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प्रश्न
Evaluate the following
cos 60° cos 45° - sin 60° ∙ sin 45°
उत्तर
cos 60° cos 45° - sin 60° ∙ sin 45° …(i)
By trigonometric ratios we know that,
`cos 60^@ = 1/2 cos 45^@ = 1/sqrt2`
`sin 60^@ = sqrt3/2 sin 45^@ = 1/sqrt2`
By substituting above value in (i), we get
`1/2. 1/sqrt2 - sqrt3/2. 1/sqrt2 => (1 - sqrt3)/(2sqrt2)`
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संबंधित प्रश्न
In the following, trigonometric ratios are given. Find the values of the other trigonometric ratios.
`sin theta = 11/5`
In the following, trigonometric ratios are given. Find the values of the other trigonometric ratios.
`cos theta = 7/25`
if `tan theta = 12/13` Find `(2 sin theta cos theta)/(cos^2 theta - sin^2 theta)`
If `sin theta = a/b` find sec θ + tan θ in terms of a and b.
if `sin theta = 3/4` prove that `sqrt(cosec^2 theta - cot)/(sec^2 theta - 1) = sqrt7/3`
Evaluate the following
sin 45° sin 30° + cos 45° cos 30°
Evaluate the Following
4(sin4 60° + cos4 30°) − 3(tan2 60° − tan2 45°) + 5 cos2 45°
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` ......`(∵ 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)`
Prove that: cot θ + tan θ = cosec θ·sec θ
Proof: L.H.S. = cot θ + tan θ
= `square/square + square/square` ......`[∵ cot θ = square/square, tan θ = square/square]`
= `(square + square)/(square xx square)` .....`[∵ square + square = 1]`
= `1/(square xx square)`
= `1/square xx 1/square`
= cosec θ·sec θ ......`[∵ "cosec" θ = 1/square, sec θ = 1/square]`
= R.H.S.
∴ L.H.S. = R.H.S.
∴ cot θ + tan θ = cosec·sec θ
(3 sin2 30° – 4 cos2 60°) is equal to ______.
