Question Number 208624 by vipin last updated on 19/Jun/24
Answered by Berbere last updated on 19/Jun/24
![(1/(cosec^− (−(√2))))=sin^(−1) (−(1/( (√2))))=−(π/4) f(x).tan^(−1) ((((√(1+x))−(√(1−x)))/( (√(1+x))+(√(1−x)))))=tan^(−1) (((1−(√((1−x)/(1+x))))/(1+(√((1−x)/(1+x))))))...E g(y)=tan^(−1) (((1−y)/(1+y)))=(π/4)−tan^(−1) (y)... prof g′(y)=((−2)/((1+y)^2 )).(1/(1+(((1−y)/(1+y)))^2 ))=−(1/(1+y^2 )) g(y)=−tan^(−1) (y)+(π/4) f(x)=−tan^(−1) ((√((1−x)/(1+x))))+(π/4) x=cos(a)⇒(√((1−x)/(1+x)))=∣tan ((a/2))∣;a∈[0,π] f(x)=(π/4)−(a/2)=(π/4)−((cos^(−1) (x))/2)](https://www.tinkutara.com/question/Q208631.png)
$$\frac{\mathrm{1}}{{cosec}^{−} \left(−\sqrt{\mathrm{2}}\right)}=\mathrm{sin}^{−\mathrm{1}} \left(−\frac{\mathrm{1}}{\:\sqrt{\mathrm{2}}}\right)=−\frac{\pi}{\mathrm{4}} \\ $$$${f}\left({x}\right).\mathrm{tan}^{−\mathrm{1}} \left(\frac{\sqrt{\mathrm{1}+{x}}−\sqrt{\mathrm{1}−{x}}}{\:\sqrt{\mathrm{1}+{x}}+\sqrt{\mathrm{1}−{x}}}\right)=\mathrm{tan}^{−\mathrm{1}} \left(\frac{\mathrm{1}−\sqrt{\frac{\mathrm{1}−{x}}{\mathrm{1}+{x}}}}{\mathrm{1}+\sqrt{\frac{\mathrm{1}−{x}}{\mathrm{1}+{x}}}}\right)…{E} \\ $$$${g}\left({y}\right)=\mathrm{tan}^{−\mathrm{1}} \left(\frac{\mathrm{1}−{y}}{\mathrm{1}+{y}}\right)=\frac{\pi}{\mathrm{4}}−\mathrm{tan}^{−\mathrm{1}} \left({y}\right)… \\ $$$${prof}\:\:{g}'\left({y}\right)=\frac{−\mathrm{2}}{\left(\mathrm{1}+{y}\right)^{\mathrm{2}} }.\frac{\mathrm{1}}{\mathrm{1}+\left(\frac{\mathrm{1}−{y}}{\mathrm{1}+{y}}\right)^{\mathrm{2}} }=−\frac{\mathrm{1}}{\mathrm{1}+{y}^{\mathrm{2}} } \\ $$$${g}\left({y}\right)=−\mathrm{tan}^{−\mathrm{1}} \left({y}\right)+\frac{\pi}{\mathrm{4}} \\ $$$${f}\left({x}\right)=−\mathrm{tan}^{−\mathrm{1}} \left(\sqrt{\frac{\mathrm{1}−{x}}{\mathrm{1}+{x}}}\right)+\frac{\pi}{\mathrm{4}} \\ $$$${x}={cos}\left({a}\right)\Rightarrow\sqrt{\frac{\mathrm{1}−{x}}{\mathrm{1}+{x}}}=\mid\mathrm{tan}\:\left(\frac{{a}}{\mathrm{2}}\right)\mid;{a}\in\left[\mathrm{0},\pi\right] \\ $$$${f}\left({x}\right)=\frac{\pi}{\mathrm{4}}−\frac{{a}}{\mathrm{2}}=\frac{\pi}{\mathrm{4}}−\frac{\mathrm{cos}^{−\mathrm{1}} \left({x}\right)}{\mathrm{2}} \\ $$$$ \\ $$$$ \\ $$$$ \\ $$