Question

if 7sin^2 theta +3 cos^2 theta = 4 then the value tan theta​

Answer 1

Answer:

$\boxed{\bold{tan \alpha = \dfrac{1}{ \sqrt{3} } \: \: \: \: or \: - \dfrac{1}{ \sqrt{3} }}}$

Step-by-step explanation:

$\bold{Given}\longrightarrow \\ 7 {sin}^{2} \alpha + 3 {cos}^{2} \alpha = 4$

$\bold{To \: find}\longrightarrow \\ value \: of \: tan \alpha$

$\bold{Concept \: used}\longrightarrow \\ \boxed{ {sin}^{2} \alpha = 1 - {cos}^{2} \alpha} \\ \boxed{tan(\pi - \alpha ) = - tan \alpha}$

$\bold{Solution}\longrightarrow \\ 7 {sin}^{2} \alpha + 3 {cos}^{2} \alpha = 4$

$= > 7(1 - {cos}^{2} \alpha ) + 3 {cos}^{2} \alpha = 4$

$= > 7 - 7 {cos}^{2} \alpha + 3 {cos}^{2} \alpha = 4$

$= > - 4 {cos}^{2} \alpha = 4 - 7$

$= > - 4 {cos}^{2} \alpha = - 3$

$= > {cos}^{2} \alpha = \dfrac{3}{4}$

$= > cos \alpha = \pm \dfrac{ \sqrt{3} }{2}$

$if \: cos \alpha = \dfrac{ \sqrt{3} }{2 } \\ = > cos \alpha = cos \dfrac{\pi}{6} \\ = > \alpha = \dfrac{\pi}{6} \\ tan \alpha = tan \dfrac{\pi}{6} \\ = > tan \alpha = \dfrac{1}{ \sqrt{3} }$

$if \: cos \alpha = - \dfrac{ \sqrt{3} }{2} \\ = > cos \alpha = cos \dfrac{5\pi}{6} \\ = > \alpha = \dfrac{5\pi}{6} \\ = > tan \alpha = tan \dfrac{5\pi}{6} \\ = > tan \alpha = tan(\pi - \dfrac{\pi}{6} ) \\ = > tan \alpha = - tan \dfrac{\pi}{6} \\ = > tan \alpha = - \dfrac{1}{ \sqrt{3} }$