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तामिळनाडू बोर्ड ऑफ सेकेंडरी एज्युकेशनएचएससी विज्ञान इयत्ता १२

Find the value of the expression in terms of x , with the help of a reference triangle sin (cos–1(1 – x))

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प्रश्न

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

sin (cos–1(1 – x))

बेरीज
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उत्तर

sin (cos–1(1 – x)) = `sin[cos^-1 ("Adj"/"Hyp")]`

`[because cos ("Adj"/"HyP") = (1 - x)/1]`

Adj = 1 – x

Hyp = 1

Opp = `sqrt(1^2 - (1 - x)^2`

= `sqrt(1 - (1 + x^2 - 2x))`

= `sqrt(1 - 1 - x^2 + 2x)`

= `sqrt(2x - x^2`

`sin("Opp"/"Hyp") = sqrt(2x - x^2)/1`

= `sqrt(2x - x^2)`

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पाठ 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
पाठ 4 Inverse Trigonometric Functions
Exercise 4.5 | Q 2. (i) | पृष्ठ १६६
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