jacqueline04r jacqueline04r
  • 07-02-2022
  • Mathematics
contestada

Question 12 please
Tan(a)/
Sec(a)-cos(a)

Question 12 please Tana Secacosa class=

Respuesta :

surjithayer surjithayer
  • 07-02-2022

Answer:

Step-by-step explanation:

[tex]\frac{tan ~\alpha }{sec~\alpha -cos~\alpha } \\=\frac{\frac{sin~\alpha }{cos~\alpha } }{\frac{1}{cos~\alpha } -cos~\alpha } \\multiply~numerator~and~denominator~by~cos~\alpha \\=\frac{sin~\alpha }{1-cos^2~\alpha }=\frac{sin~\alpha }{sin ^2~\alpha } \\=\frac{1}{sin~\alpha } \\=cosec~\alpha[/tex]

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