Question

If theta 1, theta 2 , theta 3 , theta n are in AP whose common difference is d show that sec theta 1 sec theta 2 + sec theta 2 sec theta 3 +....+ sec theta ( n-1 ) sec theta n = tan theta n - tan theta 1 / sin d.

Answer 1

θ1,θ2,θ3,....,θn are in A.P with common difference d.
∴d=θ2-θ1=θ3-θ2=....=θn-θn-1
Now, sind[secθ1secθ2+secθ2secθ3+....+secθ(n-1)secθn]
=sind[1/cosθ1cosθ2+1/cosθ2cosθ3+.....+1/cosθ(n-1)cosθn]
=sind/cosθ1cosθ2+sind/cosθ2cosθ3+.....+sind/cosθ(n-1)cosθn
=sin(θ2-θ1)/cosθ1cosθ2+sin(θ3-θ2)/cosθ2cosθ3+.....+sin(θn-θn-1)/cosθ(n-1)cosθn
=(sinθ2cosθ1-cosθ2sinθ1)/cosθ1cosθ2+(sinθ3cosθ2-cosθ3sinθ2)/cosθ2cosθ3+......+(sinθncosθn-1-cosθnsinθn-1)/cosθn-1cosθn
=sinθ2/cosθ2-sinθ1/cosθ1+sinθ3/cosθ3-sinθ2/cosθ2+........+sinθn/cosθn-sinθn-1/cosθn-1
=sinθn/cosθn-sinθ1/cosθ1
=tanθn-tanθ1
∴, secθ1secθ2+secθ2secθ3+....+secθ(n-1)secθn=(tanθn-tanθ1)/sind (Proved)

Answer 2

Answer:

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