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Find the value of the expression in terms of x, with the help of a reference triangle cos (tan–1 (3x – 1)) - Mathematics

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प्रश्न

Find the value of the expression in terms of x, with the help of a reference triangle

cos (tan–1 (3x – 1))

योग
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उत्तर

cos (tan–1 (3x – 1)) = `cos["opp"/"Hyp"]`

Let θ = tan1(3x – 1)

tan θ = 3x – 1

1 + tan2θ = 1 + (3x – 1)²

sec2θ = 9x2 – 6x + 2

sec θ = `sqrt(9x^2- 6x + 2)`

cos θ = `1/sqrt(9x^2 - 6x + 2)`

⇒ cos (tan1(3x – 1)) = `1/sqrt(9x^2 - 6x + 2)`

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  क्या इस प्रश्न या उत्तर में कोई त्रुटि है?
अध्याय 4: Inverse Trigonometric Functions - Exercise 4.5 [पृष्ठ १६६]

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सामाचीर कलवी Mathematics - Volume 1 and 2 [English] Class 12 TN Board
अध्याय 4 Inverse Trigonometric Functions
Exercise 4.5 | Q 2. (ii) | पृष्ठ १६६
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