Question : What is the value of $\frac{\sin \theta+\cos \theta}{\sin \theta-\cos \theta}+\frac{\sin \theta-\cos \theta}{\sin \theta+\cos \theta}$?

Option 1: $\frac{1}{\left(\sin ^2 \theta-\cos ^2 \theta\right)}$

Option 2: $2\left(\sin ^2 \theta-\cos ^2 \theta\right)$

Option 3: $\frac{2}{\left(\sin ^2 \theta-\cos ^2 \theta\right)}$

Option 4: $\sin ^2 \theta-\cos ^2 \theta$

Question : If $x\cos \theta -y\sin \theta =\sqrt{x^{2}+y^{2}}$ and $\frac{\cos ^2{\theta }}{a^{2}}+\frac{\sin ^{2}\theta}{b^{2}}=\frac{1}{x^{2}+y^{2}},$ then the correct relation is:

Option 1: $\frac{x^{2}}{b^{2}}-\frac{y^{2}}{a^{2}}=1$

Option 2: $\frac{x^{2}}{a^{2}}+\frac{y^{2}}{b^{2}}=1$

Option 3: $\frac{x^{2}}{b^{2}}+\frac{y^{2}}{a^{2}}=1$

Option 4: $\frac{x^{2}}{a^{2}}-\frac{y^{2}}{b^{2}}=1$

Question : If $x=\operatorname{cosec \theta}-\sin\theta$ and $y=\sec\theta-\cos\theta$, then the relation between $x$ and $y$ is:

Option 1: $x^{2}+y^{2}+3=1$

Option 2: $x^{2}y^{2}\left ( x^{2}+y^{2}+3 \right )=1$

Option 3: $x^{2}\left ( x^{2}+y^{2}-5 \right )=1$

Option 4: $y^{2}\left ( x^{2}+y^{2}-5 \right )=1$

Question : If $\sin \theta \cos \theta=\frac{1}{\sqrt{3}}$ then the value of $\left(\sin ^4 \theta+\cos ^4 \theta\right)$ is:

Option 1: $1$

Option 2: $\frac{5}{3}$

Option 3: $\frac{2}{3}$

Option 4: $\frac{1}{3}$

Question : If $\sin \theta \cos \theta=\frac{\sqrt{2}}{3}$,then the value of $\left(\sin ^6 \theta+\cos ^6 \theta\right)$ is:

Option 1: $\frac{1}{3}$

Option 2: $\frac{4}{3}$

Option 3: $\frac{2}{3}$

Option 4: $\frac{5}{3}$