Tan−1(x+3)−tan−1(x−3)=sin−1(35) then the value of x is
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give us ,
tan-¹(x+3)-tan-¹(x-3)=sin-¹(35)
there for,
tan-1(x+3)+(x-3)/1-(x+3)×(x-3) =sin-1(35)
=>tan-¹ 2x/(x²-3²)=sin-¹35
=>tan-¹2x/x²-9=sin-¹35
=>2x/x²-9=tan(sin-¹35)
i thing the qsn is not correct.
tan-¹(x+3)-tan-¹(x-3)=sin-¹(35)
there for,
tan-1(x+3)+(x-3)/1-(x+3)×(x-3) =sin-1(35)
=>tan-¹ 2x/(x²-3²)=sin-¹35
=>tan-¹2x/x²-9=sin-¹35
=>2x/x²-9=tan(sin-¹35)
i thing the qsn is not correct.
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