If x1,x2,x3,....,xn are in A.P. whose common difference is α, then the value of sinα(secx1secx2+secx2secx3+......+secxn−1secxn)=
A
sin(n−1)αcosx1cosxn
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B
sinnαcosx1cosxn
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C
sin(n−1)αcosx1cosxn
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D
sinnαcosx1cosxn
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Solution
The correct option is Asin(n−1)αcosx1cosxn We have sinαsecx1secx2+sinαsecx2secx3+.... ...+sinαsecxn−1secxn =sin(x2−x1)cosx1cosx2+sin(x3−x2)cosx2cosx3+...+sin(xn−xn−1)cosxn−1cosxn=tanx2−tanx1+tanx3−tanx2+...+tanxn−tanxn−1=tanxn−tanx1=sin(xn−x1)cosxncosx1=sin(n−1)αcosxncosx1 {(∵xn=x1+(n−1)α)}