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

Question

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

Vedantu Content Team · Accepted Answer

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.

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