Question

if 1+sin^2alpha=3sin alpha cos alpha then show that the value of tan alpha=1 or 1/2​

Answer 1

Answer:

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Answer 2

$\tan \alpha=1 or \dfrac{1}{2}$, proved.

Step-by-step explanation:

We have,

$1+\sin^2\alpha=3\sin \alpha \cos \alpha$

To prove that,  $\tan \alpha=1 or \dfrac{1}{2}$

∴ $1+\sin^2\alpha=3\sin \alpha \cos \alpha$

Dividing both sides $\cos^2 \alpha$, we get

$\dfrac{1}{\cos^2 \alpha} +\dfrac{\sin^2\alpha}{\cos^2\alpha} =\dfrac{3\sin \alpha \cos \alpha}{\cos^2 \alpha}$

⇒ $\sec^2 \alpha+\tan^2 \alpha=3\tan \alpha$

⇒ $1+\tan^2 \alpha+\tan^2 \alpha=3\tan \alpha$

⇒ $2\tan^2 \alpha-3\tan \alpha+1=0$

By factorisation method,

$2\tan^2 \alpha-2\tan \alpha-\tan \alpha+1=0$

⇒ $2\tan\alpha(\tan\alpha-1)-1(\tan\alpha-1)=0$

⇒ $(\tan\alpha-1)(2\tan\alpha-1)=0$

⇒ $\tan\alpha-1=0$ or $2\tan\alpha-1=0$

⇒ $\tan\alpha-1=0$

⇒ $\tan\alpha=1$

⇒ $2\tan\alpha-1=0$

⇒ $\tan\alpha=\dfrac{1}{2}$

Hence, $\tan \alpha=1 or \dfrac{1}{2}$, proved.