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प्रश्न
Prove that `tan [2 tan^-1 (1/2) - cot^-1 3] = 9/13`.
सिद्धांत
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उत्तर
L.H.S. = `tan [2 tan^-1 (1/2) - cot^-1 3]`
⇒ L.H.S. = `tan[tan^-1 (2 xx 1/2)/(1 - (1/2)^2) - tan^-1 1/3]`
[Because `2 tan^-1x = tan^-1 ((2x)/(1 - x^2))` if – 1 < x < 1 and `cot^-1x = tan^-1 1/x, x > 0`]
⇒ L.H.S. = `tan[tan^-1 4/3 - tan^-1 1/3]`
⇒ L.H.S. = `tan[tan^-1 ((4/3 - 1/3))/(1 + 4/3 xx 1/3)]`
⇒ L.H.S. = `tan[tan^-1 1/(13//9)]`
⇒ L.H.S. = `9/13` = R.H.S.
∴ `tan[2tan^-1 (1/2) - cot^-1 3] = 9/13`
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