Maths Answer the following questions in short.

$\therefore \sin \theta=\frac{5}{3}$

This is not possible because -1 ≤ sin θ ≤ 1 for any θ.
∴ 3 sin θ = 5 does not have any solution.

This is not possible because -1 ≤ sin θ ≤ 1 for any θ.
∴ 2 sin θ = 3 does not have any solution.

Since $\frac{1}{3} \leq \cos \theta \leq 1$ for any $\theta$

$\cos 2 \theta=\frac{1}{3}$ has solution

Here side $AB$ is the largest side. $C$ is the largest angle of $\triangle ABC$. To show that $C$ is obtuse angle.

$

\cos C =\frac{a^2+b^2-c^2}{2 a b}=\frac{2^2+3^2-4^2}{2(3)(4)}=-\frac{3}{24}=-\frac{1}{8}

$

As $\cos C$ is negative, $C$ is obtuse angle.

$\therefore \quad \triangle ABC$ is obtuse angled triangle.

Let $\cos ^{-1}\left(-\frac{1}{2}\right)=\alpha$, where $0 \leq \alpha \leq \pi$

$\therefore \cos \alpha=-\frac{1}{2}=-\cos \frac{\pi}{3}$

$\therefore \cos \alpha=\cos \left(\pi-\frac{\pi}{3}\right) \ldots[\because \cos (\pi-\theta)=-\cos \theta)$

$\therefore \cos \alpha=\cos \frac{2 \pi}{3}$

$\therefore \alpha=\frac{2 \pi}{3} \ldots\left[\because 0 \leq \frac{2 \pi}{3} \leq \pi\right]$

$\therefore$ the principal value of $\cos ^{-1}\left(-\frac{1}{2}\right)$ is $\frac{2 \pi}{3}$.

Let $\sin ^{-1}\left(\frac{1}{\sqrt{2}}\right)=\alpha$, where $\frac{-\pi}{2}<\alpha<\frac{\pi}{2}$

$\begin{aligned} & \therefore \sin \alpha=\left(\frac{1}{\sqrt{2}}\right)=\sin \frac{\pi}{4} \\ & \therefore \alpha=\frac{\pi}{4} \ldots\left[\because-\frac{\pi}{2} \leq \frac{\pi}{4} \leq \frac{\pi}{2}\right]\end{aligned}$

$\therefore$ the principal value of $\sin ^{-1}\left(\frac{1}{\sqrt{2}}\right)$ is $\frac{\pi}{4}$.

Let $\tan ^{-1}(-\sqrt{3})=\alpha$, where $\frac{-\pi}{2}<\alpha<\frac{\pi}{2}$

$\therefore \tan \alpha=-\sqrt{3}=-\tan \frac{\pi}{3}$

$\therefore \tan \alpha=\tan \left(-\frac{\pi}{3}\right) \ldots[\because \tan (-\theta)=-\tan \theta]$

$\therefore \alpha=-\frac{\pi}{3} \ldots\left[\because-\frac{\pi}{9}<\frac{-\pi}{3}<\frac{\pi}{9}\right]$

$\therefore$ the principal value of $\tan ^{-1}(-\sqrt{3})$ is $-\frac{\pi}{3}$.

Let $\tan ^{-1}(-1)=\alpha$, where $\frac{-\pi}{2}<\alpha<\frac{\pi}{2}$

$\therefore \tan \alpha=-1=-\tan \frac{\pi}{4}$

$\therefore \tan \alpha=\tan \left(-\frac{\pi}{4}\right) \ldots[\because \tan (-\theta)=-\tan \theta]$

$\therefore \alpha=-\frac{\pi}{4} \ldots\left[\because-\frac{\pi}{2}<\frac{-\pi}{4}<\frac{\pi}{2}\right]$

$\therefore$ the principal value of $\tan ^{-1}(-1)$ is $-\frac{\pi}{4}$.

Let $\operatorname{cosec}^{-1}(2)=\alpha$, where $\frac{-\pi}{2} \leq \alpha \leq \frac{\pi}{2}, \alpha \neq 0$

$\therefore \operatorname{cosec}^{-1} \alpha=2=\operatorname{cosec} \frac{\pi}{6}$

$\therefore \alpha=\frac{\pi}{6} \ldots\left[\because-\frac{\pi}{2} \leq \frac{\pi}{6} \leq \frac{\pi}{2}\right]$

$\therefore$ the principal value of $\operatorname{cosec}^{-1}(2)$ is $\frac{\pi}{6}$.

Let $\sin ^{-1}\left(\frac{1}{2}\right)=\alpha$, where $\frac{-\pi}{2} \leq \alpha \leq \frac{\pi}{2}$

$\begin{aligned} & \therefore \sin \alpha=\frac{1}{2}=\sin \frac{\pi}{6} \\ & \therefore \alpha=\frac{\pi}{6} \cdots\left[\because-\frac{\pi}{2} \leq \frac{\pi}{6} \leq \frac{\pi}{2}\right]\end{aligned}$

$\therefore$ the principal value of $\sin ^{-1}\left(\frac{1}{2}\right)$ is $\frac{\pi}{6}$.

$\theta=n \pi+\alpha, n \in Z$

Now, $\tan \theta=-1$

$\therefore \tan \theta=-\tan \frac{\pi}{4} \quad \cdots\left[\because \tan \frac{\pi}{4}=1\right]$

$\therefore \tan \theta=\tan \left(\pi-\frac{\pi}{4}\right) \ldots[\because \tan (\pi-\theta)=-\tan \theta]$

$\therefore \tan \theta=\tan \frac{3 \pi}{4}$

$\therefore$ the required general solution is

$\theta=n \pi+\frac{3 \pi}{4}, n \in Z$

$\theta=n \pi+(-1)^n \alpha, n \in Z$

Now, $\operatorname{cosec} \theta=-\sqrt{2}$

$\therefore \sin \theta=-\frac{1}{\sqrt{2}}$

$\therefore \sin \theta=-\sin \frac{\pi}{4} \quad \cdots\left[\because \sin \frac{\pi}{4}=\frac{1}{\sqrt{2}}\right]$

$\therefore \sin \theta=\sin \left(\pi+\frac{\pi}{4}\right) \quad \ldots[\because \sin (\pi+\theta)=-\sin \theta]$

$\therefore \sin \theta=\sin \frac{5 \pi}{4}$

$\therefore$ the required general solution is

$\theta=n \pi+(-1)^n\left(\frac{5 \pi}{4}\right), n \in Z$

$\theta=n \pi \pm \alpha, n \in Z$

Now, $\sec \theta=\sqrt{2} \therefore \cos \theta=\frac{1}{\sqrt{2}}$

$\therefore \cos \theta=\cos \frac{\pi}{4} \ldots\left[\cos \frac{\pi}{4}=\frac{1}{\sqrt{2}}\right]$

∴ the required general solution is

$\theta=2 \mathrm{n} \pi \pm \frac{\pi}{4}, \mathrm{n} \in Z$