Question

2tanA=√3then findsinA, cosA​

Answer 1

Answer:

secA + tanA = (2+ √5). ………………(1).

Formula:-

sec^2 A - tan^2 A = 1.

or, (sec A- tan A).(sec A + tan A)= 1.

or, (sec A - tan A).(2+√5) = 1.

or, sec A - tan A = 1/(2+√5).

or, sec A - tan A = 1×(2-√5)/(2+√5)×(2-√5). = (2-√5)/(4–5). = -2 +√5.

or, sec A - tan A = (-2+√5). …………(2).

Adding eqn.(1) and (2).

2.sec A = 2√5.

or, sec A = √5

or, cos A = 1/√5. Answer.

Answer 2

Answer:

2tanA=/3,

ஃtanA=/3/2

we know that

tanA= opp/adj

we need to find hypotenuse

by using Pythagoras theorem

opp^2+adj^2=hyp^2

=(/3)^2+(2)^2

=3+4=7

hyp^2=7

ஃhyp=/7

we know that

sinA=opp/hyp

cosA=adj/hyp

ஃsinA=/3//7

cosA=2//7

here(/)represents root.

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