If A is an acute angle of triangle ABC, the $$\frac{cos⁡A-sin⁡A+1}{cos⁡A+sin⁡A-1}$$​is equal to:

Correct option is B

Given:
A is an acute angle of triangle ABC

$$\frac{\cos A - \sin A + 1}{\cos A + \sin A - 1}$$ 

Formula Used: 

​$$\frac{\cos A}{\sin A} = \cot A \\ \ \\ 1+\cot^2A = \cosec^2A$$​​

Solution:

Divide numerator and denominator by sin A.

= $$\frac{\frac{\cos A}{\sin A} - 1 + \frac{1}{\sin A}}{\frac{\cos A}{\sin A} + 1 - \frac{1}{\sin A}}$$​​

Substitute $$\frac{\cos A}{\sin A} = \cot A \ \ \text{and} \ \ \ \frac{1}{\sin A} = \cosec A$$​ 

​$$\frac{\cot A - 1 + \cosec A}{\cot A + 1 - \cosec A}$$    .........(1)

Using the identity $$1 = \cosec^2 A - \cot^2 A$$  in the numerator;

​$$\cot A + \cosec A - 1 = (\cot A + \cosec A) - (\cosec^2 A - \cot^2 A)$$​​

$$\cot A + \cosec A - 1 = (\cot A + \cosec A) - (\cosec A - \cot A)(\cosec A + \cot A)$$ = $$(\cot A + \cosec A) (1 - \cosec A + \cot A)$$​​

Putting Simplified numerator in equation (1)  

​$$=\frac{(\cot A + \cosec A) (1 - \cosec A + \cot A)}{\cot A + 1 - \cosec A}$$​​
= $${\cot A + \cosec A}$$

Alternate Method: 

A is acute, So putting A = $$30\degree$$  

​$$\frac{\cos A - \sin A + 1}{\cos A + \sin A - 1} \\ \ \\ =\frac{\frac{\sqrt3}{2} - \frac{1}{2} + 1}{\frac{\sqrt3}{2} + \frac{1}{2} - 1} = \sqrt3+2$$ 

From option(b);

$$\cot A +\cosec A = \cot 30\degree + \cosec 30\degree = \sqrt3+2$$​ 

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