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प्रश्न
`tan^-1 1/4+tan^-1 2/9=1/2cos^-1 3/2=1/2sin^-1(4/5)`
उत्तर
LHS = `tan^-1 1/4+tan^-1 2/9`
`=tan^-1((1/4+2/9)/(1-1/4xx2/9))` `[becausetan^-1x+tan^-1y=tan^-1((x+y)/(1-xy))]`
`=tan^-1((17/36)/(34/36))`
`=tan^-1 1/2`
`=1/2cos^-1((1-1/4)/(1+1/4))` `[becausetan^-1x=1/2cos^-1((1-x^2)/(1+x^2))]`
`=1/2cos^-1((3/4)/(5/4))`
`=1/2cos^-1(3/5)`
Now,
`tan^-1 1/2=1/2sin^-1((2/2)/(1+1/4))` `[becausetan^-1x=1/2sin^-1((2x)/(1+x^2))]`
`=1/2sin^-1 (1/(5/4))`
`=1/2sin^-1(4/5)`
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