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рдкреНрд░рд╢реНрди
In the following, trigonometric ratios are given. Find the values of the other trigonometric ratios.
`sin theta = sqrt3/2`
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`sin theta = sqrt3/2`
We know `sin theta = "opposide side"/"Hyotence" = sqrt3/2`
Now consider right-angled Δle ABC
Let x = adjacent sidead
By applying Pythagoras
ЁЭР┤ЁЭР╡2 = ЁЭР┤ЁЭР╢2 + ЁЭР╡ЁЭР╢2
4 = 3+ЁЭСе2
ЁЭСе2 = 4 − 3
ЁЭСе2 = 1
ЁЭСе = 1
`cos = "opposite side"/"Hypotenuse" = 1/2`
`tan = "Oppsite side"/"hypotenuse" = sqrt3/1 = sqrt3`
`cosec theta = 1/sin theta = 1/(sqrt3/2) = 2/sqrt3`
sec = `1/cos theta = (1/1)/2 = 2`
`cot = 1/tan theta = 1/sqrt3`
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In ΔPQR, right angled at Q, PR + QR = 25 cm and PQ = 5 cm. Determine the values of sin P, cos P and tan P.
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Prove that `(sin "A" - 2sin^3 "A")/(2cos^3 "A" - cos "A") = tan "A"`
if `tan theta = 12/13` Find `(2 sin theta cos theta)/(cos^2 theta - sin^2 theta)`
If `tan theta = 24/7`, find that sin ЁЭЬГ + cos ЁЭЬГ
Evaluate the following
`2 sin^2 30^2 - 3 cos^2 45^2 + tan^2 60^@`
Find the value of x in the following :
`2sin 3x = sqrt3`
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)`
Evaluate 2 sec2 θ + 3 cosec2 θ – 2 sin θ cos θ if θ = 45°.
If sin θ – cos θ = 0, then find the value of sin4 θ + cos4 θ.
