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I got this question from a paper but can't solve it and the question paper has no solutions section.How do you prove this?

$$\displaystyle\frac{\cos A - \sin A + 1}{\cos A + \sin A - 1} = \csc A + \cot A$$ Thanks in advance.

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  • $\begingroup$ Cross multiply and simplify. Where are you stuck? $\endgroup$ – Macavity Nov 26 '14 at 14:01
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$$\mathrm{cosec} A+\mathrm{cotan} A=\frac{1}{\sin A}+\frac{\cos A}{\sin A}=\frac{1+\cos A}{\sin A}\\=\frac{1+\cos A}{\sin A}\frac{\cos A+\sin A-1}{\cos A+\sin A-1}\\=\frac{\cos A+\sin A-1+\cos^2 A+\sin A\cos A-\cos A}{\sin A\cos A+\sin^2 A-\sin A}\\\underbrace{=}_{(1)}\frac{\sin A-1+\cos^2 A+\sin A\cos A}{\sin A\cos A+\sin^2 A-\sin A}\\\underbrace{=}_{(2)}\frac{\sin A-\sin^2 A+\sin A\cos A}{\sin A\cos A+\sin^2 A-\sin A}\\\underbrace{=}_{(3)}\frac{1-\sin A+\cos A}{\cos A+\sin A-1},$$ where:

in $(1)$ we cancel $\cos A$ and $-\cos A$ in the numerator;

in $(2)$ we use $\sin^2A+\cos^2 A=1;$

in $(3)$ we divide numerator and denominator by $\sin A.$

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Hint: A good way to start would be to rewrite the right-hand side in terms of sines and cosines, and to multiply the numerator and denominator of the left-hand side by $(\cos A-\sin A-1).$

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  • $\begingroup$ @Macavity: Yes, it is. The left-hand side then readily simplifies to the form $\frac{\sin A}{1-\cos A}.$ At that point, one could either apply some half-angle formulas for tangent, or one could multiply top and bottom by $1+\cos A.$ $\endgroup$ – Cameron Buie Nov 28 '14 at 18:04
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Here I derive the rhs by simplification.

Use $\displaystyle \sin x = \frac{2\tan\frac{x}{2}}{1+\tan^2\frac{x}{2}}$ and $\displaystyle \cos x = \frac{1-\tan^2\frac{x}{2}}{1+\tan^2\frac{x}{2}}$ to write

$$\begin{align}\frac{\cos x - \sin x +1}{\cos x + \sin x -1}&=\frac{\frac{1-\tan^2\frac{x}{2}}{1+\tan^2\frac{x}{2}} - \frac{2\tan\frac{x}{2}}{1+\tan^2\frac{x}{2}} +1} {\frac{1-\tan^2\frac{x}{2}}{1+\tan^2\frac{x}{2}} + \frac{2\tan\frac{x}{2}}{1+\tan^2\frac{x}{2}} -1}\\ &=\cot \frac{x}{2}\\ &=\frac{1+\cos x}{\sin x} \end{align}$$

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Like Need help in proving that $\frac{\sin\theta - \cos\theta + 1}{\sin\theta + \cos\theta - 1} = \frac 1{\sec\theta - \tan\theta}$,

Dividing the numerator & the denominator by $\sin A,$

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

$$=\frac{\cot A+\csc A-(\csc^2A-\cot^2A)}{\cot A+1-\csc A}$$

$$=(\cot A+\csc A)\cdot\frac{\{1-(\csc A-\cot A)\}}{\cot A+1-\csc A}=?$$

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$$\begin{align} \csc A+\cot A&=\frac 1{\sin A}+\frac{\cos A}{\sin A}=\frac{1+\cos A}{\sin A}\\ =&\frac{1-\cos^2A}{\sin A(1-\cos A)}=\frac{\sin A}{1-\cos A} \end{align}$$

Therefore $$1+\cos A=k\sin A$$ and $$\sin A=k(1-\cos A)$$ for some $k$.

Now $$\frac{\cos A-\sin A+1}{\cos A+\sin A-1}=\frac{k\sin A-\sin A}{k(1-\cos A)+\cos A-1}=\frac{(k-1)\sin A}{(k-1)(1-\cos A)}$$

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$$(\cos A-\sin A+1)\sin A=\cos A\sin A-\sin^2A+\sin A$$

$$=\cos A\sin A-(1-\cos^2A)+\sin A=\sin A(1+\cos A)-(1-\cos A)(1+\cos A)$$

$$\implies(\cos A-\sin A+1)\sin A=(1+\cos A)(\sin A-1+\cos A)$$

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

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Here is a method that I use when I really get stuck. $$LHS\cdot\frac{RHS}{RHS}=\frac{LHS}{RHS}\cdot RHS$$

Now all that is left to do is to prove that $\dfrac{LHS}{RHS}=1$

Applying that method to this particular problem we have $$\begin{array}{lll} \frac{\cos A-\sin A +1}{\cos A+\sin A -1}&=&\frac{\cos A-\sin A +1}{\cos A+\sin A -1}\cdot\frac{\csc A+\cot A}{\csc A+\cot A}\\ &=&\frac{\frac{\cos A-\sin A +1}{\sin A}}{\frac{\cos A+\sin A -1}{\sin A}}\cdot\frac{\csc A+\cot A}{\csc A+\cot A}\\ &=&\frac{\cot A - 1 + \csc A}{\cot A+1 -\csc A}\cdot\frac{\csc A+\cot A}{\csc A+\cot A}\\ &=&\frac{\cot A - 1 + \csc A}{(\cot A+1 -\csc A)(\csc A+\cot A)}\cdot(\csc A+\cot A)\\ \end{array}$$ Multiplying out the denominator we have $$\cot A\csc A+\csc A\color{blue}{-\csc^2A+\cot^2A}+\cot A-\csc A\cot A$$ But rearranging the identity $$\begin{array}{lll} \cot^2A+1&=&\csc^2A\\ \color{blue}{\cot^2A-\csc^2A}&=&\color{blue}{-1}\\ \end{array}$$ our denominator becomes $$\color{green}{\csc A \color{blue}{-1}+\cot A}$$

continuing our proof $$\begin{array}{lll} &=&\frac{\cot A - 1 + \csc A}{(\cot A+1 -\csc A)(\csc A+\cot A)}\cdot(\csc A+\cot A)\\ &=&\frac{\cot A - 1 + \csc A}{\color{green}{\csc A -1 +\cot A}}\cdot(\csc A+\cot A)\\ &=&1\cdot(\csc A+\cot A)\\ &=&\csc A+\cot A \end{array}$$

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