Question

Please help me solve this question ⁉️⁉️⁉️​

Answer 1

Let the circle in the attached figure be a unit circle

and (x,y) be a point on the unit circle

From the attached figure, we get

$\sin\theta=\frac{y}{1}=y\\and\:\cos\theta=\frac{x}{1}=x$

Using Pythagoras' theorem, we get

$x^2+y^2=1$

Part 1:

$y=\pm\sqrt{1-x^2}$   [Note: $\sqrt{n^2}=|n|$]

For (x,y) to be a point in the circle in the cartesian plane, x and y should be real numbers, i.e., $(x,y)\in\mathbb{R}^2$

Here, y is real $(y\in\mathbb{R})$ only when $1-x^2\geq 0, \forall x\in\mathbb{R}$

$\implies x^2\leq 1\\Also\:x^2\geq 0\\\implies 0\leq x^2\leq 1\\\implies 0\geq -x^2\geq -1\\\implies 0\leq 1-x^2\leq 1\\\implies 0\leq \sqrt{1-x^2}\leq 1\\\implies 0\leq |\pm\sqrt{1-x^2}|\leq 1\\\implies 0\leq |y|\leq 1\\\implies 0\leq |\sin\theta|\leq 1 \:\:(1)\\\implies 0\leq \frac{1}{|cosec\theta|}\leq 1\\\implies 1\leq |cosec\theta|<\infty\:\:(2)$

Part 2:

$x=\pm\sqrt{1-y^2}$   [Note: $\sqrt{n^2}=|n|$]

For (x,y) to be a point in the circle in the cartesian plane, x and y should be real numbers, i.e., $(x,y)\in\mathbb{R}^2$

Here, x is real $(x\in\mathbb{R})$ only when $1-y^2\geq 0, \forall y\in\mathbb{R}$

$\implies y^2\leq 1\\\\Also\:y^2\geq 0\\\implies 0\leq y^2\leq 1\\\implies 0\geq -y^2\geq -1\\\implies 0\leq 1-y^2\leq 1\\\implies 0\leq \sqrt{1-y^2}\leq 1\\\implies 0\leq |\pm\sqrt{1-y^2}|\leq 1\\\implies 0\leq |x|\leq 1\\\implies 0\leq |\cos\theta|\leq 1 \:\:(3)\\\implies 0\leq |\frac{1}{\sec\theta}|\leq 1\\\implies 1\leq |\sec\theta|<\infty\:\:(4)$

Hence, shown