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 $x=a\left ( \sin\theta+\cos\theta \right )$ and $y=b\left ( \sin\theta-\cos\theta \right )$, then the value of $\frac{x^{2}}{a^{2}}+\frac{y^{2}}{b^{2}}$ is:

Option 1: 4

Option 2: 3

Option 3: 1

Option 4: 2

Question : If $\sin (x - y) = \frac{1}2$ and $\cos (x + y) = \frac{1}2$, then what is the value of $\sin x \cos x + 2\sin^2x + cos^3x \sec x$?

Option 1: $2$

Option 2: $\sqrt{2}+1$

Option 3: $1$

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

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 $5\tan\theta=4$, then $\frac{5\sin\theta-3\cos\theta}{5\sin\theta+2\cos\theta}$ is equal to:

Option 1: $\frac{2}{3}$

Option 2: $\frac{1}{4}$

Option 3: $\frac{1}{6}$

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