Question

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

Answer 1

Here is your answer:-

Answer 2

Answer:  Proved.

Step-by-step explanation:  Given that

$1+\sin^2\theta=3\sin\theta\cos\theta.$

We are to prove the following:

$\tan \theta=1~\textup{or }\dfrac{1}{2}.$

We will be using the following identities:

$(i)~\dfrac{\sin\theta}{\cos\theta}=\tan\theta,\\\\(ii)~1+\tan^2\theta=\sec^2\theta,\\\\(iii)~\dfrac{1}{\cos\theta}=\sec\theta.$

We have

$1+\sin^2\theta=3\sin\theta\cos\theta\\\\\Rightarrow \dfrac{1+\sin^2\theta}{\cos^2\theta}=3\dfrac{\sin\theta\cos\theta}{\cos^2\theta}~~~~\textup{(dividing both sides by }\cos^2\theta)\\\\\Rightarrow \dfrac{1}{\cos^2\theta}+\dfrac{\sin^2\theta}{\cos^2\theta}=3\dfrac{\sin\theta}{\cos\theta}\\\\\Rightarrow \sec^2\theta+\tan^2\theta=3\tan\theta\\\\\Rightarrow 1+\tan^2\theta+\tan^2\theta=3\tan\theta\\\\\Rightarrow 2\tan^2\theta-3\tan\theta+1=0.~~~~~~~~~~~~(A)$

Let us consider that$\tan\theta=y.$ Then, equation (A) becomes

$2y^2-3y+1=0\\\\\Rightarrow 2y^2-2y-y+1=0\\\\\Rightarrow 2y(y-1)-1(y-1)=0\\\\\Rightarrow (y-1)(2y-1)=0\\\\\Rightarrow y-1=0,~~~~~2y-1=0\\\\\Rightarrow y=1,~~~~~~~\Rightarrow y=\dfrac{1}{2}\\\\\Rightarrow \tan\theta=1,~~~~~\Rightarrow \tan\theta=\dfrac{1}{2}.$

Hence proved.