Question

Can anyone answer this

Answer 1

Answer:

1+sin^2 theta=3 sin theta cos theta (we know that sin^2 theta + cos^2 theta =1)

= ( sin^2 theta + cos^2 theta  ) + sin ^2 theta = 3 sin theta cos theta

= sin^2 theta + cos^2 theta + sin ^2 theta = 3 sin theta cos theta

= cos^2 theta + 2 sin^2 theta = 3 sin theta cos theta

On dividing by cos^2 theta, we get

= 1 + 2 tan^2 theta = 3 tan theta

Let tan theta = b

 2b^2  - 3b + 1 = 0

= (2b-1)(b-1) = 0

b = 1 or 1/2

So, tan theta = 1 or 1/2.

Hope this helps!

PLEASE MARK IT BRAINLIEST

Answer 2

Step-by-step explanation:

(1+sin²A)/cos²A= (3sinAcosA)/cos²A

sec²A+tan²A=3tanA

1+tan²A+tan²A=3tanA

2tan²A-3tanA+1=0

2tan²A-2tanA-tanA+1=0

2tanA(tanA-1)-1(tanA-1)=0

(2tanA-1)(tanA-1)=0

tanA=1/2 or tanA=1