If θ1,θ2,θ3,…,θn are in AP, whose common difference is d, then sin dsecθ1secθ2+secθ2secθ3+…+secθn−1secθn is equal to
tanθn−tanθ2
tanθn+tanθ1
tanθn−tanθ1
None of these
Since,θ1,θ2,θ3,…,θn are in AP
⇒ θ2−θ1=θ3−θ2=…=θn−θn−1=d (i)
Now, taking only first term
sindsecθ1secθ2=sindcosθ1cosθ2=sinθ2−θ1cosθ1cosθ2=sinθ2cosθ1−cosθ2sinθ1cosθ1cosθ2=sinθ2cosθ1cosθ1cosθ2−cosθ2sinθ1cosθ1cosθ2=tanθ2−tanθ1
Similarly, we can solve other terms which will be
tanθ3−tanθ2,tanθ4−tanθ3,…∴sindsecθ1secθ2+secθ2secθ3+…+secθn−1secθn=tanθ2−tanθ1+tanθ3−tanθ2 +…+tanθn−tanθn−1 =−tanθ1+tanθn=tanθn−tanθ1