Question

sin theta =5/13 to find tan theta +1/cos theta

Answer 1

Here is your answer.......

Answer 2

Question :–

$\\ { \bold{if \: \: \sin( \theta) = \dfrac{5}{13} \: , \:then \: \: find \: \: \tan( \theta) + \dfrac{1}{ \cos( \theta) } = ?}}$

ANSWER :–

GIVEN –

$\\ \: \to \: { \bold{ \: \: \sin( \theta) = \dfrac{5}{13} }}\:$

TO FIND :–

$\\ \: \to \: { \bold{ \: \: \tan( \theta) + \dfrac{1}{ \cos( \theta) } = ? }}\:$

SOLUTION :–

$\\ \: \to \: { \bold{ \: \: \sin( \theta) = \dfrac{5}{13} }}\:$

• We know that –

$\\ \: \longrightarrow \: { \bold{ \: \: \sin {}^{2} ( \theta) + \cos {}^{2} ( \theta) = 1 }}\:$

• So that –

$\\ \: \implies\: { \bold{ \cos {}^{2} ( \theta) = 1 - \sin {}^{2} ( \theta) }}\:$

$\\ \: \implies\: { \bold{ \cos {}^{2} ( \theta) = 1 - \left({ \dfrac{5}{13}} \right) ^{2} }}\:$

$\\ \: \implies\: { \bold{ \cos ( \theta) = \sqrt{ 1 - \left({ \dfrac{5}{13}} \right) ^{2} }}}\:$

$\\ \: \implies\: { \bold{ \cos ( \theta) = \sqrt{ 1 - \left({ \dfrac{25}{169}} \right) }}}\:$

$\\ \: \implies\: { \bold{ \cos ( \theta) = \sqrt{ \left({ \dfrac{169 - 25}{169}} \right) }}}\:$

$\\ \: \implies\: { \bold{ \cos ( \theta) = \sqrt{ { \dfrac{144}{169}} }}}\:$

$\\ \: \implies\: { \bold{ \cos ( \theta) = \dfrac{12}{13} }}\:$

• Now Let's find –

$\\ \: \: { \bold{ = \: \tan( \theta) + \dfrac{1}{ \cos( \theta) } }}\:$

$\\ \: \: { \bold{ = \: \dfrac{ \sin( \theta) }{ \cos( \theta)} + \dfrac{1}{ \cos( \theta) } \: \: \: \: \: \: \: \: \: \: \: \left[ \because \: \: \tan( \theta) = \dfrac{ \sin( \theta) }{ \cos( \theta) } \right]}}\:$

• Now put the values –

$\\ \: \: { \bold{ = \: \dfrac{1 + \sin( \theta) }{ \cos( \theta)} }}\:$

$\\ \: \: { \bold{ = \: \dfrac{1 + \dfrac{5}{13} }{ \dfrac{12}{13} } }}\:$

$\\ \: \: { \bold{ = \: \dfrac{ \dfrac{13 + 5}{ \cancel{13}} }{ \dfrac{12}{ \cancel{13}} } }}\:$

$\\ \: \: { \bold{ = \: \dfrac{ 18 }{12 }}}\:$

$\\ \: \: { \bold{ = \: \dfrac{ 3 }{2 }}}\:$

Hence , ${ \bold{ \: \: \tan( \theta) + \dfrac{1}{ \cos( \theta) } = \dfrac{3}{2} }}\:$