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Hint: We will begin with the left hand side of the equation (1) and then we will multiply the numerator and denominator of the term in the left hand side of the expression by \[\sec A-\tan A\] and finally we will get the answer which will be equal to the right hand side of the expression. We will use the formula \[{{\sec }^{2}}A-{{\tan }^{2}}A=1\] which will help us in proving the given expression.

Complete step-by-step answer:

It is mentioned in the question that \[\dfrac{\sec A-\tan A}{\sec A+\tan A}=1-2\sec A\tan A+2{{\tan }^{2}}A........(1)\]

Now beginning with the left hand side of the equation (1) we get,

\[\Rightarrow \dfrac{\sec A-\tan A}{\sec A+\tan A}........(2)\]

Now multiplying the numerator and denominator in equation (2) by \[\sec A-\tan A\] and hence we get,

\[\begin{align}

& \Rightarrow \dfrac{\sec A-\tan A}{\sec A+\tan A}\times \dfrac{\sec A-\tan A}{\sec A-\tan A} \\

& \Rightarrow \dfrac{{{(\sec A-\tan A)}^{2}}}{{{\sec }^{2}}A-{{\tan }^{2}}A}..........(3) \\

\end{align}\]

Now we know that \[{{\sec }^{2}}A-{{\tan }^{2}}A=1\]. So applying this in equation (3) we get,

\[\Rightarrow \dfrac{{{(\sec A-\tan A)}^{2}}}{1}..........(4)\]

Now expanding the term in equation (4) using \[{{(a-b)}^{2}}={{a}^{2}}+{{b}^{2}}-2ab\] we get,

\[\Rightarrow {{\sec }^{2}}A+{{\tan }^{2}}A-2\sec A\tan A..........(5)\]

Now again substituting \[{{\sec }^{2}}A=1+{{\tan }^{2}}A\] in equation (5) we get,

\[\begin{align}

& \Rightarrow 1+{{\tan }^{2}}A+{{\tan }^{2}}A-2\sec A\tan A \\

& \Rightarrow 1+2{{\tan }^{2}}A-2\sec A\tan A......(6) \\

\end{align}\]

Hence from equation (6) we can say that the left hand side is equal to the right hand side in equation (1). Hence we have proved the given expression.

Note: In trigonometry remembering the formulas and the identities is very important because then it becomes easy. We may get confused about how to proceed further after equation (2) but here the key is to multiply the numerator and denominator of the term in the left hand side of the expression by \[\sec A-\tan A\]. Also we will use \[{{\sec }^{2}}A-{{\tan }^{2}}A=1\] a few times to get the left hand side equal to the right hand side of the equation (1).

Complete step-by-step answer:

It is mentioned in the question that \[\dfrac{\sec A-\tan A}{\sec A+\tan A}=1-2\sec A\tan A+2{{\tan }^{2}}A........(1)\]

Now beginning with the left hand side of the equation (1) we get,

\[\Rightarrow \dfrac{\sec A-\tan A}{\sec A+\tan A}........(2)\]

Now multiplying the numerator and denominator in equation (2) by \[\sec A-\tan A\] and hence we get,

\[\begin{align}

& \Rightarrow \dfrac{\sec A-\tan A}{\sec A+\tan A}\times \dfrac{\sec A-\tan A}{\sec A-\tan A} \\

& \Rightarrow \dfrac{{{(\sec A-\tan A)}^{2}}}{{{\sec }^{2}}A-{{\tan }^{2}}A}..........(3) \\

\end{align}\]

Now we know that \[{{\sec }^{2}}A-{{\tan }^{2}}A=1\]. So applying this in equation (3) we get,

\[\Rightarrow \dfrac{{{(\sec A-\tan A)}^{2}}}{1}..........(4)\]

Now expanding the term in equation (4) using \[{{(a-b)}^{2}}={{a}^{2}}+{{b}^{2}}-2ab\] we get,

\[\Rightarrow {{\sec }^{2}}A+{{\tan }^{2}}A-2\sec A\tan A..........(5)\]

Now again substituting \[{{\sec }^{2}}A=1+{{\tan }^{2}}A\] in equation (5) we get,

\[\begin{align}

& \Rightarrow 1+{{\tan }^{2}}A+{{\tan }^{2}}A-2\sec A\tan A \\

& \Rightarrow 1+2{{\tan }^{2}}A-2\sec A\tan A......(6) \\

\end{align}\]

Hence from equation (6) we can say that the left hand side is equal to the right hand side in equation (1). Hence we have proved the given expression.

Note: In trigonometry remembering the formulas and the identities is very important because then it becomes easy. We may get confused about how to proceed further after equation (2) but here the key is to multiply the numerator and denominator of the term in the left hand side of the expression by \[\sec A-\tan A\]. Also we will use \[{{\sec }^{2}}A-{{\tan }^{2}}A=1\] a few times to get the left hand side equal to the right hand side of the equation (1).

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