Solving Trigonometric Equations: Find sin θ + cosec θ

We are given a trigonometric equation and asked to find the value of another expression involving the same trigonometric functions.

The given equation is:

\begin{equation*} -\sin \theta + \operatorname{cosec} \theta = 6 \end{equation*}

We want to find the value of:

\begin{equation*} \sin \theta + \operatorname{cosec} \theta \end{equation*}

Let's use the identity $\operatorname{cosec} \theta = \frac{1}{\sin \theta}$. The given equation can be written as:

\begin{equation*} \frac{1}{\sin \theta} - \sin \theta = 6 \end{equation*}

We are looking for the value of $\sin \theta + \frac{1}{\sin \theta}$.

Let $A = \operatorname{cosec} \theta - \sin \theta = 6$ and $B = \operatorname{cosec} \theta + \sin \theta$. We need to find the value of $B$.

We can use the algebraic identity $(a+b)^2 = (a-b)^2 + 4ab$.

Let $a = \operatorname{cosec} \theta$ and $b = \sin \theta$. Applying the identity:

\begin{equation*} (\operatorname{cosec} \theta + \sin \theta)^2 = (\operatorname{cosec} \theta - \sin \theta)^2 + 4(\operatorname{cosec} \theta)(\sin \theta) \end{equation*}

Substitute the values $A = \operatorname{cosec} \theta - \sin \theta = 6$ and $B = \operatorname{cosec} \theta + \sin \theta$ into the equation:

\begin{equation*} B^2 = A^2 + 4 \left(\frac{1}{\sin \theta}\right)(\sin \theta) \end{equation*}

Since $(\frac{1}{\sin \theta})(\sin \theta) = 1$ (provided $\sin \theta \ne 0$), the equation simplifies to:

\begin{equation*} B^2 = A^2 + 4 \end{equation*}

Substitute the given value of $A = 6$ into the equation:

\begin{equation*} B^2 = 6^2 + 4 \end{equation*}

\begin{equation*} B^2 = 36 + 4 \end{equation*}

\begin{equation*} B^2 = 40 \end{equation*}

Taking the square root of both sides:

\begin{equation*} B = \pm \sqrt{40} \end{equation*}

So, $\sin \theta + \operatorname{cosec} \theta = \pm \sqrt{40}$.

To determine the sign, let's analyze the original equation $-\sin \theta + \operatorname{cosec} \theta = 6$, which is $\operatorname{cosec} \theta - \sin \theta = 6$.

Let $x = \sin \theta$. The equation is $\frac{1}{x} - x = 6$. Multiplying by $x$ gives $1 - x^2 = 6x$, which rearranges to $x^2 + 6x - 1 = 0$.

Using the quadratic formula $x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}$, we find the possible values for $\sin \theta$:

\begin{equation*} x = \frac{-6 \pm \sqrt{6^2 - 4(1)(-1)}}{2(1)} = \frac{-6 \pm \sqrt{36 + 4}}{2} = \frac{-6 \pm \sqrt{40}}{2} = \frac{-6 \pm 2\sqrt{10}}{2} = -3 \pm \sqrt{10} \end{equation*}

Since $\sin \theta$ must be in the range $[-1, 1]$, we check the possible values:

• $\sqrt{10} \approx 3.16$.
• $\sin \theta = -3 + \sqrt{10} \approx -3 + 3.16 = 0.16$. This value is in $[-1, 1]$.
• $\sin \theta = -3 - \sqrt{10} \approx -3 - 3.16 = -6.16$. This value is not in $[-1, 1]$.

Thus, $\sin \theta = -3 + \sqrt{10}$. Since this value is positive ($\approx 0.16$), $\sin \theta > 0$.

If $\sin \theta > 0$, then $\operatorname{cosec} \theta = \frac{1}{\sin \theta}$ is also positive.

The sum $\sin \theta + \operatorname{cosec} \theta$ of two positive numbers must be positive.

Therefore, we take the positive square root:

\begin{equation*} \sin \theta + \operatorname{cosec} \theta = \sqrt{40} \end{equation*}

Revision Table: Key Trigonometric Identities

Additional Information on sin θ and cosec θ

The sine function, $\sin \theta$, is defined for all real values of $\theta$. Its range is $[-1, 1]$. This means $-1 \le \sin \theta \le 1$ for any real $\theta$.

The cosecant function, $\operatorname{cosec} \theta$, is the reciprocal of the sine function, $\operatorname{cosec} \theta = \frac{1}{\sin \theta}$. It is defined for all real values of $\theta$ where $\sin \theta \ne 0$. This means $\theta$ cannot be an integer multiple of $\pi$ ($0^\circ, 180^\circ, 360^\circ$, etc.).

The range of $\operatorname{cosec} \theta$ is $(-\infty, -1] \cup [1, \infty)$. This means $\operatorname{cosec} \theta \le -1$ or $\operatorname{cosec} \theta \ge 1$.

If $\sin \theta > 0$, then $\operatorname{cosec} \theta > 0$. This occurs when $\theta$ is in Quadrant I or II ($0^\circ < \theta < 180^\circ$ or $0 < \theta < \pi$ radians). In this case, $\sin \theta \in (0, 1]$ and $\operatorname{cosec} \theta \in [1, \infty)$. Their sum $\sin \theta + \operatorname{cosec} \theta$ will always be positive.

If $\sin \theta < 0$, then $\operatorname{cosec} \theta < 0$. This occurs when $\theta$ is in Quadrant III or IV ($180^\circ < \theta < 360^\circ$ or $\pi < \theta < 2\pi$ radians, excluding endpoints). In this case, $\sin \theta \in [-1, 0)$ and $\operatorname{cosec} \theta \in (-\infty, -1]$. Their sum $\sin \theta + \operatorname{cosec} \theta$ will always be negative.

In our problem, we found $\sin \theta = \sqrt{10} - 3 \approx 0.16$, which is positive. This confirms that $\sin \theta + \operatorname{cosec} \theta$ must be positive.