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