Question

sec²A=2+2tanA Then find tanA = ??????

Answer 1

sec² A = 2 + 2 tan A
1 + tan² A = 2 + 2 tan A
tan² A - 2 tan A + 1 - 2 = 0
tan² A - 2 tan A - 1 = 0

By using abc formula:
a = 1
b = -2
c = -1

x1,2
= [-b ± √(b² - 4ac)]/2a
= [-(-2) ± √{(-2)² - 4 . 1 . (-1)}]/(2 . 1)
= [2 ± √(4 + 4)]/2
= (2 ± √8)/2
= (2 ± 2√2)/2
= 1 ± √2

x1 = 1 + √2
x2 = 1 - √2

Answer 2

Hey there !!!!!!!

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sec²A=2+2tanA -------Equation 1

We know that  sec²A-tan²A=1

 sec²A=1+tan²A

Substituting  sec²A=1+tan²A in Equation 1

1+tan²A=2+2tanA

tan²A-2tanA+1-2=0

tan²A-2tanA-1=0---Equation 2

Equation 2 is of the form ax²+bx+c=0

where x=-b+√(b²-4ac)/2a or -b-√(b²-4ac)/2a

Comparing tan²A-2tanA-1=0 with ax²+bx+c=0

a=1 b=-2 c=-1 and x=tanA

So,

tanA=-b+√b²-4ac/2a or -b-√b²-4ac/2a

tanA= (2+√(2²-(-1*4)))/2  or (2-√(2²-(-1*4)))/2

tanA=(2+√4+4 )/2 or (2-√4+4 )/2

tanA=(2+√8)/2  or (2-√8)/2

tanA=2+2√2/2 or 2-2√2/2

tanA=1+√2 or 1-√2

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Hope this helped you...............