InterviewSolution
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InterviewSolution
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Find the value of`2cos^(-1)3/(sqrt(13))+cot^(-1)(16)/(63)+1/2cos^(-1)7/(25)` |
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Answer» `2cos^-1 (3/sqrt13) + cot^-1 (16/63) +1/2cos^-1 (7/25)` Now, `cos^-1(3/sqrt13) = tan^-1(2/3)` `:. 2cos^-1(3/sqrt13) = 2 tan^-1(2/3) = tan^-1((2*(2/3))/(1-(2/3)^2)) = tan^-1((4/3)/(5/9)) = tan^-1(12/5)` `cot^-1(16/63) = tan^-1(63/16)` Let `1/2cos^-1(7/25) = tan^-1x` `=> cos^-1(7/25) = 2tan^-1x` `=> tan^-1(24/7) = tan^-1((2x)/(1-x^2))` `=>(2x)/(1-x^2) = 24/7` Solving it , we get , ` x =3/4` `:. 1/2cos^-1(7/25) = tan^-1(3/4)` Putting these values in the given equation, `tan^-1(12/5)+tan^-1(63/16)+tan^-1(3/4)` `=(tan^-1(12/5)+tan^-1(3/4))+tan^-1(63/16)` As, `12/5 gt 0, 3/4 gt 0 and 12/5*3/4 gt 1`, So, it becomes, `=pi+ tan^-1((12/5+3/4)/(1-(12/5)(3/4)))+tan^-1(63/16)` `=pi- tan^-1(63/16)+tan^-1(63/16)` `=pi` `:. 2cos^-1 (3/sqrt13) + cot^-1 (16/63) +1/2cos^-1 (7/25) = pi` |
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