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Find the value of `4 tan^-1 (1/5) - tan^-1 (1/239) ` |
Answer» Here, we will use the following properties, `tan^(-1)(x)+tan^(-1)(y) = tan^(-1)((x+y)/(1-xy))` `tan^(-1)(x)-tan^(-1)(y) = tan^(-1)((x-y)/(1+xy))` Now, `4tan^(-1)(1/5) = 2(tan^(-1)(1/5)+tan^(-1)(1/5))` `=2(tan^(-1)((2/5)/(1-1/25)))` `=2(tan^(-1)(5/12))` `=tan^(-1)(5/12)+tan^(-1)(5/12)` `= tan^(-1) (((5/12)+(5/12))/(1- (5/12)(5/12)))` `=tan^(-1)((10/12)/(119/144))` `4tan^(-1)(1/5)= tan^(-1) (120/119)` Now, `4tan^(-1)(1/5) - tan^(-1)(1/239) = tan^(-1) (120/119)-tan^(-1)(1/239)` `= tan^(-1)((120/119-1/239)/(1+(120/119)(1/239)))` `=tan^(-1)((120*239-119)/(119*239+120))` `=tan^(-1)(28561/28561) = tan^(-1)(1) = pi/4` |
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