We have an answer to "Solve the equation 1−sin⁡x1+sin⁡x=(sec⁡x−tan⁡x)2"

Nannie Mack

Answered question

2020-10-28

Solve the equation

\frac{1 - \sin x}{1 + \sin x} = (\sec x - \tan x)^{2}

hosentak

Skilled2020-10-29Added 100 answers

(\sec (x) - \tan (x))^{2} = (\frac{1}{\cos (x)} - \frac{\sin (x)}{\cos (x)})^{2}

= (\frac{1 - \sin (x)}{\cos (x)})^{2}

= \frac{(1 - \sin (x))^{2}}{c o s^{2} (x)}

= \frac{(1 - \sin (x))^{2}}{1 - s i n^{2} (x)}

= \frac{- (1 - \sin (x))^{2}}{(\sin (x) + 1) (\sin (x) - 1)}

= \frac{1 - \sin (x)}{\sin (x) + 1}

= \frac{1 - \sin (x)}{1 + \sin (x)}

Therefore

\frac{1 - \sin (x)}{1 + \sin (x)} = (\sec (x) - \tan (x))^{2}

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