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प्रश्न
Solve the following equation for x:
`2tan^-1(sinx)=tan^-1(2sinx),x!=pi/2`
उत्तर
`2tan^-1(sinx)=tan^-1(2sinx),x!=pi/2`
`=>tan^-1((2sinx)/(1-sin^2x))=tan^-1(2sinx)` `[because2tan^-1x=tan^-1((2x)/(1-x^2))]`
`=>(2sinx)/(1-sin^2x)=2sinx`
`=>2sinx=2sinx-2sin^3x`
`=>2sin^2x=0`
`=>sinx=0`
`=>x=0`
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