if 1 + sin^2 theta = 3 sin theta×cos theta then prove that tan theta = 1/2 or 1
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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!
= ( 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!
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