Question

If θ₁, θ₂, θ₃, ..., θ_n are in A.P., whose common difference is d, show that secθ₁ secθ₂ + secθ₂ secθ₃ + ... + secθ_n–1 . secθ_n = `(tan theta_n - tan theta_1)/sin d`

Answer 1

Since θ₁, θ₂, θ₃, ..., θ_n are in A.P.

∴ θ₂ – θ₁ = θ₃ – θ₂ = ... = θ_n – θ_n-1  = d

Now we have to prove that

secθ₁ secθ₂ + secθ₂ secθ₃ + ... + secθ_n–1 . secθ_n = `(tan theta_n - tan theta_1)/sind`  L.H.S.

⇒ `sin d/sin d [sec theta_1 * sec theta_2 + sec theta_2 sec theta_3 + ... + sec theta_(n - 1) *  sec theta_n]`

Taking only `(sind[sec theta_1 * sec theta_2])/sind = (sind[1/cos theta_1 * 1/cos theta_2])/sind`

= `(sin(theta_2 - theta_1))/sind * 1/(costheta_1 costheta_2)`

= `1/sind [(sin theta_2 cos theta_1 - cos theta_2 sin theta_1)/(cos theta_1 cos theta_2)]`

= `1/sind [(sin theta_2 cos theta_1)/(cos theta_1 cos theta_2) - (cos theta_2 sin theta_1)/(cos theta_1 cos theta_2)]`

= `1/sind [tan theta_2 - tan theta_1]`

Similarly we can solve other terms which will be

`1/sind [tan theta_3 - tan theta_2]`

And `1/sind [tan theta_4 - tan theta_3]`

Here L.H.S. = `1/sind [tan theta_2 - tan theta_1 + tan theta_3 - tan theta_2 + ... + tan theta_n - tan theta_(n - 1)]`

= `1/sind [- tan theta_1 + tan theta_n]`

= `(tan theta_n - tan theta_1)/sind`  R.H.S.

L.H.S. = R.H.S.

Hence proved.