Question

answer....it....fast.....

Answer 1

45 degree in think it's right

Answer 2

$\mathsf{Given : \;\;\dfrac{sin^2\theta - 5sin\theta + 3}{cos^2\theta} = 1}$

$\implies \mathsf{sin^2\theta - 5sin\theta + 3 = cos^2\theta}$

$\textsf{We know that : \boxed{\mathsf{cos^2\theta = 1 - sin^2\theta}}}$

$\implies \mathsf{sin^2\theta - 5sin\theta + 3 = 1 - sin^2\theta}$

$\implies \mathsf{2sin^2\theta - 5sin\theta + 2 = 0}$

$\implies \mathsf{2sin^2\theta - 4sin\theta - sin\theta + 2 = 0}$

$\implies \mathsf{2sin\theta(sin\theta - 2) - (sin\theta - 2) = 0}$

$\implies \mathsf{(sin\theta - 2) (2sin\theta - 1) = 0}$

$\implies \mathsf{sin\theta = 2\;\;(or)\;\;sin\theta = \dfrac{1}{2} }$

$\bigstar\;\;\mathsf{We\;know \;that : Range \;of \;sin\theta\;is \;[-1 , 1]}$

$\bigstar\;\;\mathsf{That\;is : -1 \leq sin\theta \leq 1}$

$\mathsf{\implies sin\theta\;cannot\;be\;equal\;to\;2\;for\;any\;value\;of\;\theta}$

$\mathsf{\implies sin\theta \neq 2}$

$\mathsf{\implies sin\theta\;should\;be\;equal\;to\;\dfrac{1}{2}}$

$\mathsf{\implies sin\theta = \dfrac{1}{2}}$

$\mathsf{\implies \theta = sin^{-1}\bigg(\dfrac{1}{2}\bigg)}$

$\mathsf{\implies \theta = 30^{\circ}}$