# Prove that \[\dfrac{{{\cot }^{2}}A\left( \sec A-1 \right)}{1+\sin A}={{\sec }^{2}}A\left( \dfrac{1-\sin A}{1+\sec A} \right)\].

Last updated date: 29th Mar 2023

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Answer

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Hint: Take LHS individually and by using Trigonometric Identities Trigonometric ratios simplify it. Similarly, take RHS and simplify it.

“Complete step-by-step answer:”

Given the \[LHS=\dfrac{{{\cot }^{2}}A\left( \sec A-1 \right)}{1+\sin A}\]

We know that \[\cot A=\dfrac{\cos A}{\sin A}\]and \[\sec A=\dfrac{1}{\cos A}\]

\[\therefore LHS=\dfrac{\dfrac{{{\cos }^{2}}A}{{{\sin }^{2}}A}\left[ \dfrac{1}{\cos A}-1 \right]}{1+\sin A}=\dfrac{\dfrac{{{\cos }^{2}}A}{{{\sin }^{2}}A}\left[ \dfrac{1-\cos A}{\cos A} \right]}{1+\sin A}\]

Cancel out \[\cos A\]from numerator and denominator.

\[LHS=\dfrac{\dfrac{\cos A}{{{\sin }^{2}}A}\left( 1-\cos A \right)}{1+\sin A}\]

We know, \[{{\sin }^{2}}A+{{\cos }^{2}}A=1\]

\[\begin{align}

& \Rightarrow {{\sin }^{2}}A=1-{{\cos }^{2}}A \\

& 1-{{\cos }^{2}}A=\left( 1-\cos A \right)\left( 1+\cos A \right) \\

& \therefore LHS=\dfrac{\dfrac{\cos A}{\left( 1-{{\cos }^{2}}A \right)}\left( 1-\cos A \right)}{1+\sin A} \\

& =\dfrac{\cos A\left( 1-\cos A \right)}{\left( 1-\cos A \right)\left( 1+\cos A \right)\left( 1+\sin A \right)} \\

\end{align}\]

Cancel out \[\left( 1-\cos A \right)\]from numerator and denominator.

\[=\dfrac{\cos A}{\left( 1+\cos A \right)\left( 1+\sin A \right)}-\left( 1 \right)\]

Now take \[RHS={{\sec }^{2}}A\left( \dfrac{1-\sin A}{1+\sec A} \right)\]

We know, \[\sec A=\dfrac{1}{\cos A}\]

\[\begin{align}

& \Rightarrow RHS=\dfrac{1}{{{\cos }^{2}}A}\left[ \dfrac{1-\sin A}{1+\dfrac{1}{\cos A}} \right] \\

& =\dfrac{1}{{{\cos }^{2}}A}\left[ \dfrac{1-\sin A}{\dfrac{\cos A+1}{\cos A}} \right] \\

& =\dfrac{1}{{{\cos }^{2}}A}\left[ \dfrac{\cos A\left( 1-\sin A \right)}{1+\cos A} \right] \\

\end{align}\]

We know, \[{{\sin }^{2}}A+{{\cos }^{2}}A=1\]

\[\begin{align}

& \Rightarrow {{\cos }^{2}}A=1-{{\sin }^{2}}A \\

& \left( 1-{{\sin }^{2}}A \right)=\left( 1-\sin A \right)\left( 1+\sin A \right) \\

& \therefore RHS=\dfrac{1}{\left( 1-\sin A \right)\left( 1+\sin A \right)}\left[ \dfrac{\cos A\left( 1-\sin A \right)}{1+\cos A} \right] \\

\end{align}\]

Cancel out \[\left( 1-\sin A \right)\] from numerator and denominator.

\[RHS=\dfrac{\cos A}{\left( 1+\cos A \right)\left( 1+\sin A \right)}-(2)\]

Now (1) = (2)

Which shows that LHS = RHS.

Note: By solving LHS and RHS, we simplify both to \[\dfrac{\cos A}{\left( 1+\cos A \right)\left( 1+\sin A \right)}\]which shows that LHS = RHS.

“Complete step-by-step answer:”

Given the \[LHS=\dfrac{{{\cot }^{2}}A\left( \sec A-1 \right)}{1+\sin A}\]

We know that \[\cot A=\dfrac{\cos A}{\sin A}\]and \[\sec A=\dfrac{1}{\cos A}\]

\[\therefore LHS=\dfrac{\dfrac{{{\cos }^{2}}A}{{{\sin }^{2}}A}\left[ \dfrac{1}{\cos A}-1 \right]}{1+\sin A}=\dfrac{\dfrac{{{\cos }^{2}}A}{{{\sin }^{2}}A}\left[ \dfrac{1-\cos A}{\cos A} \right]}{1+\sin A}\]

Cancel out \[\cos A\]from numerator and denominator.

\[LHS=\dfrac{\dfrac{\cos A}{{{\sin }^{2}}A}\left( 1-\cos A \right)}{1+\sin A}\]

We know, \[{{\sin }^{2}}A+{{\cos }^{2}}A=1\]

\[\begin{align}

& \Rightarrow {{\sin }^{2}}A=1-{{\cos }^{2}}A \\

& 1-{{\cos }^{2}}A=\left( 1-\cos A \right)\left( 1+\cos A \right) \\

& \therefore LHS=\dfrac{\dfrac{\cos A}{\left( 1-{{\cos }^{2}}A \right)}\left( 1-\cos A \right)}{1+\sin A} \\

& =\dfrac{\cos A\left( 1-\cos A \right)}{\left( 1-\cos A \right)\left( 1+\cos A \right)\left( 1+\sin A \right)} \\

\end{align}\]

Cancel out \[\left( 1-\cos A \right)\]from numerator and denominator.

\[=\dfrac{\cos A}{\left( 1+\cos A \right)\left( 1+\sin A \right)}-\left( 1 \right)\]

Now take \[RHS={{\sec }^{2}}A\left( \dfrac{1-\sin A}{1+\sec A} \right)\]

We know, \[\sec A=\dfrac{1}{\cos A}\]

\[\begin{align}

& \Rightarrow RHS=\dfrac{1}{{{\cos }^{2}}A}\left[ \dfrac{1-\sin A}{1+\dfrac{1}{\cos A}} \right] \\

& =\dfrac{1}{{{\cos }^{2}}A}\left[ \dfrac{1-\sin A}{\dfrac{\cos A+1}{\cos A}} \right] \\

& =\dfrac{1}{{{\cos }^{2}}A}\left[ \dfrac{\cos A\left( 1-\sin A \right)}{1+\cos A} \right] \\

\end{align}\]

We know, \[{{\sin }^{2}}A+{{\cos }^{2}}A=1\]

\[\begin{align}

& \Rightarrow {{\cos }^{2}}A=1-{{\sin }^{2}}A \\

& \left( 1-{{\sin }^{2}}A \right)=\left( 1-\sin A \right)\left( 1+\sin A \right) \\

& \therefore RHS=\dfrac{1}{\left( 1-\sin A \right)\left( 1+\sin A \right)}\left[ \dfrac{\cos A\left( 1-\sin A \right)}{1+\cos A} \right] \\

\end{align}\]

Cancel out \[\left( 1-\sin A \right)\] from numerator and denominator.

\[RHS=\dfrac{\cos A}{\left( 1+\cos A \right)\left( 1+\sin A \right)}-(2)\]

Now (1) = (2)

Which shows that LHS = RHS.

Note: By solving LHS and RHS, we simplify both to \[\dfrac{\cos A}{\left( 1+\cos A \right)\left( 1+\sin A \right)}\]which shows that LHS = RHS.

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