Prove that `1/(secA-1) + 1/(SecA+1) = 2 cosec A.cotA`

Answer

`a rarr p, q; b rarr r, s; c rarr p,q; d rarr p,q`

Explanation

O is circumcentre, then in `DeltaOMB, OM = R cos A` (distance circumcentre from BC)
Similary, distance of circumcentre from AC and AB are `R cos B and R cos C`
Applying sine rule in triangle OBC, we have
`2R_(1) = (a)/(sin 2A) = (a)/(2 sin A cos A) = (R)/(cos A)`
or `R_(1) = (R)/(2 cos A)`
Similarly, `R_(2) = (R)/(cos R) and R_(3) = (R)/(cos C)`
[where `R_(1), R_(2), R_(3)` are circumcenter of `DeltaOBC, DeltaOCA, and DeltaOAB`, respectively]
H is orthocentre, then in `DeltaAFH`,
`AAH = (AF)/(cos angleFAH) = (b cos A)/(cos (0^(@) -B)) = 2R cos A`
Similarly, `BH = 2R cos B, CH = 2R cos C`
Alos, `HF = AF tan angleFAH`
`=b cos A cot B = 2R cos A cos B`
Similarly, `HE = 2R cos A cos C and HD = 2R cos B cos C`

I is incentre, then in `DeltaIDB, BI = (ID)/(sin angleIBD) = (r)/(sin (B//2))`
Similarly, `CI = (r)/(sin(C//2)) and AI = (r)/(sin(A//2))`
Also in `DeltaIBI_(1)`,
`II_(1) = (BI)/(cos angleBII_(1)) = (r//[sin (B//2)])/(cos ((pi)/(2) -(C)/(2))) = (r)/(sin (B//2) sin (C//2))`
Similarly, `II_(2) = (r)/(sin (A//2) sin (C//2))`
and `II_(3) = (r)/(sin(A//2) sin (B//2))`