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If `Cot Theta = 3/4` Prove that `Sqrt((Sec Theta - Cosec Theta)/(Sec Theta +Cosec Theta)) = 1/Sqrt7` - Mathematics

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if `cot theta = 3/4` prove that `sqrt((sec theta - cosec theta)/(sec theta +cosec theta)) = 1/sqrt7`

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`cot theta = "ЁЭСОЁЭССЁЭСЧЁЭСОЁЭСРЁЭСТЁЭСЫЁЭСб ЁЭСаЁЭСЦЁЭССЁЭСТ"/"ЁЭСЬЁЭСЭЁЭСЭЁЭСЬЁЭСаЁЭСЦЁЭСбЁЭСТ ЁЭСаЁЭСЦЁЭССЁЭСТ"`

Let x be the hypotenuse by applying Pythagoras theorem.

ЁЭР┤ЁЭР╢2 = ЁЭР┤ЁЭР╡2 + ЁЭР╡ЁЭР╢2

ЁЭСе2 = 16 + 9

`x^2 = 25 => x = 5`

`sec theta = (AC)/(AB) = 5/4`

`cosec theta = (AC)/(AB) = 5/4`

On substituting in equation we get

`sqrt((sec theta - cosec theta)/(sec theta + cosec theta)) = sqrt((5/3 - 5/4)/(5/3 + 5/4))`

`= sqrt(((20 - 15)/12)/((20 + 15)/12)) = sqrt(5/35) = 1/sqrt7`

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рдЕрдзреНрдпрд╛рдп 10: Trigonometric Ratios - Exercise 10.1 [рдкреГрд╖реНрда реирел]

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рдЖрд░рдбреА рд╢рд░реНрдорд╛ Mathematics [English] Class 10
рдЕрдзреНрдпрд╛рдп 10 Trigonometric Ratios
Exercise 10.1 | Q 26 | рдкреГрд╖реНрда реирел

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= `square/square + square/square`  ......`[тИ╡ cot θ = square/square, tan θ = square/square]`

= `(square + square)/(square xx square)`  .....`[тИ╡ square + square = 1]`

= `1/(square xx square)`

= `1/square xx 1/square`

= cosec θ·sec θ  ......`[тИ╡ "cosec"  θ = 1/square, sec θ = 1/square]`

= R.H.S.

∴ L.H.S. = R.H.S.

∴ cot θ + tan θ = cosec·sec θ


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