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Question
Evaluate the following
cos 60° cos 45° - sin 60° ∙ sin 45°
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Solution
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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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)`
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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
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