\text{.H}}{\text{.S}}\] and then equate it to proof the question.

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