Question

Tan square theta /1+Tan square theta

Answer 1

Answer:

$\huge \purple{{ \sin {}^{2} (\theta)}}$

Step-by-step explanation:

$\frac{\tan {}^{2} (\theta) }{1 + \tan {}^{2} (\theta) } \\ \\ = \frac{ \tan {}^{2} (\theta) }{ \sec {}^{2} ( \theta) } \\ \\ = \frac{ \sin {}^{2} (\theta) }{ \cos {}^{2} (\theta). \frac{1}{ \cos {}^{2} (\theta)} } \\ \\ = \: \: \: \: \: \: \: \: \: \: \: \sin {}^{2} (\theta)$

Answer 2

$\huge \green{ \boxed{ \bf \purple{ {sin}^{2} \theta}}}$

Step-by-step explanation:

$\bf \frac{\tan^{2}\theta}{1 + { \tan}^{2} \theta}$

We know that,

1 + tan²θ = sec²θ

So,

$\bf \frac{ {tan}^{2} \theta}{1 + { tan}^{2} \theta} \\ \\ = > \bf \frac{tan {}^{2} \theta}{sec {}^{2} \theta} \\ \\ \bf = > \frac{tan \theta}{sec \theta} \times \frac{tan \theta}{sec \theta}$

We should remember that,

$\large \frac{tan \theta}{sec \theta} = sin \theta$

So,

$\bf\frac{tan \theta}{sec \theta} \times \frac{tan \theta}{sec \theta} \\ \\ = > \bf sin \theta \times sin \theta \\ \\ = > \red{ \boxed{ \bf \blue{ sin {}^{2} \theta}}}$

Hence,

The answer is sin²θ.

$\:$

Hope it helps.

#BeBrainly :-)