If `x , y , z` are in A.P., then `(sinx-sinz)/(cosz-cosx)` is equal to `tany` (b) `coty` (c) `siny` (d) `coty`

Explanation

As `x,y and z` are in A.P.

`:. 2y = z+x`

`=>y = (z+x)/2`

Now, `(sinx-sinz)/(cosz-cosx) = (2sin((x-z)/2)cos((x+z)/2))/(2sin((x-z)/2)sin((z+x)/2))`

`= cosy/siny = cot y`

`:. (sinx-sinz)/(cosz-cosx) = coty`