Question

If 7sin^2a +3cos^2a=4, then prove that seca + coseca=2/3^1/2+2

Answer 1

$given \: \: 7 { \sin}^{2} a + 3 { \cos }^{2}a = 4 \\ 4 { \sin }^{2} a + 3 {sin}^{2} a + 3 {cos}^{2} a = 4 \\ 4 {sin}^{2} a = 4 - 3 \times 1 = 1 \\ {sin}^{2} a = \frac{1}{4} \\ sina = + \: or \: - \frac{1}{\sqrt{2} }$
$cosa = \sqrt{1 - {sin}^{2} a} = + \: or \: - \frac{ \sqrt{3} }{2}$
$seca + coseca = \frac{1}{cosa} + \frac{1}{sina} \\ taking \: positive \: values \: (by \: \\ assuming \: a \: is \: in \: first \: quadrant) \\ = \frac{2}{ \sqrt{3} } + 2$

Answer 2

Answer:

$sec \: a + cosec \: a = \frac{2}{ \sqrt{3} } + 2 \\ \:$

Step-by-step explanation:

If

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

To prove:

$sec \: a + cosec \: a = \frac{2}{ \sqrt{3} } + 2$

As we know that

${sin}^{2} a = 1 - {cos}^{2} a$

put this value in given condition

$7 {sin}^{2} a + 3 {cos}^{2} a = 4 \\ \\ 7 (1 - {cos}^{2} a) + 3 {cos}^{2} a = 4 \\ \\ 7 - 7 {cos}^{2} a + 3 {cos}^{2} a = 4 \\ \\ - 4{cos}^{2} a = - 3 \\ \\ {cos}^{2} a = \frac{3}{4} \\ \\ cos \: a = \frac{ \sqrt{3} }{2} \\ \\ so \\ \\ sec \: a = \frac{2}{ \sqrt{3} }$

put the value of cos a in given condition to find the value of sin a

$7 {sin}^{2} a + 3 \times \frac{3}{4} = 4 \\ \\ 7 {sin}^{2} a = 4 - \frac{9}{4} \\ \\ 7 {sin}^{2} a = \frac{7}{4} \\ \\ {sin}^{2} a = \frac{1}{4} \\ \\ sin \: a = \frac{1}{2} \\ \\ so \\ \\ cosec \: a = 2$

Thus

$sec \: a + cosec \: a = \frac{2}{ \sqrt{3} } + 2$

Hence proved.

Hope it helps you.

Note*: Assuming all values in first Quadrant.