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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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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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рдкрд╛рда 10: Trigonometric Ratios - Exercise 10.1 [рдкреГрд╖реНрда реирек]

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рдЖрд░.рдбреА. рд╢рд░реНрдорд╛ Mathematics [English] Class 10
рдкрд╛рда 10 Trigonometric Ratios
Exercise 10.1 | Q 15 | рдкреГрд╖реНрда реирек

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