Indian Institute of Foreign Trade 2008 Solved Paper

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Question : 46 of 150

Marks: +1, -0

\cot^{-1}[{√{1-\sinα}+√{1+\sinα}}/{√{1-\sinα}-√{1+\sinα}}]=

$2π-a$
$π-1/2a$
$1/2a-3π$
None of the above

Solution:

\cot^{-1}[{√{\sin^{2}(a/2)+\cos^{2}(a/2)-2\sin(a/2)\cos(a/2)}+√{\sin^{2}(a/2)+\cos^{2}(a/2)+2\sin(a/2)\cos(a/2)}}/{√{\sin^{2}(a/2)+\cos^{2}(a/2)-2\sin(a/2)\cos(a/2)}-√{\sin^{2}(a/2)+\cos^{2}(a/2)+2\sin(a/2)\cos(a/2)}}]

\cot^{-1}[{\cos(a/2)-\sin(a/2)+\cos(a/2)+\sin(a/2)}/{\cos(a/2)-\sin(a/2)+\cos(a/2)-\sin(a/2)}]

\cot^{-1}[{-2\cos(a/2)}/{2\sin(a/2)}]

\cot[\cot(−a/2)]

We know that

\cot^{-1}[-x]=π-\cot^{-1}[x]

Therefore,

\cot^{-1}[\cot(-a/2)]=π-\cot^{-1}[\cot(a/2)]=π-1/2a

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