Question

(secA-TanA)=1/3, then find cotA​

Answer 1

$\sec(a) = \frac{1}{3} + \tan(a) \\ squaring \: both \: the \: sides \\ { \sec(a) }^{2} = \frac{1}{9} + { \tan(a) }^{2} + \frac{2}{3} \tan(a) \\ { \sec(a) }^{2} - { \tan(a) }^{2} = \frac{1}{9} + \frac{2}{3} \tan(a) \\ \frac{ - 2}{3} tan(a) = \frac{1}{9} \\ \tan(a) = \frac{1}{9} \times \frac{3}{ - 2} \\ tan(a) = \frac{1}{ - 6} \\ so \\ \cot(a) = - 6$

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Answer 2

Answer:

Step-by-step explanation:secA-tanA=1/3

SecA=1/3+tanA

Squaring on both sides

Sec2A=1/9+tan2A+2/3tanA

Sec2A-tan2A=1/9+2/3tanA

1=1/9+2/3tanA

1-1/9=2/3tanA

8/9=2/3tanA

8/9*2/3=tanA

4/3=tanA

CotA=3/4