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प्रश्न
Evaluate:
`sin(tan^-1x+tan^-1 1/x)` for x < 0
उत्तर
`sin(tan^-1x+tan^-1 1/x)=sin[tan^-1(-x)+tan^-1(-1/x)]` `[thereforex<0]`
`=sin[-tan^-1(x)-tan^-1(1/x)]`
`=sin{-[tan^-1(x)+tan^-1(1/x)]}`
`=sin[-(tan^-1x+cot^-1x)]` `[thereforetan^-1 1/x=cot^-1x]`
`=-sin(tan^-1x+cot^-1x)`
`=-sin(pi/2)` `[thereforetan^-1x+cot^-1x=pi/2]`
= -1
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