2 years ago

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

Mathematics

1 answer:

stepladder [879]2 years ago

4 0

Answer:

Step-by-step explanation:

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

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