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# Prove that $\dfrac{{{\cot }^{2}}A\left( \sec A-1 \right)}{1+\sin A}={{\sec }^{2}}A\left( \dfrac{1-\sin A}{1+\sec A} \right)$. Verified
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Hint: Take LHS individually and by using Trigonometric Identities Trigonometric ratios simplify it. Similarly, take RHS and simplify it.

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.
Last updated date: 01st Oct 2023
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