If `theta_1,theta_2,theta_3, ,theta_n` are in AP, whose common difference is `d` , show that `sectheta_1s e ctheta_2+s e ctheta_2sectheta_3+....+sectheta_(n-1)sectheta_n=(t a ntheta_n-t a ntheta_1)/(sind)`

Explanation

theta1,theta2,theta3,....,thetan are in A.P with common difference d.

`∴d=theta2-theta1=theta3-theta2=....=thetan-thetan-1`

Now,
`sind[sectheta1sectheta2+sectheta2sectheta3+....+sectheta(n-1)secthetan]`

`=sind[1/costheta1costheta2+1/costheta2costheta3+.....+1/costheta(n-1)costhetan]`

`=sind/costheta1costheta2+sind/costheta2costheta3+.....+sind/costheta(n-1)costhetan`

`=sin(theta2-theta1)/costheta1costheta2+sin(theta3-theta2)/costheta2costheta3+.....+sin(thetan-thetan-1)/costheta(n-1)costhetan`

`=(sintheta2costheta1-costheta2sintheta1)/costheta1costheta2+(sintheta3costheta2-costheta3sintheta2)/costheta2costheta3+......+(sinthetancosthetan-1-costhetansinthetan-1)/costhetan-1costhetan`

`=sintheta2/costheta2-sintheta1/costheta1+sintheta3/costheta3-sintheta2/costheta2+........+sinthetan/costhetan-sinthetan-1/costhetan-1`

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