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मान निकालें `tan[(1)/(2)(sin^(-1)""(2x)/(1+x^(2))+cos^(-1)""(1-y^(2))/(1+y^(2)))]` |
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Answer» हम जानते है कि `sin^(-1)"" (2x)/(1+x^(2))=2tan^(-1)x` तथा `cos^(-1)""(1-y^(2))/(1+y^(2))=2tan^(-1)y` अब `L.H.S. = tan(tan^(-1)x+tan^(-1)y)` ` = tan[tan^(-1)((x+y)/(1-xy))] =(x+y)/(1-xy) " "[because tan(tan^(-1)x) =x]` |
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