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प्रश्न
Find the value of the following: `tan(1/2)[sin^(-1)((2x)/(1+x^2))+cos^(-1)((1-y^2)/(1+y^2))],|x| <1,y>0 and xy <1`
उत्तर
We have to find the value of `tan(1/2)[sin^(-1)((2x)/(1+x^2))+cos^(-1)((1-y^2)/(1+y^2))]`
We know that: `sin^(-1)(2x)/(1+x^2)=2tan^(-1)x for |x| ≤ 1 …… (1)`
`cos^(-1)(1-y^2)/(1+y^2)=2tan^(-1)y for y > 0 …… (2)`
`Now sin^(-1)((2x)/(1+x^2)) + cos^(-1)((1-y^2)/(1+y^2))=2tan^(-1)x+2tan^(-1)y`
`tan(1/2)[sin^(-1)((2x)/(1+x^2))+cos^(-1)((1-y^2)/(1+y^2))]=tan(1/2)(2tan^(-1)x+2tan^(-1)y)=tan(tan^(-1)x+tan^(-1)y)`
Since, ` tan^(−1)x + tan^(−1)y = tan^(−1)((x+y)/(1-xy)) for xy < 1`
`therefore tan(1/2)[sin^(-1)((2x)/(1+x^2))+cos^(-1)((1-y^2)/(1+y^2))]=tan(tan^(−1)((x+y)/(1-xy)))=(x+y)/(1-xy)`
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