Question

if sin theta=1/root 3, then find the value of (tan theta + 1/cos theta)^2 + (tan theta - 1/cos theta)^2

Answer 1

Given :

$\bullet \sf{sin \theta = \dfrac{1}{ \sqrt{3} } }$

To find :

$\bullet \sf{(tan \theta + \dfrac{1}{cos \theta})^2 \: + \: (tan \theta - \dfrac{1}{cos \theta} )^2 = ? }$

Solution :

$\implies\sf{ { \big(tan \theta + \dfrac{1}{cos \theta} \big)}^{2} + { \big(tan \theta - \dfrac{1}{cos \theta} \big) }^{2} }$

$\red{putting \: tan \theta = \dfrac{sin \theta}{cos \theta} }$

$\implies\sf{ { \big( \dfrac{sin \theta}{cos \theta} + \dfrac{1}{cos \theta} \big)}^{2} + { \big( \dfrac{sin \theta}{cos \theta} - \dfrac{1}{cos \theta} \big) }^{2} }$

$\implies\sf{ { \big( \dfrac{sin \theta + 1}{cos \theta} \big)}^{2} + { \big( \dfrac{sin \theta - 1}{cos \theta} \big) }^{2} }$

$\implies\sf{ \dfrac{(sin \theta + 1)^{2} }{ {cos}^{2} \theta} + \dfrac{(sin \theta - 1)^{2} }{ {cos}^{2} \theta } }$

$\implies\sf{ \dfrac{(sin \theta + 1)^{2} + {(sin \theta - 1)}^{2} }{ {cos}^{2} \theta} }$

$\red{using \: algebraic \: identity}$

$\red{ \star \: {(a + b)}^{2} = {a}^{2} + {b}^{2} + 2ab \: \: and }$

$\red{ \star \: {(a - b)}^{2} = {a}^{2} + {b}^{2} - 2ab }$

$\implies\sf{ \dfrac{(sin^{2} \theta + 1 + 2sin \theta) + (sin^{2} \theta + 1 - 2sin \theta)}{ {cos}^{2} \theta} }$

$\implies\sf{ \dfrac{2sin^{2} \theta + 2 }{ {cos}^{2} \theta} }$

$\red{using \: trigonometric \: identity}$

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

$\implies\sf{ \dfrac{2sin^{2} \theta + 2 }{ 1 - {sin}^{2} \theta} }$

$\red{putting \: value \: of \: sin \theta}$

$\implies\sf{ \dfrac{2( \dfrac{1}{ \sqrt{3}} )^{2} + 2 }{ 1 - {( \dfrac{1}{ \sqrt{3} } )}^{2} } }$

$\implies\sf{ \dfrac{2( \dfrac{1}{3} ) + 2 }{ 1 - {( \dfrac{1}{ 3})} } }$

$\implies\sf{ \dfrac{ \dfrac{2}{3} + 2 }{ 1 - \dfrac{1}{ 3} } }$

$\implies\sf{ \dfrac{ \: \: \: \: \dfrac{2 + 6}{3} \: \: \: \: }{ \dfrac{3 - 1}{ 3} } }$

$\implies\sf{ \dfrac{ \: \: \: \dfrac{8}{3} \: \: \: }{ \dfrac{2}{ 3} } }$

$\implies \sf{ \dfrac{8}{2} = 4}$

therefore,

$\red{ \bigstar} \boxed{\sf{ { \big(tan \theta + \dfrac{1}{cos \theta} \big)}^{2} + { \big(tan \theta - \dfrac{1}{cos \theta} \big) }^{2} = 4 } }$