Solving Trigonometric Equations

The problem asks us to find the value of \((\tan θ + \operatorname{cosec} θ)\) given a trigonometric equation involving \(\cos θ\), \(\cot θ\), and \(\sin θ\).

The given equation is:

\(\rm \frac{\cos^2 θ}{\cot^2 θ + \sin^2 θ -1} = 3\)

We are given that \(0^\circ < θ < 90^\circ\), which means θ is in the first quadrant. In the first quadrant, all trigonometric ratios are positive.

Simplifying the Denominator

Let's simplify the denominator of the given fraction: \(\cot^2 θ + \sin^2 θ -1\).

We know the identity: \(\sin^2 θ + \cos^2 θ = 1\). Rearranging this gives \(\sin^2 θ - 1 = -\cos^2 θ\).

Substitute this into the denominator:

Denominator \( = \cot^2 θ + (\sin^2 θ -1)\)

Denominator \( = \cot^2 θ - \cos^2 θ\)

We also know the identity: \(\cot θ = \frac{\cos θ}{\sin θ}\), so \(\cot^2 θ = \frac{\cos^2 θ}{\sin^2 θ}\).

Substitute this into the denominator:

Denominator \( = \frac{\cos^2 θ}{\sin^2 θ} - \cos^2 θ\)

Factor out \(\cos^2 θ \) from the denominator:

Denominator \( = \cos^2 θ \left(\frac{1}{\sin^2 θ} - 1\right)\)

Simplify the expression in the parenthesis:

\(\frac{1}{\sin^2 θ} - 1 = \frac{1 - \sin^2 θ}{\sin^2 θ}\)

Using the identity \(1 - \sin^2 θ = \cos^2 θ\), we get:

\(\frac{1 - \sin^2 θ}{\sin^2 θ} = \frac{\cos^2 θ}{\sin^2 θ} = \cot^2 θ\)

So, the denominator simplifies to:

Denominator \( = \cos^2 θ \cdot \cot^2 θ\)

Solving for θ

Now substitute the simplified denominator back into the original equation:

\(\rm \frac{\cos^2 θ}{\cos^2 θ \cdot \cot^2 θ} = 3\)

Since \(0^\circ < θ < 90^\circ\), \(\cos θ \neq 0\). So, we can cancel \(\cos^2 θ\) from the numerator and denominator:

\(\frac{1}{\cot^2 θ} = 3\)

We know that \(\frac{1}{\cot θ} = \tan θ\), so \(\frac{1}{\cot^2 θ} = \tan^2 θ\).

\(\tan^2 θ = 3\)

Taking the square root of both sides:

\(\tan θ = \pm \sqrt{3}\)

Since \(0^\circ < θ < 90^\circ\), \(\tan θ\) must be positive. Therefore:

\(\tan θ = \sqrt{3}\)

We know that \(\tan 60^\circ = \sqrt{3}\). Since θ is in the first quadrant, the value of θ is \(60^\circ\).

Evaluating the Expression (tan θ + cosec θ)

Now we need to find the value of \((\tan θ + \operatorname{cosec} θ)\) for \(\theta = 60^\circ\).

• \(\tan 60^\circ = \sqrt{3}\)
• \(\sin 60^\circ = \frac{\sqrt{3}}{2}\)
• \(\operatorname{cosec} 60^\circ = \frac{1}{\sin 60^\circ} = \frac{1}{\frac{\sqrt{3}}{2}} = \frac{2}{\sqrt{3}}\)

Now add these values:

\(\tan θ + \operatorname{cosec} θ = \sqrt{3} + \frac{2}{\sqrt{3}}\)

To add these, find a common denominator, which is \(\sqrt{3}\):

\(\sqrt{3} + \frac{2}{\sqrt{3}} = \frac{\sqrt{3} \cdot \sqrt{3}}{\sqrt{3}} + \frac{2}{\sqrt{3}} = \frac{3}{\sqrt{3}} + \frac{2}{\sqrt{3}}\)

\(= \frac{3 + 2}{\sqrt{3}} = \frac{5}{\sqrt{3}}\)

To rationalize the denominator, multiply the numerator and the denominator by \(\sqrt{3}\):

\(\frac{5}{\sqrt{3}} \times \frac{\sqrt{3}}{\sqrt{3}} = \frac{5\sqrt{3}}{3}\)

So, the value of \((\tan θ + \operatorname{cosec} θ)\) is \(\frac{5\sqrt{3}}{3}\).

Summary of Steps

1. Start with the given trigonometric equation.
2. Simplify the denominator using trigonometric identities.
3. Substitute the simplified denominator back into the equation and solve for \(\tan^2 θ\).
4. Find the value of \(\tan θ\) considering the given range of θ.
5. Determine the value of θ.
6. Evaluate the required expression \((\tan θ + \operatorname{cosec} θ)\) using the value of θ and standard trigonometric values.

Revision Table: Key Trigonometric Identities

Additional Information: Trigonometric Ratios in the First Quadrant

For angles \( θ \) in the first quadrant (\(0^\circ < θ < 90^\circ\)), all basic trigonometric ratios (sin, cos, tan, cosec, sec, cot) are positive. This is because in the first quadrant, both the x and y coordinates of a point on the unit circle are positive. The trigonometric ratios are defined based on the ratios of these coordinates and the radius (which is always positive).

For example:

• \(\sin θ = \frac{\text{y-coordinate}}{\text{radius}} > 0\)
• \(\cos θ = \frac{\text{x-coordinate}}{\text{radius}} > 0\)
• \(\tan θ = \frac{\text{y-coordinate}}{\text{x-coordinate}} > 0\)

This information was crucial in determining that \(\tan θ = \sqrt{3}\) (positive value) after solving \(\tan^2 θ = 3\).