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if x1, x2, x3......xn in AP then prove thatsecx1.secx2 + secx2.secx3 + secx3.secx4................................secxn-1.secxn = (tanxn-tanx1)/sin(x2-x1)
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]/sindNow substitute r=1,2,3,....,n-1 and add, we will get(tanxn – tanx1)/sind=(tanxn – tanx1)/sin(x2 – x1)
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