Question

if 3 cos squared theta + 7 sin squared theta is equal to 4 show that cot theta is equal to underroot3

Answer 1

${ 3\cos }^{2} \theta + 7 { \sin}^{2} \theta = 4 \\ { 3\cos }^{2} \theta + 3 { \sin}^{2} \theta + 4{ \sin}^{2} \theta = 4 \\ 3( {\cos }^{2} \theta +{ \sin}^{2} \theta) + 4{ \sin}^{2} \theta = 4 \\ 3 \times 1 + 4{ \sin}^{2} \theta = 4 \\ 3 + 4{ \sin}^{2} \theta = 4 \\ 4{ \sin}^{2} \theta = 4 - 3 = 1 \\ { \sin}^{2} \theta = \frac{1}{4} \\ { \sin}\theta = \sqrt{ \frac{1}{4} } = \frac{1}{2} \\ \cos \theta = \sqrt{1 - { \sin }^{2} \theta} = \sqrt{1 - { (\frac{1}{2} })^{2} } \\ \cos \theta = \sqrt{1 - \frac{1}{4} } = \sqrt{ \frac{3}{4} } = \frac{ \sqrt{3} }{2} \\ now \\ \cot\theta = \frac{ \cos \theta} { \sin\theta }= \frac{ \frac{ \sqrt{3} }{2} } { \frac{1}{2} } \\ \cot \theta = \sqrt{3}$