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`Sin^-1 5/13+Cos^-1 3/5=Tan^-1 63/16` - Mathematics

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प्रश्न

`sin^-1  5/13+cos^-1  3/5=tan^-1  63/16`

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उत्तर

LHS = `sin^-1  5/13+cos^-1  3/5`

`=sin^-1  5/13+cos^-1  3/5`

`=sin^-1  5/13+sin^-1sqrt(1-(3/5)^2)`         `[because sin^-1x=cos^-1sqrt(1-x^2)]`

`=sin^-1  5/13+sin^-1  4/5`

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

`=sin^-1(5/13xx3/5+4/5xx12/13)`

`=sin^-1(3/13+48/65)`

`=sin^-1(63/65)`

`=tan^-1((63/65)/sqrt(1-(63/65)^2))`     `[becausesin^-1x=tan^-1(x/sqrt(1-x^2))]`

`=tan^-1((63/65)/(16/65))`

`=tan^-1(63/16)=`  RHS

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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 2.2 | पृष्ठ ८९

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