SSC CGL 11 Aug 2017 Shift 1 Paper

Question : 71 of 100

Marks: +1, -0

\cot A = n /( n + 1)

and

\cot B = 1 / (2n + 1),

then what is the value of

\cot(A + B) ?

Solution:

\cot A = n /( n + 1)

\cot B = 1 / (2n + 1)

\cot(A + B) ={\cot A \cot B -1} /{\cot A + \cot B}

={n /( n + 1) ×1 / (2n + 1) -1}/{n /( n + 1)+1 / (2n + 1)}

={ n - ( n + 1) (2n + 1) }/{n (2 n + 1) + n + 1}

={n - 2 n^2 - n - 2 n - 1}/{2 n^2 + n + n + 1}

{-2 n^ 2 - 2 n - 1}/{2 n^2 + 2n + 1 }

-({2 n^2 + 2n + 1 })/({2 n^2 + 2n + 1 })=-1

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