# Prove the following equation:

$\left( {1 + {{\tan }^2}A} \right)\left( {1 + \dfrac{1}{{{{\tan }^2}A}}} \right) = \dfrac{1}{{{{\sin }^2}A - {{\sin }^4}A}}$.

Answer

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Use formulae $1 + {\tan ^2}A = {\sec ^2}A$ and $\tan A = \dfrac{{\sin A}}{{\cos A}}$ to solve the given equation.

From the question, left hand side is:

$

\Rightarrow {\text{L}}{\text{.H}}{\text{.S}}{\text{.}} = \left( {1 + {{\tan }^2}A} \right)\left( {1 + \dfrac{1}{{{{\tan }^2}A}}} \right), \\

\Rightarrow {\text{L}}{\text{.H}}{\text{.S}}{\text{.}} = \left( {1 + {{\tan }^2}A} \right)\left( {\dfrac{{1 + {{\tan }^2}A}}{{{{\tan }^2}A}}} \right) \\

$

We know that $1 + {\tan ^2}A = {\sec ^2}A$, putting this we’ll get:

$ \Rightarrow {\text{L}}{\text{.H}}{\text{.S}}{\text{.}} = {\text{se}}{{\text{c}}^2}A\left( {\dfrac{{{\text{se}}{{\text{c}}^2}A}}{{{{\tan }^2}A}}} \right)$

Further, we know that $\tan A = \dfrac{{\sin A}}{{\cos A}}$ and $\sec A = \dfrac{1}{{\cos A}}$. Using these formulae, we’ll get:

\[

\Rightarrow {\text{L}}{\text{.H}}{\text{.S}}{\text{.}} = \dfrac{1}{{{{\cos }^2}A}}\left( {\dfrac{{\dfrac{1}{{{{\cos }^2}A}}}}{{\dfrac{{{{\sin }^2}A}}{{{{\cos }^2}A}}}}} \right), \\

\Rightarrow {\text{L}}{\text{.H}}{\text{.S}}{\text{.}} = \dfrac{1}{{{{\cos }^2}A{{\sin }^2}A}}, \\

\]

Putting ${\cos ^2}A = 1 - {\sin ^2}A$, well get:

\[

\Rightarrow {\text{L}}{\text{.H}}{\text{.S}}{\text{.}} = \dfrac{1}{{{{\sin }^2}A\left( {1 - {{\sin }^2}A} \right)}}, \\

\Rightarrow {\text{L}}{\text{.H}}{\text{.S}}{\text{.}} = \dfrac{1}{{{{\sin }^2}A - {{\sin }^4}A}}, \\

\Rightarrow {\text{L}}{\text{.H}}{\text{.S}}{\text{. = R}}{\text{.H}}{\text{.S}}{\text{.}} \\

\]

Thus, this is the required proof.

Note: We can also start with the right hand side and prove that it is coming equal to the left hand side.

From the question, left hand side is:

$

\Rightarrow {\text{L}}{\text{.H}}{\text{.S}}{\text{.}} = \left( {1 + {{\tan }^2}A} \right)\left( {1 + \dfrac{1}{{{{\tan }^2}A}}} \right), \\

\Rightarrow {\text{L}}{\text{.H}}{\text{.S}}{\text{.}} = \left( {1 + {{\tan }^2}A} \right)\left( {\dfrac{{1 + {{\tan }^2}A}}{{{{\tan }^2}A}}} \right) \\

$

We know that $1 + {\tan ^2}A = {\sec ^2}A$, putting this we’ll get:

$ \Rightarrow {\text{L}}{\text{.H}}{\text{.S}}{\text{.}} = {\text{se}}{{\text{c}}^2}A\left( {\dfrac{{{\text{se}}{{\text{c}}^2}A}}{{{{\tan }^2}A}}} \right)$

Further, we know that $\tan A = \dfrac{{\sin A}}{{\cos A}}$ and $\sec A = \dfrac{1}{{\cos A}}$. Using these formulae, we’ll get:

\[

\Rightarrow {\text{L}}{\text{.H}}{\text{.S}}{\text{.}} = \dfrac{1}{{{{\cos }^2}A}}\left( {\dfrac{{\dfrac{1}{{{{\cos }^2}A}}}}{{\dfrac{{{{\sin }^2}A}}{{{{\cos }^2}A}}}}} \right), \\

\Rightarrow {\text{L}}{\text{.H}}{\text{.S}}{\text{.}} = \dfrac{1}{{{{\cos }^2}A{{\sin }^2}A}}, \\

\]

Putting ${\cos ^2}A = 1 - {\sin ^2}A$, well get:

\[

\Rightarrow {\text{L}}{\text{.H}}{\text{.S}}{\text{.}} = \dfrac{1}{{{{\sin }^2}A\left( {1 - {{\sin }^2}A} \right)}}, \\

\Rightarrow {\text{L}}{\text{.H}}{\text{.S}}{\text{.}} = \dfrac{1}{{{{\sin }^2}A - {{\sin }^4}A}}, \\

\Rightarrow {\text{L}}{\text{.H}}{\text{.S}}{\text{. = R}}{\text{.H}}{\text{.S}}{\text{.}} \\

\]

Thus, this is the required proof.

Note: We can also start with the right hand side and prove that it is coming equal to the left hand side.

Last updated date: 02nd Oct 2023

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