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प्रश्न
Show that `cos^-1 (12/13) + sin^-1 (3/5) = sin^-1 (56/65)`
उत्तर
LHS = `cos^-1 (12/13) + sin^-1 (3/5)`
`= sin^-1 sqrt(1 - (12/13)^2) + sin^-1 (3/5)`
`[∵ cos^-1 x = sin^-1 sqrt(1 - x^2)]`
`= sin^-1 sqrt((169 - 144)/169) + sin^-1 (3/5)`
`= sin^-1 (5/13) + sin^-1 (3/5)`
`= sin^-1 [5/13 sqrt(1 - (3/5)^2) + 3/5 sqrt(1 - (5/13)^2)]`
`[∵ sin^-1 x + sin^-1 y = sin^-1 (xsqrt(1 - y^2) + ysqrt(1 - x^2))]`
`= sin^-1 [5/13 xx sqrt((25 - 9)/25) + 3/5 sqrt((169- 25)/169)]`
`= sin^-1 [5/13 xx 4/5 + 3/5 xx 12/13]`
`= sin^-1 [4/13 + 36/65]`
`= sin^-1 [(20 + 36)/65]`
`= sin^-1 (56/65)` = RHS
Hence proved.
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