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प्रश्न
Evaluate:
`cos(tan^-1 3/4)`
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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