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`Sin^-1 4/5+2tan^-1 1/3=Pi/2` - Mathematics

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प्रश्न

`sin^-1  4/5+2tan^-1  1/3=pi/2`

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उत्तर

LHS = `sin^-1  4/5+2tan^-1  1/3`

`=sin^-1  4/5+tan^-1{(2xx1/3)/(1-(1/3)^2)}`     `[because2tan^-1x=tan^-1{(2x)/(1-x^2)}`

`=sin^-1  4/5+tan^-1{(2/3)/(8/9)}`

`=sin^-1  4/5+tan^-1  3/4`

`=sin^-1  4/5+cos^-1  1/(sqrt(1+9/16)`      `[becausetan^-1x=cos^-1  1/sqrt(1+x^2)]`


`=sin^-1  4/5+cos^-1  1/(5/4)`

`=sin^-1  4/5+cos^-1  4/5`

`=pi/2=`RHS

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पाठ 4: Inverse Trigonometric Functions - Exercise 4.14 [पृष्ठ ११५]

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आरडी शर्मा Mathematics [English] Class 12
पाठ 4 Inverse Trigonometric Functions
Exercise 4.14 | Q 2.05 | पृष्ठ ११५

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