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Question
If A = 30° and B = 60°, verify that: `(sin("A" -"B"))/(sin"A" . sin"B")` = cotB - cotA
Solution
A = 30° and B = 60°
L.H.S.
= `(sin("A" -"B"))/(sin"A" . sin"B")`
= `(sin(30° - 60°))/(sin30° xx sin60°)`
= `(-sin"30°)/(sin30° xx sin60°)`
= `(-(1)/(2))/((1)/(2) xx sqrt(3)/(2)`
= `-(2)/sqrt(3)`
R.H.S.
= cotB - cotA
= cot60° - cot30°
= `(1)/sqrt(3) - sqrt(3)`
= `(1 - 3)/sqrt(3)`
= `-(2)/sqrt(3)`
⇒ `(sin("A" -"B"))/(sin"A" . sin"B")` = cotB - cotA.
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