`cos^-1(4/5)+sin^-1(5/13) =tan^-1(56/33)`

1.	`cos^-1(4/5)+sin^-1(5/13) =tan^-1(56/33)`
Answer» `cos^-1 4/5 = theta` `cos theta = 4/5` `sin theta = 3/5` `theta = sin^-1 ( 3/5)` `sin^-1 (3/5) + sin^-1(5/13)` `sin^-1 x + sin^-1 y = sin^-1 ( x sqrt(1-y^2) + ysqrt(1 - x^2))` `sin^-1 ( 3/5sqrt(1 - (5/13)^2 ) + 5/3 sqrt(1 - (3/5)^2 ) )` `sin^-1(3/5 sqrt(144/169) + 5/13sqrt(16/25))` `sin^-1(3/5 xx 12/13 + 5/13 xx 4/5` `sin^-1((36+20)/65)= sin^-1(56/65) = phi` `sin phi = 56/65` as `ab^2 = ac^2 - bc^2` `= 65^2 - 56^2 ` `= (121 xx 9)= 1089= 33^2` so`ab = 33` `tan phi = 56/33` `phi = tan^-1 (56/33)` Answer

Discussion

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