Prove that: `sin^(-1)((63)/(65))=sin^(-1)(5/(13))+cos^(-1)(3/5)`

1.	Prove that: `sin^(-1)((63)/(65))=sin^(-1)(5/(13))+cos^(-1)(3/5)`
Answer» RHS=`sin^(-1)"(5)/(13)+cos^(-1)""(3)/(5)` `=tan^(-1)""((5)/(13))/(sqrt(1-((5)/(13))^(2)))+tan^(-1)""(sqrt(1-((3)/(5))^(2)))/((3)/(5)) " "( :. sin^(-1)x=tan^(-1)""(x)/(sqrt(1-x^(2))) " and " cos^(-1)x=tan^(-1)""(sqrt(1-x^(2)))/(x))` `=tan^(-1)""(5)/(12)+tan^(-1)""(4)/(3)` `=tan^(-1)""((5)/(12)+(4)/(3))/(1-(5)/(12)xx(4)/(3))=tan^(-1)""((15+48)/(36))/((36-20)/(36))` `tan^(-1)""(63)/(16)`= LHS Hence Proved.

Discussion

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