NIMCET 2012 Question Paper with solutions

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Question : 4 of 120

Marks: +1, -0

The value of

cot^{-1}

(21) +

cot^{-1}

(13) +

cot^{-1}

(-8) is

0
π
∞
$π/2$

Solution:

⇒

tan^{-1}(1/{21})

+

tan^{-1}(1/{13})

+

cot^{-1}

(-8)

(cot^{-1} = tan^{-1}1/x)

⇒

tan^{-1}\{{1/{21}+1/{13}}/{1-1/{21} 1/{13}}\}

+

cot^{-1}

(-8)

[\Since tan^{-1}x+tan^{-1}y=tan^{-1}({x+y}/{1-xy})]

⇒

tan^{-1}({34}/{272})

+

tan^{-1}(-1/8)

=

tan^{-1}({34}/{272})

+ π -

tan^{-1}(1/8)

⇒ π +

tan^{-1}\{{{34}/{272}-1/8}/{1+{34}/{272} 1/8}\}

=

tan^{-1}\{{34-34}/{2210}\}

+ π

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