tan, cot, sec

[red and white kop!]

solve this equation for A in the range 0<A<360
tanA + cotA = 2secA
i found A= 30, 150, 210, 330 degrees and all of these results fit the tangent function i solved for. but the answer includes only A = 30 and 150 degrees. why is this? :x

 

[Aladdin]

\(\displaystyle tanA=\frac{sinA}{cosA}\)

\(\displaystyle cotA=\frac{cosA}{sinA}\)

Substitute :

\(\displaystyle \frac{sinA}{cosA}+\frac{cosA}{sinA}=2secA\) ....

What does secA= ???(you actually know it right ?)

Same denominator .... & solve, Remember sin^2+cos^2=1

 

[red and white kop!]

ok i understand now, its bcos i squared the equation at a certain point. thanks for your helpfulness
kop rules

 

[Deleted member 4993]

solve this equation for A in the range 0<A<360
tanA + cotA = 2secA

tanA + 1/tanA = 2 secA

sec[sup:27x5m8o9]2[/sup:27x5m8o9]A/tanA = 2 secA

secA = 2 tanA

1 = 2 sinA

How did you solve to get 210 and 330 in your solution?

i found A= 30, 150, 210, 330 degrees and all of these results fit the tangent function i solved for. but the answer includes only A = 30 and 150 degrees. why is this? :x