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Question
Show that `cos(2tan^-1 1/7) = sin(4tan^-1 1/3)`
Solution
L.H.S. `cos(2tan^-1 1/7)`
= `cos[cos^-1 (1 - 1/49)/(1 + 1/49)]` .....`[because 2tan^-1x = cos^-1 (1 - x^2)/(1 + x^2)]`
= `cos[cos^-1 48/50]`
= `cos[cos^-1 24/25]`
= `24/25`
R.H.S `sin[4 tan^-1 1/3]`
= `sin[2tan^-1 ((2 xx 1/3)/(1 - 1/9))]` .....`[because 2tan^-1x = tan^-1 (2x)/(1 - x^2)]`
= `sin[2tan^-1 ((2/3)/(8/9))]`
= `sin[2tan^-1 3/4]`
= `sin[sin^-1 (2 xx 3/4)/(1 + 9/16)]` ......`[because 2tan^-1x = sin^-1 (2x)/(1 + x^2)]`
= `sin[sin^-1 24/25]`
⇒ `24/25`
L.H.S. = R.H.S.
Hence poved.
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