Which of the following equation has no solutions

\sqrt{3} sin ϑ - cos ϑ = 2

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cos θ + sin θ = \sqrt{2}

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cos θ sec θ = 1

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csc θ - sec θ = csc θ . sec θ

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Solution

The correct option is A $\sqrt{3} sin ϑ - cos ϑ = 2$
$\begin{matrix} A) \sqrt{3 sin θ} - cos θ = 2 \Rightarrow {(\sqrt{3 sin θ})}^{2} = {(2 + cos θ)}^{2} \Rightarrow \sqrt{3 sin θ} = 2 + cos θ \Rightarrow {(\sqrt{3 sin θ})}^{2} = {(2 + cos θ)}^{2} \Rightarrow 3 {sin}^{2} θ = 4 + {cos}^{2} θ + 4 cos θ \Rightarrow 3 (1 - {cos}^{2} θ) = 4 + {cos}^{2} θ + 4 cos θ \Rightarrow 3 - 3 {cos}^{2} θ - 4 - {cos}^{2} θ - 4 cos θ = 0 \Rightarrow - 4 {cos}^{2} θ - 4 cos θ - 1 = 0 \Rightarrow {cos}^{2} θ + cos θ + \frac{1}{4} = 0 cos = - 1 \pm \frac{\sqrt{{(1)}^{2} - 4 \times 1} \times \frac{1}{4}}{2 \times 1} \Rightarrow cos θ = cos (π = \frac{π}{3}) \Rightarrow cos θ = cos (\frac{2 π}{3}) \Rightarrow θ = 2 n π \pm 2 \frac{π}{3} (B) cos θ + sin θ = \sqrt{2} \Rightarrow \frac{1}{\sqrt{2} cos θ + \frac{1}{\sqrt{2}} sin θ} = 1 \Rightarrow sin \frac{π}{4} cos θ + cos \frac{π}{4} sin θ \Rightarrow sin (θ + \frac{π}{4}) = 1 \Rightarrow θ + \frac{π}{4} = \frac{π}{4} + 2 π n \Rightarrow θ = \frac{π}{4} + 2 π n (C) H a s n o s o l u t i o n (D) cos e c θ - s e c θ \Rightarrow cos e c θ \cdot sec θ \Rightarrow \frac{1}{sin θ} - \frac{1}{cos θ} = \frac{1}{sin θ cos θ} \Rightarrow cos θ - sin θ = 1 \Rightarrow (cos θ - sin θ) = 1 \Rightarrow {cos}^{2} θ + {sin}^{2} θ - 2 cos θ sin θ = 1 \Rightarrow 1 - {sin}^{2} θ = 1 \Rightarrow {sin}^{2} θ = 0 \Rightarrow 2 θ = 0 + 2 π n \Rightarrow θ = π n \end{matrix}$

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(iv) $\frac{s e c θ - 1}{s e c θ + 1} = {(\frac{s i n θ}{1 + c o s θ})}^{2}$

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