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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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рдЙрддреНрддрд░
`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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State whether the following are true or false. Justify your answer.
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If Cosec A = 2 find `1/(tan A) + (sin A)/(1 + cos A)`
Evaluate the following
sin 45° sin 30° + cos 45° cos 30°
Evaluate the Following:
`tan 45^@/(cosec 30^@) + sec 60^@/cot 45^@ - (5 sin 90^@)/(2 cos 0^@)`
Find the value of x in the following :
`2 sin x/2 = 1`
If sin 2A = `1/2` tan² 45° where A is an acute angle, then the value of A is ______.
If cos A = `4/5`, then the value of tan A is ______.
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)`
What will be the value of sin 45° + `1/sqrt(2)`?
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