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Hint: In order to solve this question easily we will transform the given terms in of sin and cos. In this question we have to prove that the left - hand side is equal to the right - hand side.

Complete step-by-step answer:

Now, by using trigonometric identities we will easily solve the given problem. We know that $\sin {\text{A = }}\dfrac{1}{{\cos ec{\text{A}}}}$, ${\text{cosA = }}\dfrac{1}{{sec{\text{A}}}}$, ${\text{cotA = }}\dfrac{{\cos {\text{A}}}}{{\sin {\text{A}}}}$, ${\text{tanA = }}\dfrac{{\sin {\text{A}}}}{{\cos {\text{A}}}}$. By using these properties, we will solve this question. Now, putting these values in the given question, we get

L. H. S = $\left( {1 + {\text{ }}\dfrac{{\cos {\text{A}}}}{{\sin {\text{A}}}}{\text{ - }}\dfrac{1}{{\sin {\text{A}}}}} \right)\left( {1{\text{ + }}\dfrac{{\sin {\text{A}}}}{{\cos {\text{A}}}}{\text{ + }}\dfrac{1}{{\cos {\text{A}}}}} \right)$

\[ \Rightarrow \] L. H. S = \[\left( {\dfrac{{\sin {\text{A + cosA - 1}}}}{{\sin {\text{A}}}}} \right)\left( {\dfrac{{\cos {\text{A + sinA + 1}}}}{{\cos {\text{A}}}}} \right)\]

Now we can see that in the above equation we can use the property \[{{\text{a}}^2} - {{\text{b}}^2} = ({\text{a - b)(a + b)}}\]

So, applying this property we get

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

By solving further, we get

\[ \Rightarrow \] L. H. S = \[\left( {\dfrac{{({{\sin }^2}{\text{A + co}}{{\text{s}}^2}{\text{A + 2cosA sinA) }} - {\text{ }}1}}{{\sin {\text{A cosA}}}}} \right)\]

\[ \Rightarrow \] L. H. S = \[\left( {\dfrac{{2\sin {\text{A cosA}}}}{{\sin {\text{A cosA}}}}} \right)\] as ${\sin ^2}{\text{A + }}{\text{co}}{{\text{s}}^2}{\text{A = 1}}$

\[ \Rightarrow \] L. H. S = 2 = R. H. S

Hence proved.

Note: To solve questions which include trigonometric terms it is suggested that you should simplify the given term by converting it into sin or cos whichever is possible. Converting in sin or cos simplify the term and you can easily solve the given term. Use identity ${\sin ^2}{\text{A + }}{\text{co}}{{\text{s}}^2}{\text{A = 1}}$ properly after converting the trigonometric term.

Complete step-by-step answer:

Now, by using trigonometric identities we will easily solve the given problem. We know that $\sin {\text{A = }}\dfrac{1}{{\cos ec{\text{A}}}}$, ${\text{cosA = }}\dfrac{1}{{sec{\text{A}}}}$, ${\text{cotA = }}\dfrac{{\cos {\text{A}}}}{{\sin {\text{A}}}}$, ${\text{tanA = }}\dfrac{{\sin {\text{A}}}}{{\cos {\text{A}}}}$. By using these properties, we will solve this question. Now, putting these values in the given question, we get

L. H. S = $\left( {1 + {\text{ }}\dfrac{{\cos {\text{A}}}}{{\sin {\text{A}}}}{\text{ - }}\dfrac{1}{{\sin {\text{A}}}}} \right)\left( {1{\text{ + }}\dfrac{{\sin {\text{A}}}}{{\cos {\text{A}}}}{\text{ + }}\dfrac{1}{{\cos {\text{A}}}}} \right)$

\[ \Rightarrow \] L. H. S = \[\left( {\dfrac{{\sin {\text{A + cosA - 1}}}}{{\sin {\text{A}}}}} \right)\left( {\dfrac{{\cos {\text{A + sinA + 1}}}}{{\cos {\text{A}}}}} \right)\]

Now we can see that in the above equation we can use the property \[{{\text{a}}^2} - {{\text{b}}^2} = ({\text{a - b)(a + b)}}\]

So, applying this property we get

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

By solving further, we get

\[ \Rightarrow \] L. H. S = \[\left( {\dfrac{{({{\sin }^2}{\text{A + co}}{{\text{s}}^2}{\text{A + 2cosA sinA) }} - {\text{ }}1}}{{\sin {\text{A cosA}}}}} \right)\]

\[ \Rightarrow \] L. H. S = \[\left( {\dfrac{{2\sin {\text{A cosA}}}}{{\sin {\text{A cosA}}}}} \right)\] as ${\sin ^2}{\text{A + }}{\text{co}}{{\text{s}}^2}{\text{A = 1}}$

\[ \Rightarrow \] L. H. S = 2 = R. H. S

Hence proved.

Note: To solve questions which include trigonometric terms it is suggested that you should simplify the given term by converting it into sin or cos whichever is possible. Converting in sin or cos simplify the term and you can easily solve the given term. Use identity ${\sin ^2}{\text{A + }}{\text{co}}{{\text{s}}^2}{\text{A = 1}}$ properly after converting the trigonometric term.

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