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Question
Prove that `2sin^-1 3/5 - tan^-1 17/31 = pi/4`
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Solution
Let `sin^-1 3/5` = θ
Then sin θ = `3/5`
Where θ ∈ `[(-pi)/2, pi/2]`
Thus tan θ = `3/4`
Which gives θ = `tan^-1 3/4`.
Therefore, `2sin^-1 3/5 - tan^-1 17/31`
= `2theta - tan^-1 17/31`
= `2tan^-1 3/4 - tan^-1 17/31`
= `tan^-1 ((2 * 3/4)/(1 - 9/16)) - tan^-1 17/31`
= `tan^-1 24/7 - tan^-1 17/31`
= `tan^-1 ((24/7 - 17/31)/(1 + 24/7 * 17/31))`
= `pi/4`
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