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प्रश्न
Prove that `cot(pi/4 - 2cot^-1 3)` = 7
योग
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उत्तर
L.H.S. `cot(pi/4 - 2cot^-1 3)`
= `cot[tan^-1(1) - 2 tan^-1 1/3]` ......`[because cot^-1x = tan^-1 1/x]`
= `cot[tan^-1(1) - tan^-1 (2 xx 1/3)/(1 - (1/3)^2)]` ......`[because 2tan^-1x = tan^-1 (2x)/(1 - x^2)]`
= `cot[tan^-1(1) - tan^-1 (2/3)/(8/9)]`
= `cot[tan^-1(1) - tan^-1 3/4]`
= `cot[tan^-1 ((1 - 3/4)/(1 + 1 xx 3/4))]`
= `cot[tan^-1 ((1/4)/(7/4))]`
= `cot[tan^-1 1/7]` ......`[because tan^-1 1/x = cot^-1x]`
= `cot[cot^-1 (7)]`
= 7 R.H.S
Hence Proved.
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