Question Number 202041 by hardmath last updated on 19/Dec/23 | ||
![]() | ||
$$\mathrm{Simplify}:\:\:\:\frac{\sqrt{\mathrm{2}}\:−\:\mathrm{sin}\alpha\:−\:\mathrm{cos}\alpha}{\mathrm{sin}\alpha\:−\:\mathrm{cos}\alpha} \\ $$ | ||
Answered by cortano12 last updated on 19/Dec/23 | ||
![]() | ||
$$\:=\:\frac{\sqrt{\mathrm{2}}−\sqrt{\mathrm{2}}\:\left(\frac{\mathrm{1}}{\:\sqrt{\mathrm{2}}}\:\mathrm{sin}\:\alpha+\frac{\mathrm{1}}{\:\sqrt{\mathrm{2}}}\:\mathrm{cos}\:\alpha\right)}{\:\sqrt{\mathrm{2}}\:\left(\frac{\mathrm{1}}{\:\sqrt{\mathrm{2}}}\:\mathrm{sin}\:\alpha−\frac{\mathrm{1}}{\:\sqrt{\mathrm{2}}}\:\mathrm{cos}\:\alpha\right)} \\ $$$$\:=\:\frac{\mathrm{1}−\mathrm{cos}\:\left(\alpha−\mathrm{45}°\right)}{\mathrm{sin}\:\:\left(\alpha−\mathrm{45}°\right)} \\ $$$$\:=\:\mathrm{csc}\:\left(\alpha−\mathrm{45}°\right)−\mathrm{cot}\:\left(\alpha−\mathrm{45}°\right) \\ $$$$ \\ $$ | ||
Commented by hardmath last updated on 20/Dec/23 | ||
![]() | ||
$${thankyou}\:{Ser} \\ $$ | ||