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Question Number 130589 by bramlexs22 last updated on 27/Jan/21

 Prove the identity tan^(−1) (x)+cot^(−1) (x)=π/2

$$\:\mathrm{Prove}\:\mathrm{the}\:\mathrm{identity}\:\mathrm{tan}^{−\mathrm{1}} \left(\mathrm{x}\right)+\mathrm{cot}^{−\mathrm{1}} \left(\mathrm{x}\right)=\pi/\mathrm{2} \\ $$

Answered by EDWIN88 last updated on 27/Jan/21

 If f(x)=tan^(−1) (x)+cot^(−1) (x)  then f ′(x)=(1/(1+x^2 )) − (1/(1+x^2 )) = 0  for ∀ values of x. Therefore   f(x)= C , a constant.  To determine the value of C  we put x=1 since we can evaluate  f(1) exactly. Then C=f(1)=   tan^(−1) (1)+cot^(−1) (1)=(π/4)+(π/4)=(π/2)

$$\:{If}\:{f}\left({x}\right)=\mathrm{tan}^{−\mathrm{1}} \left({x}\right)+\mathrm{cot}^{−\mathrm{1}} \left({x}\right) \\ $$$${then}\:{f}\:'\left({x}\right)=\frac{\mathrm{1}}{\mathrm{1}+{x}^{\mathrm{2}} }\:−\:\frac{\mathrm{1}}{\mathrm{1}+{x}^{\mathrm{2}} }\:=\:\mathrm{0} \\ $$$${for}\:\forall\:{values}\:{of}\:{x}.\:{Therefore}\: \\ $$$${f}\left({x}\right)=\:{C}\:,\:{a}\:{constant}. \\ $$$${To}\:{determine}\:{the}\:{value}\:{of}\:{C} \\ $$$${we}\:{put}\:{x}=\mathrm{1}\:{since}\:{we}\:{can}\:{evaluate} \\ $$$${f}\left(\mathrm{1}\right)\:{exactly}.\:{Then}\:{C}={f}\left(\mathrm{1}\right)= \\ $$$$\:\mathrm{tan}^{−\mathrm{1}} \left(\mathrm{1}\right)+\mathrm{cot}^{−\mathrm{1}} \left(\mathrm{1}\right)=\frac{\pi}{\mathrm{4}}+\frac{\pi}{\mathrm{4}}=\frac{\pi}{\mathrm{2}} \\ $$

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