Question Number 160933 by blackmamba last updated on 09/Dec/21
![In the given equation below , applying the formula for the derivative of inverse trigonometric functions , what is the ′′u ′′ from the given function. y = cosec^(−1) [ sin (((1+sin x)/(cos x)))]](Q160933.png)
$$\:{In}\:{the}\:{given}\:{equation}\:{below}\:,\:{applying} \\ $$$${the}\:{formula}\:{for}\:{the}\:{derivative}\:{of} \\ $$$$\:{inverse}\:{trigonometric}\:{functions}\:, \\ $$$$\:{what}\:{is}\:{the}\:''{u}\:''\:{from}\:{the}\:{given}\:{function}. \\ $$$$\:{y}\:=\:\mathrm{cosec}^{−\mathrm{1}} \left[\:\mathrm{sin}\:\left(\frac{\mathrm{1}+\mathrm{sin}\:{x}}{\mathrm{cos}\:{x}}\right)\right] \\ $$
Answered by bobhans last updated on 10/Dec/21
![y=cosec^(−1) [sin (((1+sin x)/(cos x)))] cosec y= sin (((1+sin x)/(cos x))) (−cosec y cot y )y′= (((1+sin x)/(cos^2 x))) cos (((1+sin x)/(cos x))) y′ = (((1+sin x)/(cos^2 x)))tan (((1+sin x)/(cos x)))tan (cosec^(−1) [sin (((1+sin x)/(cos x)))])](Q160991.png)
$$\:\mathrm{y}=\mathrm{cosec}^{−\mathrm{1}} \left[\mathrm{sin}\:\left(\frac{\mathrm{1}+\mathrm{sin}\:\mathrm{x}}{\mathrm{cos}\:\mathrm{x}}\right)\right] \\ $$$$\:\mathrm{cosec}\:\mathrm{y}=\:\mathrm{sin}\:\left(\frac{\mathrm{1}+\mathrm{sin}\:\mathrm{x}}{\mathrm{cos}\:\mathrm{x}}\right)\: \\ $$$$\:\:\left(−\mathrm{cosec}\:\mathrm{y}\:\mathrm{cot}\:\mathrm{y}\:\right)\mathrm{y}'=\:\left(\frac{\mathrm{1}+\mathrm{sin}\:\mathrm{x}}{\mathrm{cos}\:^{\mathrm{2}} \mathrm{x}}\right)\:\mathrm{cos}\:\left(\frac{\mathrm{1}+\mathrm{sin}\:\mathrm{x}}{\mathrm{cos}\:\mathrm{x}}\right) \\ $$$$\:\:\mathrm{y}'\:=\:\left(\frac{\mathrm{1}+\mathrm{sin}\:\mathrm{x}}{\mathrm{cos}\:^{\mathrm{2}} \mathrm{x}}\right)\mathrm{tan}\:\left(\frac{\mathrm{1}+\mathrm{sin}\:\mathrm{x}}{\mathrm{cos}\:\mathrm{x}}\right)\mathrm{tan}\:\left(\mathrm{cosec}^{−\mathrm{1}} \left[\mathrm{sin}\:\left(\frac{\mathrm{1}+\mathrm{sin}\:\mathrm{x}}{\mathrm{cos}\:\mathrm{x}}\right)\right]\right) \\ $$