Question

Prove the following expression: \[\left( \text{cosec}\,A-\sin \,A \right)\,\left( \sec \,A-\cos A \right)=\dfrac{1}{\tan \,A+\cot \,A}\]

Answer 1

Hint: We will first convert cosec in terms of sin and sec in terms of cos. Then we will use simple trigonometric relations to prove this question. We will start and work with the L.H.S part and then finally we will prove it equal to R.H.S.

Complete step-by-step answer:
We will begin with left hand side of the given question,
 So L.H.S \[\,=\left( \text{cosec}\,A-\sin \,A \right)\,\left( \sec \,A-\cos A \right).........(1)\]
By first converting cosec in terms of sin and sec in terms of cos in equation (1) we get,
 \[\,\Rightarrow \left( \dfrac{1}{\sin A}-\sin \,A \right)\,\left( \dfrac{1}{\cos A}-\cos A \right)........(2)\]
Now solving equation (2) by taking sin A and cos A as L.C.M we get,
 \[\,\Rightarrow \left( \dfrac{1-{{\sin }^{2}}A}{\sin A} \right)\,\left( \dfrac{1-{{\cos }^{2}}A}{\cos A} \right)........(3)\]
We know that \[{{\sin }^{2}}A+{{\cos }^{2}}A=1\] so now applying this relationship between \[{{\sin }^{2}}A\] and \[{{\cos }^{2}}A\] in equation (3) we get,
 \[\,\Rightarrow \left( \dfrac{{{\cos }^{2}}A}{\sin A} \right)\,\left( \dfrac{{{\sin }^{2}}A}{\cos A} \right)........(4)\]
Cancelling the similar terms in equation (4) we get,
 \[\,\Rightarrow \left( \dfrac{\cos A\,\sin A}{1} \right)\,........(5)\]
We know that \[{{\sin }^{2}}A+{{\cos }^{2}}A=1\] so using this in equation (5) we get,
 \[\,\Rightarrow \left( \dfrac{\cos A\,\sin A}{{{\sin }^{2}}A+{{\cos }^{2}}A} \right)\,........(6)\]
We will now divide the numerator and the denominator in equation (6) by \[\cos A.\sin A\] and so we get,
 \[\,\Rightarrow \left( \dfrac{\dfrac{\cos A\,\sin A}{\cos A\,\sin A}}{\dfrac{{{\sin }^{2}}A+{{\cos }^{2}}A}{\cos A\,\sin A}} \right)\,........(7)\]
Now canceling similar terms from equation (7) we get,
 \[\,\Rightarrow \dfrac{1}{\dfrac{{{\sin }^{2}}A+{{\cos }^{2}}A}{\cos A\,\sin A}}........(8)\]
Now rearranging the denominator in equation (8) we get,
 \[\,\Rightarrow \dfrac{1}{\dfrac{{{\sin }^{2}}A}{\cos A\,\sin A}+\dfrac{{{\cos }^{2}}A}{\cos A\,\sin A}}........(9)\]
Cancelling the similar terms in equation (9) we get,
 \[\,\Rightarrow \dfrac{1}{\dfrac{\sin A}{\cos A\,}+\dfrac{\cos A}{\sin A}}........(10)\]
We know that \[\dfrac{\sin A}{\cos A\,}=\tan A\] and \[\dfrac{\cos A}{\sin A\,}=\cot A\] so now applying these formulas in equation (10) we get,
 \[\,\Rightarrow \dfrac{1}{\tan A+\cot A}\]
This is equal to R.H.S and hence we have proved this.

Note: The initial thing to do in these types of questions is to simplify it in basic terms like we did by converting cosec into sin and sec into cos. The next important thing is that we should know all the basic trigonometric formulas and relationships. We can get confused on how to continue after equation (5) but we know this relation \[{{\sin }^{2}}A+{{\cos }^{2}}A=1\] and hence in place of 1 we can substitute \[{{\sin }^{2}}A+{{\cos }^{2}}A\].