Question #480b6

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Answer 1 · 2016-08-12T16:24:56

$1$

Use $sin x = cos (pi/2-x), cos x = sin (pi/2-x)$ ,

$tan x = cot (pi/2-x) and cot x = tan(pi/2-x)$ .

Also, use $f^(-1)f(X)=X.$

The given expression is

$(sin^(-1)sin(pi/2-x)+cos^(-1)sin(pi/2-x))/(tan^(-1)tan(pi/2-x)+cot^(-1)cot(pi/2-x))$

$=((pi/2-x)+(pi/2-x))/((pi/2-x)+(pi/2-x))$

$=1$

Here, as $x to pi/2$ , the limit is also 1..