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рдкреНрд░рд╢реНрди
In the following, one of the six trigonometric ratios is given. Find the values of the other trigonometric ratios.
`cot theta = 12/5`
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рдЙрддреНрддрд░
`cot alpha = "ЁЭСОЁЭССЁЭСЧЁЭСОЁЭСРЁЭСТЁЭСЫЁЭСб ЁЭСаЁЭСЦЁЭССЁЭСТ"/"ЁЭСЬЁЭСЭЁЭСЭЁЭСЬЁЭСаЁЭСЦЁЭСбЁЭСТ ЁЭСаЁЭСЦЁЭССЁЭСТ" = 12/5`
Now consider a right-angled Δle ABC,
By applying Pythagoras theorem
ЁЭР┤ЁЭР╢2 = ЁЭР┤ЁЭР╡2 + ЁЭР╡ЁЭР╢2
ЁЭСе2 = 25 + 144
`x^2 = 169 = sqrt169`
ЁЭСе = 13
`tan theta = 1/cot theta = (1/12)/5 = 5/12`
`sin theta = "ЁЭСЬЁЭСЭЁЭСЭЁЭСЬЁЭСаЁЭСЦЁЭСбЁЭСТ ЁЭСаЁЭСЦЁЭССЁЭСТ"/"тДОЁЭСжЁЭСЭЁЭСЬЁЭСбЁЭСТЁЭСЫЁЭСвЁЭСаЁЭСТ" = 5/13`
`cos theta = "ЁЭСОЁЭССЁЭСЧЁЭСОЁЭСРЁЭСТЁЭСЫЁЭСб ЁЭСаЁЭСЦЁЭССЁЭСТ"/"тДОЁЭСжЁЭСЭЁЭСЬЁЭСбЁЭСТЁЭСЫЁЭСвЁЭСаЁЭСТ" = 12/13`
`cosec theta = 1/sin theta = 1/(5/13) = 13/5`
`sec theta = 1/cos theta = 1/(12/13) = 13/12`
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If ∠A and ∠B are acute angles such that cos A = cos B, then show that ∠A = ∠B.
State whether the following are true or false. Justify your answer.
sec A = `12/5` for some value of angle A.
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Prove that sec θ + tan θ = `cos θ/(1 - sin θ)`.
Proof: L.H.S. = sec θ + tan θ
= `1/square + square/square`
= `square/square` ......`(тИ╡ sec θ = 1/square, tan θ = square/square)`
= `((1 + sin θ) square)/(cos θ square)` ......[Multiplying `square` with the numerator and denominator]
= `(1^2 - square)/(cos θ square)`
= `square/(cos θ square)`
= `cos θ/(1 - sin θ)` = R.H.S.
∴ L.H.S. = R.H.S.
∴ sec θ + tan θ = `cos θ/(1 - sin θ)`
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