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प्रश्न
`sin^-1 5/13+cos^-1 3/5=tan^-1 63/16`
उत्तर
LHS = `sin^-1 5/13+cos^-1 3/5`
`=sin^-1 5/13+cos^-1 3/5`
`=sin^-1 5/13+sin^-1sqrt(1-(3/5)^2)` `[because sin^-1x=cos^-1sqrt(1-x^2)]`
`=sin^-1 5/13+sin^-1 4/5`
`=sin^-1[5/13sqrt(1-(4/5)^2)+4/5sqrt(1-(5/13)^2)]` `[because sin^-1x+sin^-1y=sin^-1(xsqrt(1-y^2)+ysqrt(1-x^2))]`
`=sin^-1(5/13xx3/5+4/5xx12/13)`
`=sin^-1(3/13+48/65)`
`=sin^-1(63/65)`
`=tan^-1((63/65)/sqrt(1-(63/65)^2))` `[becausesin^-1x=tan^-1(x/sqrt(1-x^2))]`
`=tan^-1((63/65)/(16/65))`
`=tan^-1(63/16)=` RHS
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