Question

7 sin²θ + 3 cos²θ = 4. Find the value of Secθ + cosecθ.

Answer 1

Step-by-step explanation:

This is the solution .Reply if it is ok .Best of luck .

Answer 2

Answer:

$\large\bold\red{\sec( \theta ) + \csc( \theta ) = \frac{2}{ \sqrt{3} }}$±$\large\bold\red{2}$

Step-by-step explanation:

Given,

$7 { \sin }^{2} \theta + 3 { \cos}^{2} \theta = 4$

But,

We know that,

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

Therefore,

Putting the values,

We get,

$= > 7 { \sin }^{2} \theta + 3(1 - { \sin}^{2} \theta) = 4 \\ = > 7 { \sin}^{2} \theta + 3 - 3 { \sin }^{2} \theta = 4 \\ = > 4 { \sin}^{2} \theta = 4 - 3 \\ = > 4 { \sin }^{2} \theta = 1 \\ = > { \sin }^{2} \theta = \frac{1}{4}$

Therefore,

We get,

$= > \sin( \theta ) =$±$\frac{1}{2}$

But,

We know that,

$\bold{\csc( \theta ) = \frac{1}{ \sin( \theta ) } }$

Therefore,

We get,

$= > \csc( \theta ) =$±$2 \: \: \: \: .............(i)$

Now,

We know that,

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

Putting the values,

We get,

$= > \cos( \theta ) = \sqrt{1 - {( \frac{1}{2} )}^{2} } \\ = > \cos( \theta ) = \sqrt{1 - \frac{1}{4} } \\ = > \cos( \theta ) = \sqrt{ \frac{4 - 1}{4} } \\ = > \cos( \theta ) = \sqrt{ \frac{3}{4} } \\ = > \cos( \theta ) = \frac{ \sqrt{3} }{2}$

But,

We know that,

$\sec( \theta ) = \frac{1}{ \cos( \theta ) }$

Therefore,

We get,

$\sec( \theta ) = \frac{2}{ \sqrt{3} } \: \: \: \: \: \: ...............(ii)$

Now,

From Equation $(i)$ and $(ii)$,

We get,

$= > \sec( \theta ) + \csc( \theta ) = \frac{2}{ \sqrt{3} } +$(±$2)$

$\large\bold{=>\sec( \theta ) + \csc( \theta ) = \frac{2}{ \sqrt{3} }}$±$\large\bold{2}$