Question

If cosec A + sec A = cosec B + sec B prove that cot`(("A + B"))/2` = tan A tan B.

Answer 1

Given that cosec A + sec A = cosec B + sec B

`1/(sin "A") + 1/(cos "A") = 1/(sin "B") + 1/(cos "A")`

`1/(sin "A") - 1/(sin "B") = 1/(cos "B") - 1/1/(cos "A")`

Arrange T-ratios of the sine and cosine in the separate side

∴ `(sin "B" - sin "A")/(sin "A" sin "B") = (cos "A" - cos "B")/(cos "A" cos "B")`

∴ `(sin "B" - sin "A")/(cos "A" - cos "B")` = tan A tan B

`[∵ sin "C" - sin "D" = 2 cos (("C + D")/2) sin (("C - D")/2)]`

∴ `(2 cos (("B + A")/2) sin (("B - A")/2))/(- 2 sin (("A + B")/2) sin(("A - B")/2))` = tan A tan B

∴ `(2 cos (("A + B")/2) sin (("- A + B")/2))/(- 2 sin (("A + B")/2) sin(("A - B")/2))` = tan A tan B

∴ `(-2 cos (("A + B")/2) sin (("A - B")/2))/(- 2 sin (("A + B")/2) sin(("A - B")/2))` = tan A tan B

∴ `cot (("A + B")/2)` = tan A tan B