Let $α \in (0, \frac{π}{2})$ be fixed. If the integral $\int \frac{tan x + tan α}{tan x - tan α} d x = A (x) cos 2 α + B (x) sin 2 α + C,$ where $C$ is a constant of integration, then the functions $A (x)$ and $B (x)$ are respectively:

x - α

and

{log}_{e} | sin (x - α) |

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x + α

and

{log}_{e} | sin (x - α) |

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x - α

and

{log}_{e} | cos (x - α) |

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x + α

and

{log}_{e} | sin (x + α) |

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Solution

The correct option is A $x - α$ and ${log}_{e} | sin (x - α) |$
$\int \frac{tan x + tan α}{tan x - tan α} d x$
$\Rightarrow \int \frac{sin x cos α + cos x sin α}{sin x cos α - cos x sin α} d x = \int \frac{sin (x + α)}{sin (x - α)} d x = \int \frac{sin (x - α + 2 α)}{sin (x - α)} d x = \int \frac{sin (x - α) cos 2 α}{sin (x - α)} d x + \int \frac{cos (x - α) sin 2 α}{sin (x - α)} d x = x cos 2 α + sin 2 α log | sin (x - α) | + c$

Here, option $(a)$ and $(b)$ both are correct so it is bonus.

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