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If `cot theta = 1/sqrt3` show that `(1 - cos^2 theta)/(2 - sin^2 theta) = 3/5`
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`cot theta = 1/sqrt3 (1 - cos^2 theta)/(2 - sin^2 theta) = 3/5`
`cot theta = "ЁЭСОЁЭССЁЭСЧЁЭСОЁЭСРЁЭСТЁЭСЫЁЭСб ЁЭСаЁЭСЦЁЭССЁЭСТ"/"ЁЭСЬЁЭСЭЁЭСЭЁЭСЬЁЭСаЁЭСЦЁЭСбЁЭСТ ЁЭСаЁЭСЦЁЭССЁЭСТ" = 1/sqrt3`
Let x be the hypotenuse
By applying Pythagoras
ЁЭР┤ЁЭР╢2 = ЁЭР┤ЁЭР╡2 + ЁЭР╡ЁЭР╢2
`x^2 = (sqrt3)^2 + 1`
`x^2 = 3 + 1`
ЁЭСе2 = 3 + 1 ⇒ ЁЭСе = 2
`cos theta = (BC)/(AC) = 1/2`
`sin theta = (AB)/(AC) = sqrt3/2`
`(1 - cos^2 theta)/(2 - sin^2 theta) => (1 - (1/2)^2)/(2 - (sqrt3)/2)^2`
`=> (1 - 1/4)/(2 - 3/4) => (3/4)/(5/4`
`= 3/5`
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