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Question

# If x1,x2,x3,…xn in AP, whose common difference is θ, then the value of sinθ(secx1secx2+secx2secx3+……..+secxn−1secxn) is

A
sinnθcosx1cosxn
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B
sin(n1)θcosx1cosxn
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C
sinnθcosx1cosxn
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D
cos(n1)θcosx1cosxn
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Solution

## The correct option is B sin(n−1)θcosx1cosxnTo find: sinθ(secx1secx2+secx2secx3+⋯+secxn−1secxn)It is given that x1,x2,............xn are in A.P Whose common difference is θSince θ is the common difference Therefore ,x2−x1=θx3−x2=θ..xn−xn−1=θNow,sinθ(secx1secx2+secx2secx3+⋯+secxn−1secxn)This can be written as=sinθ(1cosx1cosx2+1cosx2cosx3+⋯1cosxn−1cosxn)=(sinθcosx1cosx2+sinθcosx2cosx3+⋯sinθcosxn−1cosxn)From using above equations, we get=(sin(x2−x1)cosx1cosx2+sin(x3−x2)cosx2cosx3+⋯sin(xn−xn−1)cosxn−1cosxn)As we know that,tanA−tanB=sin(A−B)cosAcosB∴ On using this, we get =(sin(x2−x1)cosx1cosx2+sin(x3−x2)cosx3cosx2+⋯sin(xn−xn−1)cosxncosxn−1)=tanx2−tanx1+tanx3−tanx2+⋯+tanxn−tanxn−1Finally, we get=tanxn−tanx1=(sin(xn−x1)cosxncosx1)Also , xn can be written as xn=(n−1)θ+x1On putting this value, we get=(sin((n−1)θ+x1−x1)cosxncosx1)=(sin(n−1)θcosxncosx1) Hence, option B.

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