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Question: Prove that \(\dfrac{{\cos A - \sin A + 1}}{{\cos A + \sin A - 1}} = \cos ecA + \cot A\)...

Prove that cos⁡A−sin⁡A+1cos⁡A+sin⁡A−1=cos⁡ecA+cot⁡A\dfrac{{\cos A - \sin A + 1}}{{\cos A + \sin A - 1}} = \cos ecA + \cot A

Explanation

Solution

In the given question the solution or which we have to prove involves the trigonometric functions that are cos⁡ec\cos ec and cot⁡\cot , if we have a look on both of the trigonometric function then we will find that they have same denominator that is a sine function so we can divide the numerator and the denominator by the sine function and simplify to get the required result.

Complete step-by-step answer:
Since, as discussed in the hint part that the trigonometric function involves the trigonometric function which have sine function as the denominator so let us divide the numerator and denominator with sine function as stated below
⇒\Rightarrow cos⁡A−sin⁡A+1cos⁡A+sin⁡A−1=cos⁡A−sin⁡A+1sin⁡Acos⁡A+sin⁡A−1sin⁡A\dfrac{{\cos A - \sin A + 1}}{{\cos A + \sin A - 1}} = \dfrac{{\dfrac{{\cos A - \sin A + 1}}{{\sin A}}}}{{\dfrac{{\cos A + \sin A - 1}}{{\sin A}}}}
Now, separating the sine which we have divided in both numerator as well as denominator
=cos⁡Asin⁡A−sin⁡Asin⁡A+1sin⁡Acos⁡Asin⁡A+sin⁡Asin⁡A−1sin⁡A= \dfrac{{\dfrac{{\cos A}}{{\sin A}} - \dfrac{{\sin A}}{{\sin A}} + \dfrac{1}{{\sin A}}}}{{\dfrac{{\cos A}}{{\sin A}} + \dfrac{{\sin A}}{{\sin A}} - \dfrac{1}{{\sin A}}}}
Now, substituting the trigonometric relation that are
cos⁡Asin⁡A=cot⁡A\dfrac{{\cos A}}{{\sin A}} = \cot A
And
1sin⁡A=cos⁡ecA\dfrac{1}{{\sin A}} = \cos ecA
If we use the above relation , we get
=cot⁡A−1+cos⁡ecAcot⁡A+1−cos⁡ecA= \dfrac{{\cot A - 1 + \cos ecA}}{{\cot A + 1 - \cos ecA}}
Now we can see that the answer involves cos⁡ecA+cot⁡A\cos ecA + \cot A that is present in numerator in the solution that means we need to substitute some value of 11 that will be same as that of a denominator so that it cancels out one another and we can get the required solution
So,
cos⁡ec2A−cot⁡2A=1\cos e{c^2}A - {\cot ^2}A = 1
Using the above in place of 11 which is present in numerator
=cot⁡A+cos⁡ecA−(cos⁡ec2A−cot⁡2A)cot⁡A+1−cos⁡ecA= \dfrac{{\cot A + \cos ecA - (\cos e{c^2}A - {{\cot }^2}A)}}{{\cot A + 1 - \cos ecA}}
Now, we need to form the numerator such a way that can cancel the denominator
Taking common such a way that
⇒(cot⁡A+cos⁡ecA)(1−cos⁡ecA+cot⁡A)=cot⁡A−cos⁡ecAcot⁡A+cot⁡2A+cos⁡ecA−cos⁡ec2A+cos⁡ecAcot⁡A\Rightarrow \left( {\cot A + \cos ecA} \right)(1 - \cos ecA + \cot A) = \cot A - \cos ecA\cot A + {\cot ^2}A + \cos ecA - \cos e{c^2}A + \cos ecA\cot A
The Right hand side can be simplified further
⇒cot⁡A−cos⁡ecAcot⁡A+cot⁡2A+cos⁡ecA−cos⁡ec2A+cos⁡ecAcot⁡A=cot⁡A+cos⁡ecA−(cos⁡ec2A−cot⁡2A)\Rightarrow \cot A - \cos ecA\cot A + {\cot ^2}A + \cos ecA - \cos e{c^2}A + \cos ecA\cot A = \cot A + \cos ecA - (\cos e{c^2}A - {\cot ^2}A)
Hence, the above is the prove that we can write the numerator as the above format
Now, replacing the numerator we get
=(cot⁡A+cos⁡ecA)(1−cos⁡ecA+cot⁡A)cot⁡A+1−cos⁡ecA= \dfrac{{\left( {\cot A + \cos ecA} \right)(1 - \cos ecA + \cot A)}}{{\cot A + 1 - \cos ecA}}
Now, it is clear that the numerator involves the same term which is present in denominator and it cancels one another
So, the required result is

cos⁡A−sin⁡A+1cos⁡A+sin⁡A−1=cos⁡ecA+cot⁡A\dfrac{{\cos A - \sin A + 1}}{{\cos A + \sin A - 1}} = \cos ecA + \cot A

Note: In such a question where the trigonometric function which we desire is missing in such case we need to substitute or we need to find the relation with that specific trigonometric function that we need, in the given solution we substitute the relation in place of 11 to get the required result.