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if `cot theta = 3/4` prove that `sqrt((sec theta - cosec theta)/(sec theta +cosec theta)) = 1/sqrt7`
рдЙрддреНрддрд░
`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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cos 60° cos 45° - sin 60° тИЩ sin 45°
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Prove that: cot θ + tan θ = cosec θ·sec θ
Proof: L.H.S. = cot θ + tan θ
= `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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