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Maths-

General

Easy

Question

If $1 plus sin squared space theta equals 3 Sin backslash$ $t h e t a$ $Cos space theta$ then the solution set in $open parentheses 0 comma pi over 2 close parentheses$ is

$\{\frac{π}{4}, {Cos}^{- 1} \frac{1}{3}\}$
$\{\frac{π}{4}, {Tan}^{- 1} \frac{1}{2}\}$
$\{\frac{π}{3}, {Tan}^{- 1} \frac{1}{3}\}$
$\{\frac{π}{6}, {Sin}^{- 1} \frac{1}{3}\}$

The correct answer is: $open curly brackets pi over 4 comma Tan to the power of negative 1 end exponent space 1 half close curly brackets$

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