If `a_(1), a_(2), ……., a_(n)` are in A.P. with common differece `d != 0`, then `(sin d) [sec a_(1) sec a_(2) + sec a_(2) sec a_(3) + .... + sec a_(n - 1) sec a_(n)]` is equal to

Options

A. `cosec a_(n)-cosec a`
B. `cot a_(n)-cot a`
C. `sec a_(n)- sec a_(1)`
D. `tan a_(n)- tan a_(1)`

Answer

D

Explanation

As `a_(1),a_(2),a_(3),…,a_(n-1),a_(n)` are in A.P., hence
`d=a_(2)-a_(1)=a_(3)-a_(2)=…=a_(n)-a_(n-1)`
`sind[seca_(1)seca_(2)+seca_(2)seca_(3)+…+seca_(n-1)seca_(n)]`
`=(sin(a_(2)-a_(1)))/(cosa_(1)cosa_(2))+(sin(a_(3)-a_(2)))/(cosa_(2)cosa_(3))+...+(sin(a_(n)-a_(n-1)))/(cosa_(n-1)cosa_(n))`
`=(tana_(2)-tana_(1))+(tana_(3)-tana_(2))+...+(tana_(n)-tana_(n-1))`
`=tana_(n)-tana_(1)`