Question

Sin square theta plus 1 by 1 plus tan square theta

Answer 1

As we know that,

$\purple{ \tt \bullet \qquad sin^{2} \theta + cos^{2} \theta = 1}$

$\purple{ \tt \bullet \qquad tan^{2} \theta = \dfrac{ {sin}^{2} \theta}{ {cos}^{2} \theta} }$

So,

$\tt\Longrightarrow \dfrac{sin^2\theta+1}{1+tan^2\theta}$

$\tt\Longrightarrow \dfrac{sin^2\theta+1}{1+ \frac{ {sin}^{2} \theta}{ {cos}^{2} \theta } }$

$\tt\Longrightarrow \dfrac{sin^2\theta+1}{ \frac{ {cos}^{2} \theta + {sin}^{2} \theta}{ {cos}^{2} \theta } }$

$\tt\Longrightarrow \dfrac{sin^2\theta+1}{ \frac{1}{ {cos}^{2} \theta } }$

$\tt\Longrightarrow (1 + sin^2\theta) {cos}^{2} \theta$

$\tt\Longrightarrow (1 + sin^2\theta) (1 - {sin}^{2} \theta)$

$\tt\Longrightarrow (1 - {sin}^{4} \theta)$

$\tt\Longrightarrow ( {cos}^{4} \theta)$

$\green{ \tt\Longrightarrow {cos}^{4} \theta}$