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рдкреНрд░рд╢реНрди
In the following, trigonometric ratios are given. Find the values of the other trigonometric ratios.
`cos A = 4/5`
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рдЙрддреНрддрд░
We know that `cos theta = "adjacent side"/"hypotenuse"`
Let us consider a right-angled ΔABC
Let opposite side BC = x.
By applying Pythagoras theorem, we get
ЁЭР┤ЁЭР╢2 = ЁЭР┤ЁЭР╡2 + ЁЭР╡ЁЭР╢2
25 = x + 16
x = 25 - 16 = 9
x = `sqrt9 = 3`
We know that `cosA = 4/5`
`sin A = "opposite side"/"hypotenuse" = 3/5`
`tan A = "opposite side"/"adjacent side" = 3/4`
`cosec A = 1/(sin A) = (1/3)/5 = 5/3`
`sec A = 1/(cos A) = (1/4)/5 = 5/4`
`cot A = 1/(tan A) =(1/3)/4 = 4/3`
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Given sec θ = `13/12`, calculate all other trigonometric ratios.
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- sin A cos C + cos A sin C
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`4/(cot^2 30^@) + 1/(sin^2 60^@) - cos^2 45^@`
Evaluate the Following:
`(tan^2 60^@ + 4 cos^2 45^@ + 3 sec^2 30^@ + 5 cos^2 90)/(cosec 30^@ + sec 60^@ - cot^2 30^@)`
Find the value of x in each of the following :
cos x = cos 60º cos 30º + sin 60º sin 30º
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If x sin (90° – θ) cot (90° – θ) = cos (90° – θ), then x is equal to ______.
If cos A = `4/5`, then the value of tan A is ______.
If sin A = `1/2`, then the value of cot A is ______.
Prove the following:
If tan A = `3/4`, then sinA cosA = `12/25`
Find the value of sin 0° + cos 0° + tan 0° + sec 0°.
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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