Arithmetic progression and trigonometry.!

If A1, A2, A3......... An are in A.P. whose common difference is d, show that sin d ( sec A1 sec A2 + secA2 secA3 + ...... upto n terms) = tan An - tanA1




If u can't give the whole solution. Atleast give me some hint to proceed. Please.!

d = (A2 - A1) = (A3 - A4) = ....... = (An - An-1)

sin d (sec A1 sec A2 + secA2 secA3 + .... + sec An-1 sec An )
= sin d (1/ cosA1 * 1/cosA2 + 1/ cosA2 * 1/cosA3+ .... + 1/cosAn_1 *1/ cosAn )
= sin d / cosA1cosA2 + sin d/cosA2cosA3+ .... + sin d/cosAn_1cosAn )
= sin (A2-A1) / cosA1cosA2 + sin(A3-A2)/cosA2cosA3+ .... + sin(An - An_1)/cosAn_1cosAn )


= (sinA2cosA1 - cosA2sinA1) / cosA1cosA2 + (sinA3cosA2 - cosA3sinA2) / cosA2cosA3 .. + sin(An - An_1)/cosAn_1cosAn )
= sinA2/cosA2 - sinA1/cosA1 + sinA3/cosA3 - sinA2/cosA2 +......+ sinAn/cosAn - sinAn_1 / cosAn_1

if you use a lot of terms and rearrange like terms (e.g sinA2/cosA2 and -sinA2/cosA2 = 0), the terms in the middle will cancel out
= sinA2/cosA2 - sinA1/cosA1 + sinA3/cosA3 - sinA2/cosA2 +......+ sinAn/cosAn - sinAn_1 / cosAn_1
= - sinA1/cosA1+.......+ sinAn/cosAn
= -tanA1 + tanAn
= tanAn - tanA1
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