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`2sin^-1 3/5-tan^-1 17/31=Pi/4` - Mathematics

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प्रश्न

`2sin^-1  3/5-tan^-1  17/31=pi/4`

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उत्तर

LHS = `2sin^-1  3/5-tan^-1  17/31`

`=2tan^-1  (3/4)/sqrt(1-9/25)-tan^-1  17/31`      `[becausesin^-1x=tan^-1  x/sqrt(1-x^2)]`

`=2tan^-1  (3/5)/(4/5)-tan^-1  17/31`

`=2tan^-1  3/4-tan^-1  17/31`

`=tan^-1{(2xx3/4)/(1-(3/4)^2)}-tan^-1  17/31`    `[because2tan^-1x=tan^-1{(2x)/(1-x^2)}]`

`=tan^-1{(3/2)/(7/16)}-tan^-1  17/31`

`=tan^-1  24/7-tan^-1  17/31`

`=tan^-1((24/7-17/31)/(1+24/7xx17/31))`      `[becausetan^-1x-tan^-1y=tan^-1((x+y)/(1+xy))]`

`=tan^-1((625/217)/(625/217))`

`=tan^-1 1=pi/4=`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 2.06 | पृष्ठ ११५

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