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`Sin^-1 63/65=Sin^-1 5/13+Cos^-1 3/5` - Mathematics

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प्रश्न

`sin^-1  63/65=sin^-1  5/13+cos^-1  3/5`

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उत्तर

RHS

      `sin^-1  5/13+cos^-1  3/5`

`=sin^-1  5/13+sin^-1  4/5`        `[because cos^-1x=sin^-1sqrt(1-x^2)]`

`=sin^-1{5/13sqrt(1-(4/5)^2)+4/5sqrt(1-(5/13)^2)}`

`=sin^-1{5/13xx3/5+4/5xx12/13}`

`=sin^-1{15/65+48/65}`

`=sin^-1  63/65=`LHS

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पाठ 4: Inverse Trigonometric Functions - Exercise 4.12 [पृष्ठ ८९]

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आरडी शर्मा Mathematics [English] Class 12
पाठ 4 Inverse Trigonometric Functions
Exercise 4.12 | Q 2.1 | पृष्ठ ८९

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