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Question
Prove the following result-
`tan^-1 63/16 = sin^-1 5/13 + cos^-1 3/5`
Solution
Let a = `sin^-1 5/13` b = `cos^-1 3/5`
Let a = `sin^-1 5/13`
We know that
cos a = `sqrt(1 - sin^2 "a")`
`= sqrt(1 - (5/13)^2)`
`= sqrt(144/169)`
`= 12/13`
Let b = `cos^-1 3/5`
cos b = `3/5`
We know that
sin b = `sqrt(1 - cos^2 "b")`
`= sqrt(1 - (3/5)^2)`
`= sqrt(16/25)`
`= 4/5`
Now,
tan a = `(sin a)/(cos a)`
`= (5/13)/(12/13)`
`= 5/13 xx 13/12`
`= 5/12`
tan b = `(sin b)/(cos b)`
`= (4/5)/(3/5)`
`= 4/5 xx 5/3`
`= 4/3`
Now we know that
tan (a + b) = `(tan a + tan b)/(1 - tan a tan b)`
Putting tan a = `5/12` and tan b = `4/3`
tan (a + b) = `(5/12 + 4/3)/(1 - 5/12 xx 4/3)`
tan (a + b) = `((5 xx 3 + 4 xx 12)/36)/(1 - 20/36)`
= `((15 + 48)/36)/((36 - 20)/36)`
= `(63/36)/(16/36)`
= `63/36 xx 36/16`
= `63/16`
Thus, tan (a + b) = `63/16`
a + b = `tan^-1 (63/16)`
Putting values of a and b
`sin^-1 5/13 + cos^-1 3/5 = tan^-1 (63/16)`
Hence L.H.S = R.H.S
Hence Proved.
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