Sum
Prove that sin `[tan^-1 ((1 - x^2)/(2x)) + cos^-1 ((1 - x^2)/(1 + x^2))]` = 1
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Solution
L.H.S. = sin `[tan^-1 ((1 - x^2)/(2x)) + cos^-1 ((1 - x^2)/(1 + x^2))]`
Substituting x = tan θ, we get
L.H.S. = sin `[tan^-1 ((1 - tan^2theta)/(2tantheta)) + cos^-1 ((1 - tan^2theta)/(1 + tan^2theta))]`
= `sin[tan^-1 (1/tan 2theta) + cos^(-1) (cos 2theta)]`
= `sin[tan^-1 (cot 2theta) + cos^-1 (cos 2theta)]`
= `sin[tan^-1 {tan (pi/2 - 2theta)} + 2theta]`
= `sin(pi/2 - 2theta + 2theta)`
= `sin(pi/2)`
= 1
= R.H.S.
Concept: Inverse Trigonometric Functions
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