If tan^-1 ⁡ ( 11+1.2 )+ tan^-1 ⁡ ( 11+2.3 )+ …+ tan^-1 ( 11+n (n+1 ) )= tan^-1 ⁡ [x ], then x is equal to

Question

Accepted Answer

Correct answer: D. (d) Given tan^-1 ⁡ ( 11+1.2 )+ tan^-1 ⁡ ( 11+2.3 )+ … + tan^-1 ⁡ ( 11+n (n+1 ) )= tan^-1 ⁡ (x ) ⇒ tan^-1 ⁡ ( 2-11+1.2 )+ tan^-1 ⁡ ( 3-21+2.3 )+ … + tan^-1 ⁡ ( n+1-n1+n (n+1 ) )= tan^-1 ⁡ (x ) ⇒ tan^-1 ⁡ (2 )- tan^-1 ⁡ (1 )+ tan^-1 ⁡ (3 )- tan^-1 ⁡ (2 )+ … + tan^-1 ⁡ (n+1 )- tan^-1 ⁡ (n )= tan^-1 ⁡ (x ) ⇒ tan^-1 ⁡ (n+1 )- tan^-1 ⁡ (1 )= tan^-1 ⁡ (x ) ( ∵ tan^-1 ⁡ (A )- tan^-1 ⁡ (B )= tan^-1 ⁡ ( A-B1+A B )) ⇒ tan^-1 ⁡ ( nn+2 )= tan^-1 ⁡ (x ) ⇒x= nn+2

$If tan^{- 1} (\frac{1}{1 + 1.2}) + tan^{- 1} (\frac{1}{1 + 2.3}) + \dots + tan^{- 1} (\frac{1}{1 + n ( n + 1 )}) = tan^{- 1} [x], then x is equal to$

Solution

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