If `A ,\ B ,\ C` are in A.P. then `(sin A-sin C)/(cos C-cos A)=` `adot\ t a n B` b. `cot B` c. `t a n2B` d. none of these

Explanation

Since `A_{1}, B` and `C` are in `A_{1} P_{1}`




`B-A=C-B `


`text { or, } 2 {~B}={A}+{C} `


`frac{sin A-sin C}{cos C-cos A} `


`=frac{2 sin (frac{A-C}{2}) cos (frac{A+C}{2})}{-2 sin (frac{C+A}{2}) sin (frac{C-A}{2})}[because sin A-sin B=2 sin (frac{A-B}{2}) cos (frac{A+B}{2}) text { and } cos A-cos B=-2 sin (frac{A+B}{2}) cos (frac{A-B}{2})] `


`=frac{sin (frac{A-C}{2}) cos (frac{A+C^{2}}{2})}{-sin (frac{A+C}{2}) sin (frac{C-A}{2})} `


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