Question
Prove that `tan {pi/4 + 1/2 cos^(-1) a/b} + tan {pi/4 - 1/2 cos^(-1) a/b} = (2b)/a`
Solution
Let `cos^(-1) (a/b) = 0`
Then `cos theta = a/b`
L.H.S:
`tan {pi/4 + 1/4 cos^(-1) a/b} + tan (pi/4 - 1/2cos^(-1) a/b)`
= `tan (pi/4 + theta/2) + tan (pi/4 - theta/2)`
`= (1 + tan theta/2)/(1 - tan theta/2) + (1 - tan theta/2)/(1 + tan theta/2)`
`((1 + tan theta/2)^2 + (1 - tan theta/2)^2)/(1 - tan^2 theta/2)`
= `2((1 + tan^2 theta/2)/(1 - tan^2 theta/2))`
= `2/(cos theta)` `[∵ cos 2 theta = (1 - tan^2 theta)/(1 + tan^2 theta)]`
= `(2b)/a`
=RHS
LHS = RHS
Hence Proved
Is there an error in this question or solution?
Solution Prove that Tan {Pi/4 + 1by2 Cos^(-1) Abyb} + Tan {Piby4 - 1by2 Cos^(-1) Abyb} = (2b)Bya` Concept: Properties of Inverse Trigonometric Functions.
