Question

Prove that: \[\dfrac{\cos A}{1-\tan A}+\dfrac{\sin A}{1-\cot A}=\sin A+\cos A\]

Answer 1

Hint: We will begin with the left hand side of the equation (1) and then we will convert all the cot and tan terms in sin and cos. When we will proceed further we will get \[{{\sin }^{2}}A-{{\cos }^{2}}A\] which we will expand using the formula \[{{a}^{2}}-{{b}^{2}}=(a+b)(a-b)\] and finally we will get the answer which will be equal to the right hand side of the expression.

Complete step-by-step answer:

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

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

\[\Rightarrow \dfrac{\cos A}{1-\tan A}+\dfrac{\sin A}{1-\cot A}.........(2)\]

Now simplifying by converting all the tan and cot terms in sin and cos in equation (2) and hence we get,

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

Now we will take the LCM of both the terms in equation (3) and simplifying we will get,

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

Now rearranging equation (4) we get,

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

Now again taking the LCM in equation (5) we get,

\[\Rightarrow \dfrac{{{\sin }^{2}}A-{{\cos }^{2}}A}{\sin A-\cos A}.........(6)\]

Now we know that \[{{a}^{2}}-{{b}^{2}}=(a+b)(a-b)\]. So applying this formula in equation (6) we get,

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

Now cancelling similar terms in equation (7) we get,

\[\Rightarrow \sin A+\cos A.........(8)\]

Hence from equation (8) 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 (4) but here the key is to make the denominator of both the terms same and hence we rearrange by taking out minus from the first term.