If f x -0 x tan -1 t dt -0tan x -1 t dt- if 0- x -2- thenf -4 is equal toA- -2B- -4C- -2D- -4

The correct option is A π/2f ( π/4 )=∫_0^cot ( π/4 ) tan^ 1t dt + ( π/4 )=∫_0^tan ( π/4 ) cot^ 1t dt = ∫_0^1 tan^ 1t dt + ∫_0^1cot^ 1 dt =∫_0^1(tan^ 1t+cot ...

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Byju's Answer

Standard XII

Mathematics

Algebra of Derivatives

If f x =∫0 x ...

Question

If $f (x) = \int_{0}^{c o t x} t a n^{- 1} (t) d t + \int_{0}^{t a n x} c o t^{- 1} t d t, i f 0 < x < \frac{π}{2}, t h e n f (\frac{π}{4})$ is equal to

\frac{π}{2}

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- \frac{π}{2}

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\frac{π}{4}

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- \frac{π}{4}

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Solution

The correct option is A $\frac{π}{2}$
$f (\frac{π}{4}) = \int_{0}^{c o t (\frac{π}{4})} t a n^{- 1} t d t + (\frac{π}{4}) = \int_{0}^{t a n (\frac{π}{4})}$ $c o t^{- 1} t d t = \int_{0}^{1} t a n^{- 1} t d t + \int_{0}^{1} c o t^{- 1} d t$
$= \int_{0}^{1} (t a n^{- 1} t + c o t^{- 1} t) d t = \frac{π}{2} \int_{0}^{1} d t = \frac{π}{2}$

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If f(x)=∫cot x0tan−1(t)dt+∫tan x0cot−1t dt,if0<x<π2,thenf(π4) is equal to

The correct option is A π2f(π4)=∫cot(π4)0tan−1t dt+(π4)=∫tan(π4)0 cot−1t dt=∫10tan−1t dt+∫10cot−1 dt =∫10(tan−1t+cot−1t)dt=π2∫10dt=π2

If $f (x) = \int_{0}^{c o t x} t a n^{- 1} (t) d t + \int_{0}^{t a n x} c o t^{- 1} t d t, i f 0 < x < \frac{π}{2}, t h e n f (\frac{π}{4})$ is equal to