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Mathematics Question on homogeneous differential equation

Solution of the equation sec⁡θ−cosec θ=43\sec \theta - cosec \:\theta=\frac{4}{3} is

A

θ=nπ2+(−1)n2sin⁡−134\theta=\frac{n\pi}{2}+\frac{(-1)^n}{2}\sin^{-1}\frac{3}{4}

B

θ=nπ+(−1)nsin⁡−134\theta=n\pi+(-1)^n \sin^{-1}\frac{3}{4}

C

θ=nπ±sin⁡−1(34)\theta=n\pi\pm \sin^{-1}\left(\frac{3}{4}\right)

D

None of these.

Answer

θ=nπ2+(−1)n2sin⁡−134\theta=\frac{n\pi}{2}+\frac{(-1)^n}{2}\sin^{-1}\frac{3}{4}

Explanation

Solution

We have 1cos⁡ θ−1sin⁡ θ=43\frac{1}{\cos \, \theta} -\frac{1}{\sin \, \theta} = \frac{4}{3} ⇒\Rightarrow sin⁡ θ−cos⁡ θ=43sin⁡ θ cos⁡ θ\sin\, \theta - \cos \,\theta = \frac{4}{3} \sin\, \theta \,\cos \,\theta ⇒ 3(sin⁡ θ−cos⁡ θ)=2 sin⁡ θ\Rightarrow\, 3\left(\sin\, \theta - \cos \,\theta\right) = 2\, \sin \,\theta ⇒ 9(sin⁡2 θ+cos⁡2 θ−2 sin⁡ θ cos⁡ θ)=4 sin⁡2 2 θ\Rightarrow \,9\left(\sin^{2} \,\theta +\cos^{2}\,\theta - 2\, \sin\, \theta\, \cos \, \theta\right) = 4\, \sin^{2}\, 2\,\theta ⇒ 9(1−sin⁡ 2θ)=4sin⁡2 θ\Rightarrow \,9\left(1 - \sin\, 2\theta\right) = 4 \sin^{2} \,\theta ⇒ sin⁡ 2 θ=34\Rightarrow\, \sin\, 2 \, \theta = \frac{3}{4} or sin⁡ 2 θ=−3\sin \,2 \, \theta = - 3 But sin⁡ 2 θ=−3\sin \, 2 \, \theta = - 3 is not possible. ∴\therefore sin⁡ 2θ=34\sin\, 2\theta = \frac{3}{4} ∴\therefore 2θ=nπ+(−1)nsin⁡−1342\theta = n\pi +\left(-1\right)^{n } \sin^{-1} \frac{3}{4} ⇒θ=nπ2+(−1)n2sin(34),n∈I\Rightarrow \theta = \frac{n\pi}{2 } + \frac{\left(-1\right)^{n}}{2} sin \left(\frac{3}{4}\right), n \in I