Question

if 1 + sin squared theta is equal to 3 sin theta cos theta then prove that tan theta is equal to 1 or half​

Answer 1

this is your proof this was also confused me in my 10th class

Answer 2

$1+sin^2 \theta = 3sin\theta cos\ \theta \text{ then } tan\ \theta = \frac{1}{2}\ or\ 1$

Solution:

Given that, we have to prove:

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

To prove:

$tan\ \theta = \frac{1}{2}\ or\ 1$

From given,

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

$Divide\ by\ cos^2 \theta\ on\ both\ sides$

$\frac{1}{cos^2 \theta} + \frac{sin^2 \theta}{cos^2 \theta} = \frac{3sin\theta cos\theta}{cos^2 \theta}\\\\\frac{1}{cos^2 \theta} + \frac{sin^2 \theta}{cos^2 \theta} = \frac{3sin\theta}{cos\ \theta}$

We know that,

$\frac{1}{cos \theta} = sec\ \theta \text{ and } \frac{sin \theta}{cos \theta} = tan \theta$

Therefore,

$sec^2 \theta + tan^2 \theta = 3tan \theta\\\\We\ know\ that\\\\1 + tan^2 \theta = sec^2 \theta\\\\Therefore\\\\1 + tan^2 \theta + tan^2 \theta = 3 tan\theta\\\\1 + 2tan^2 \theta - 3tan \theta = 0\\\\2tan^2 \theta - 3tan \theta + 1 = 0\\\\Split\ the\ middle\ term\\\\2tan^2 \theta -2tan \theta - tan\ \theta + 1 = 0$

$2tan \theta (tan \theta - 1) -(tan \theta - 1) = 0\\\\(2 tan \theta - 1)(tan \theta - 1) = 0\\\\Equate\ to\ 0\\\\2 tan \theta - 1 = 0\\\\2tan \theta = 1\\\\\boxed{tan\ \theta = \frac{1}{2}}\\\\Also,\\\\tan \theta - 1 = 0\\\\\boxed{tan \theta =1}$

Thus proved

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