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प्रश्न
Show that `2tan^-1 (-3) = (-pi)/2 + tan^-1 ((-4)/3)`
योग
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उत्तर
L.H.S. `2tan^-1 (-3) = -2tan^-1 (3)`
= `- cos^-1 [(1- (3)^2)/(1 + (3)^2)]` ......`[because 2tan^-1x = cos^-1 ((1 - x^2)/(1 + x^2))]`
= `-cos^-1 ((1 - 9)/(1 + 9))`
= `- cos^-1 ((-8)/10)`
= `- cos^-1 ((-4)/5)`
= `- [pi - cos^-1 (4/5)]`
= `- pi + cos^-1 4/5`
= `- pi + tan^-1 (3/4)` ......`[because cos^-1 4/5 = tan^-1 3/4]`
= `- pi + pi/2 - cot^-1 (3/4)` ......`[tan^-1x = pi/2 - cot^-1x]`
= `(-pi)/2 - cot^-1 (3/4)`
= `(-pi)/2 - tan^-1 (4/3)` .......`[because tan^-1x = cot^-1 1/x]`
= `(-pi)/2 + tan^-1 (- 4/3)` R.H.S
Hence proved.
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