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Hint: We will begin with the left hand side of the expression and then we will solve it by first converting all the terms in sin and cos and then we will take the LCM and finally we will get the answer which will be equal to the right hand side of the expression after we convert and simplify it in terms of sin and cos. We will use the formula \[{{\sin }^{2}}A+{{\cos }^{2}}A=1\] which will help us in proving the given expression.

__Complete step-by-step answer:__

It is mentioned in the question that \[(\text{cosec}A-\sin A)(\sec A-\cos A)=\dfrac{1}{\tan A+\cot A}.........(1)\]

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

\[\Rightarrow (\text{cosec}A-\sin A)(\sec A-\cos A).........(2)\]

Now converting sec, cosec, tan and cot in equation (2) in terms of cos and sin and hence we get,

\[\Rightarrow \left( \dfrac{1}{\sin A}-\sin A \right)\left( \dfrac{1}{\cos A}-\cos A \right).....(3)\]

Taking the LCM in equation (3) and simplifying we get,

\[\Rightarrow \left( \dfrac{1-{{\sin }^{2}}A}{\sin A} \right)\left( \dfrac{1-{{\cos }^{2}}A}{\cos A} \right).......(4)\]

Now we know that \[{{\sin }^{2}}A+{{\cos }^{2}}A=1\]. So applying this in equation (4) we get,

\[\Rightarrow \left( \dfrac{{{\cos }^{2}}A}{\sin A} \right)\left( \dfrac{{{\sin }^{2}}A}{\cos A} \right).......(5)\]

Now cancelling similar terms in the numerator and denominator in equation (5) we get.

\[\Rightarrow \sin A\cos A.......(5)\]

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

\[\Rightarrow \dfrac{1}{\tan A+\cot A}........(6)\]

Now converting all the terms in equation (6) in sin and cos we get,

\[\Rightarrow \dfrac{1}{\dfrac{\sin A}{\cos A}+\dfrac{\cos A}{\sin A}}........(7)\]

Now taking the LCM and rearranging in equation (7) we get,

\[\Rightarrow \dfrac{\sin A\cos A}{{{\sin }^{2}}A+{{\cos }^{2}}A}........(8)\]

Now we know that \[{{\sin }^{2}}A+{{\cos }^{2}}A=1\]. So applying this in equation (8) we get,

\[\Rightarrow \sin A\cos A........(9)\]

Hence from equation (5) and equation (9) 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 have to keep in mind that in equation (4) we are using \[{{\sin }^{2}}A+{{\cos }^{2}}A=1\] to substitute cos in terms of sin and sin in terms of cos.

It is mentioned in the question that \[(\text{cosec}A-\sin A)(\sec A-\cos A)=\dfrac{1}{\tan A+\cot A}.........(1)\]

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

\[\Rightarrow (\text{cosec}A-\sin A)(\sec A-\cos A).........(2)\]

Now converting sec, cosec, tan and cot in equation (2) in terms of cos and sin and hence we get,

\[\Rightarrow \left( \dfrac{1}{\sin A}-\sin A \right)\left( \dfrac{1}{\cos A}-\cos A \right).....(3)\]

Taking the LCM in equation (3) and simplifying we get,

\[\Rightarrow \left( \dfrac{1-{{\sin }^{2}}A}{\sin A} \right)\left( \dfrac{1-{{\cos }^{2}}A}{\cos A} \right).......(4)\]

Now we know that \[{{\sin }^{2}}A+{{\cos }^{2}}A=1\]. So applying this in equation (4) we get,

\[\Rightarrow \left( \dfrac{{{\cos }^{2}}A}{\sin A} \right)\left( \dfrac{{{\sin }^{2}}A}{\cos A} \right).......(5)\]

Now cancelling similar terms in the numerator and denominator in equation (5) we get.

\[\Rightarrow \sin A\cos A.......(5)\]

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

\[\Rightarrow \dfrac{1}{\tan A+\cot A}........(6)\]

Now converting all the terms in equation (6) in sin and cos we get,

\[\Rightarrow \dfrac{1}{\dfrac{\sin A}{\cos A}+\dfrac{\cos A}{\sin A}}........(7)\]

Now taking the LCM and rearranging in equation (7) we get,

\[\Rightarrow \dfrac{\sin A\cos A}{{{\sin }^{2}}A+{{\cos }^{2}}A}........(8)\]

Now we know that \[{{\sin }^{2}}A+{{\cos }^{2}}A=1\]. So applying this in equation (8) we get,

\[\Rightarrow \sin A\cos A........(9)\]

Hence from equation (5) and equation (9) 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 have to keep in mind that in equation (4) we are using \[{{\sin }^{2}}A+{{\cos }^{2}}A=1\] to substitute cos in terms of sin and sin in terms of cos.

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