Question

if 1 + sin² A = 3 sin A cos A , prove that tan A = 1 or 1/2​

Answer 1

$\huge \red{\mathfrak{solution} }$

Given :

$\star \bold{ 1 + {sin}^{2} \theta \: = 3 sin \theta \: cos \theta \: }$

To prove :

$\star \bold{ \: tan \theta \: = 1 \: or \: \frac{1}{2} }$

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

Divide both sides by

${cos}^{2} \theta$

$\rightarrow \: \frac{1 + {sin}^{2} \theta}{ {cos}^{2} \theta } = \frac{3sin \theta cos \theta}{ {cos}^{2} \theta } \\ \\ \rightarrow \: {sec}^{2} \theta + {tan}^{2} \theta \: = 3tan \theta \\ \\ \rightarrow \: 1 + {tan}^{2} \theta \: + {tan}^{2} \theta = 3tan \theta \\ \\ \rightarrow \: 2 {tan}^{2} \theta - 3 \tan \theta \: + 1 = 0$

$\bold{let \: tan \theta \: = x}$

$\star \: 2 {x}^{2} - 3x + 1 = 0 \\ \\ \star \: 2 {x}^{2} - 2x - x + 1 = 0 \\ \\ \star \: 2x(x - 1) - 1(x - 1) = 0 \\ \\ \star \: (2x - 1)(x - 1) = 0$

Now ,

$\leadsto \: 2x - 1 = 0 \\ \\ \leadsto \: x = \frac{1}{2}$

And

$\implies \: x - 1 = 0 \\ \\ \implies \: x = 1$

Hence ,the values of X or tan@ is 1 and 1/2

$\huge{ \green{\mathfrak{proved}}}$

Formula:

$\star \bold{ {sec}^{2} \theta \: - {tan}^{2} \theta \: = 1}$

$\star \bold { tan \theta \: = \frac{sin \theta}{cos \theta} }$

Answer 2

Refer to the attachment⭑⭑♡⭑