Question

if x1,x2, x3,......., xn are in AP, then the value if 1/x1x2 +1/x2x3 1/x3x+.......+1/xn-1xn is

Answer 1

secxr.secxr+1

=1/(cosxr.cosxr+1)

= sind/(sind.cosxr.cosxr+1)

(where d is difference of given AP)

=sin(xr+1-xr)/(sind.cosxr.cosxr+1)

= (sinxr+1cosxr – cosxr+1sinxr)/(sind.cosxr.cosxr+1)

=[(sinxr+1cosxr)/(sind.cosxr.cosxr+1)] – [(cosxr+1sinxr)/(sind.cosxr.cosxr+1)]

=[(sinxr+1)/(sind.cosxr+1)] – [(sinxr)/(sind.cosxr.)]

=[tanxr+1 – tanxr]/sind

Now substitute r=1,2,3,....,n-1 and add, we will get

(tanxn – tanx1)/sind

=(tanxn – tanx1)/sin(x2 – x1)