If x1,x2,x3,...,xn are in A.P whose common difference is α, then the value of sinα(secx1secx2+secx2secx3+...+secxn-1secxn) is
[sin(n–1)α](cosx1cosxn)
[sinnα](cosx1cosxn)
Explanation for the correct option :
It is given that, x1,x2,x3,...,xn are in A.P with d=α.
⇒x2-x1=x3-x2=......=xn-xn-1=α
sinα(secx1secx2+secx2secx3+...+secxn-1secxn)=sinx2-x1cosx1cosx2+sinx3-x2cosx2cosx3+....+sinxn-xn-1cosxn-1cosxnUsingsin(A-B)=sinAcosB-sinBcosA=sinx2cosx1-sinx1cosx2cosx1cosx2+sinx3cosx2-sinx2cosx3cosx2cosx3+.....+sinxncosxn-1-sinxn-1cosxncosxn-1cosxn=tanx2-tanx1+tanx3-tanx2+......+tanxn-tanxn-1=tanxn-tanx1=sinxncosxn-sinx1cosx1=sinxncosx1-sinx1cosxncosxncosx1=sinxn-x1cosxncosx1=sinn-1αcosxncosx1
Hence, option A is correct.