If sec a =5/4 then evaluate tan a/(1+tan2a)
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Let's take a triangle ABC right angled at B,
Now sec a = Hypotenuse/ Base = AC/AB = 5/4
Using Pythagoras theorem,
H² = P² + B²
P² = H² - B²
P² = 25 - 16
P = √9 = 3
so BC = 3
now tan a = Perpendicular/Base = BC/AB
tan a = 3/4
and 1+ tan²a = sec²a
putting values we get,
tan a/(1+tan²a)
=tan a / (sec²a)
= (3/4)/ (5/4)²
= 12/25
Now sec a = Hypotenuse/ Base = AC/AB = 5/4
Using Pythagoras theorem,
H² = P² + B²
P² = H² - B²
P² = 25 - 16
P = √9 = 3
so BC = 3
now tan a = Perpendicular/Base = BC/AB
tan a = 3/4
and 1+ tan²a = sec²a
putting values we get,
tan a/(1+tan²a)
=tan a / (sec²a)
= (3/4)/ (5/4)²
= 12/25
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