मराठी

Evaluate: `Cos(Tan^-1 3/4)` - Mathematics

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प्रश्न

Evaluate:

`cos(tan^-1  3/4)`

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उत्तर

We have

`cos(tan^-1  3/4)=cos[1/2cos^-1((1-(3/4)^2)/(1+(3/4)^2))]`   `[therefore 2tan^-1x+cos^-1((1-x^2)/(1+x^2))]`

`=cos[1/2cos^-1(7/25)]`

Let

`y=cos^-1(7/25)`

`=>cosy=7/25`

Now,

`cos[1/2cos^-1(7/25)]=cos[1/2y]`

`=sqrt((cosy+1)/2)`    `[thereforecos2x=2cos^2x-1]`

`=sqrt((7/25+1)/2)`

`=sqrt(32/50)`

`=4/5`

`therefore cos[tan^-1(3/4)]=4/5`

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पाठ 4: Inverse Trigonometric Functions - Exercise 4.09 [पृष्ठ ५८]

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आरडी शर्मा Mathematics [English] Class 12
पाठ 4 Inverse Trigonometric Functions
Exercise 4.09 | Q 2.3 | पृष्ठ ५८

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