Question

if
$7sin^{2}theta + 3cos^{2}theta = 4$
then find sec theta+ cosec theta
​

Answer 1

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Given :-

7 sin²θ + 3cos²θ = 4

To find :-

secθ + cosecθ

Solution :-

$7 {sin}^{2} θ + 3 {cos}^{2} θ = 4 \\ 7 {sin}^{2} θ + 3(1 - {sin}^{2} θ) = 4 \\ 7 {sin}^{2} θ + 3 - 3 {sin}^{2} θ = 4 \\ 4 {sin}^{2} θ = 4 - 3 \\ 4 {sin}^{2} θ = 1 \\ {sin}^{2} θ = \frac{1}{4} \\ sinθ = \frac{1}{2} \\ \large\bold\red{ = > \: θ = 30} \\ so \: secθ \: + cosecθ \\ = sec30 + cosec30 \\ = \frac{2}{ \sqrt{3} } + \frac{2}{1} \\ = \frac{2 + 2 \sqrt{3} }{ \sqrt{3} } \times \frac{ \sqrt{3} }{ \sqrt{3} } \\ = \frac{2 \sqrt{3} + 6 }{3}$

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Answer 2

Answer:

this answer is correct...................