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Find the value of 4tan-1 15-tan-1 1239 - Mathematics

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Question

Find the value of `4tan^-1  1/5 - tan^-1  1/239`

Sum
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Solution

`4tan^-1  1/5 - tan^-1  1/239`

= `2(2tan^-1  1/5) - tan^-1  1/239`

= `2tan^-1  (2/5)/(1 - (1/5)^2) - tan^-1  1/239`  .....`(because 2tan^-1x = tan^-1  (2x)/(1 - x^2))`

= `2tan^-1  (2/5)/(24/25) - tan^-1  1/239`

= `2tan^-1  5/12 - tan^-1  1/239`

= `2tan^-1  (2/5)/(1 - (1/5)^2) - tan^-1  1/239` .....`(because 2tan^-1x = tan^-1  (2x)/(1 - x^2))`

= `2tan^-1  (2/5)/(24/25) - tan^-1  1/239`

= `2tan^-1  5/12 - tan^-1  1/239`

= `tan^-1  (2*5/12)/(1 - (5/12)^2) - tan^-1  1/239`  ......`(because 2tan^-1x = tan^-1  (2x)/(1 - x^2))`

= `tan^-1  (144 xx 5)/(119 xx 6) - tan^-1  1/239`

= `tan^-1  120/119 - tan^-1  1/239`

= `tan^-1  (120/119 - 1/239)/(1 + 120/119 * 1/239)`  ......`(because tan^-1x - tan^-1y = tan^-1  (x - y)/(1 + xy))`

= `tan^-1  (120 xx 239 - 119)/(119 xx 239 + 120)`

= `tan^-1  (28680 - 119)/(28441 + 120)`

= `tan^-1  28561/28561`

= `tan^-1 1 = pi/4`

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Chapter 2: Inverse Trigonometric Functions - Exercise [Page 36]

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NCERT Exemplar Mathematics [English] Class 12
Chapter 2 Inverse Trigonometric Functions
Exercise | Q 17 | Page 36

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