Question

if 3cos theta = 5sin theta , find (tan^2 theta + 1)/(sin^2 theta + cos^2 theta)​

Answer 1

Given:-

• $3cos \theta = 5sin \theta$

To Find:-

• $\frac{(tan^2 \theta + 1)}{(sin^2 \theta + cos^2 \theta)}$

SOLUTION:-

$3cos \theta = 5sin \theta$

$= > cos \theta = \frac{5}{3} \times sin \theta$

Now, Squaring Both sides

$= > { \cos }^{2} \theta = { (\frac{5}{3} )}^{2} \times { \sin}^{2} \theta$

We Know that

$\: \bf\underline \red{\underline{{ \cos }^{2} \theta = 1 - { \sin}^{2} \theta}}$

So,

$= > 1 - {{ \sin}^{2} \theta} = { (\frac{5}{3} )}^{2} \times { \sin}^{2} \theta$

$= > 1 - {{ \sin}^{2} \theta} = \frac{25}{9} \times { \sin}^{2} \theta$

$= > 1 = \frac{25}{9} { \sin }^{2}\theta + { \sin }^{2}\theta$

$= > 1 = \frac{25}{9} { \sin }^{2}\theta + \frac{9}{9} { \sin }^{2}\theta$

$= > 1 = \frac{34}{9} { \sin }^{2}\theta$

$= > { \sin }^{2} \theta = \frac{9}{34}$

$\therefore \sin\theta = \frac{3}{ \sqrt{34} }$

${ \cos }^{2} \theta = 1 - { \sin}^{2} \theta$

${ \cos }^{2} \theta = 1 - \frac{ {3}^{2} }{ { \sqrt{34} }^{2} }$

$= > { \cos }^{2} \theta = 1 - \frac{9}{34} = \frac{34 - 9}{34}$

$= > { \cos }^{2} \theta = \frac{25}{34}$

$= > \therefore{ \cos }\theta = \frac{5}{ \sqrt{34} }$

Now,

$\frac{(tan^2 \theta + 1)}{(sin^2 \theta + cos^2 \theta)}$

$= > \frac{( \frac{ {\sin }^{2}\theta }{ { \cos }^{2} \theta}) + 1}{( \frac{3}{ \sqrt{34} } + \frac{5}{ \sqrt{34} } )}$

$= > \frac{ \frac{ \frac{3}{ \sqrt{34} } }{ \frac{5}{ \sqrt{34} } } + 1}{ \frac{8}{ \sqrt{34} } }$

$= > \frac{ \frac{3}{5} + 1}{ \frac{8}{ \sqrt{34} } }$

$= > \frac{ \frac{8}{5} }{ \frac{8}{ \sqrt{34} } }$

$= > \frac{ \sqrt{34} }{5}$