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

Question

06-02-2022
Mathematics

Answered

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

Sagot :

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surjithayer surjithayer · Answer 1 · 2022-02-06T20:44:06-05:00

surjithayer surjithayer

06-02-2022

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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Sagot :

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