Proving$\frac{(\sec{v}-\tan{v})^2+1}{\csc{v}(\sec{v}-\tan{v})}=2\tan{v}$

Question

Please show me how to prove this identity:

Original: ((sec v - tan v)^2 +1)/csc v ( sec v - tan v) = 2tan v

$$\frac{(\sec{v}-\tan{v})^2+1}{\csc{v}(\sec{v}-\tan{v})}=2\tan{v}$$

Thank you

Answer 1

$\frac{(\sec{v}-\tan{v})^2+1}{\csc{v}(\sec{v}-\tan{v})}$

$=\frac{(\sec{v})^2-2\frac{\sin{v}}{(\cos{v})^2}+(\tan{v})^2+1}{\csc{v}\sec{v}-\frac{1}{\cos{v}}}$

$=\frac{2\sec^2{v}-2\frac{\sin{v}}{\cos^2{v}}}{\frac{1}{\cos{v}.\sin{v}}-\frac{\sin{v}}{\cos{v}.\sin{v}}}$

$=\frac{2\frac{1-\sin{v}}{\cos^2{v}}}{\frac{1-\sin{v}}{\cos{v}.\sin{v}}}$

$=\frac{2\frac{1-\sin{v}}{\cos^2{v}}}{\frac{1-\sin{v}}{\cos{v}.\sin{v}}}$

$=\frac{2\cos{v}.\sin{v}}{\cos^2{v}}$

$=2\tan{v}$

Answer 2

$$\dfrac{(\sec t-\tan t)^2+1}{\sec t-\tan t}=\sec t-\tan t+\dfrac1{\sec t-\tan t}=2\sec t$$

$$2\tan t\csc t=?$$

Answer 3

You have to show (cross multiplied)

$$\frac{(\sec{v}-\tan{v})^2+1}{(\sec{v}-\tan{v})}=\frac{2}{\cos{v}}$$

the shorthand you can guess

$$(\sec{v}-\tan{v})= \frac{1-s}{c}$$

so you have to show

$$ \frac{1-s}{c}+ \frac{c}{1-s} = \frac{1+s^2-2s +c^2}{c(1-s)} = \frac{2(1-s)}{c(1-s)} = \frac{2}{c} $$