Understanding the Trigonometry Problem in ΔPQR

The question asks us to evaluate a trigonometric expression involving angles P and R in a right-angled triangle ΔPQR, where ∠Q = 90°. We are given that cot R = \(\frac{1}{3}\).

In a right-angled triangle, the trigonometric ratios relate the angles to the lengths of the sides. Since ∠Q = 90°, PR is the hypotenuse, PQ is the side opposite to angle R and adjacent to angle P, and QR is the side opposite to angle P and adjacent to angle R.

Determining Side Lengths using cot R

We are given cot R = \(\frac{1}{3}\). In a right triangle, cotangent of an angle is defined as the ratio of the adjacent side to the opposite side.

For angle R:

• Adjacent side to R is QR.
• Opposite side to R is PQ.

So, cot R = \(\frac{QR}{PQ} = \frac{1}{3}\).

We can assume the lengths of the sides are proportional to this ratio. Let QR = \(k\) and PQ = \(3k\) for some positive constant \(k\).

Now, we need to find the length of the hypotenuse PR using the Pythagorean theorem:

$PR^2 = PQ^2 + QR^2$

$PR^2 = (3k)^2 + (k)^2$

$PR^2 = 9k^2 + k^2$

$PR^2 = 10k^2$

$PR = \sqrt{10k^2} = k\sqrt{10}$

So, the side lengths are PQ = \(3k\), QR = \(k\), and PR = \(k\sqrt{10}\).

Calculating Trigonometric Ratios for Angles R and P

Now, we can calculate the required trigonometric ratios for both angle R and angle P using the side lengths we found.

Trigonometric Ratios for Angle R:

• sin R = \(\frac{\text{Opposite}}{\text{Hypotenuse}} = \frac{PQ}{PR} = \frac{3k}{k\sqrt{10}} = \frac{3}{\sqrt{10}}\)
• cos R = \(\frac{\text{Adjacent}}{\text{Hypotenuse}} = \frac{QR}{PR} = \frac{k}{k\sqrt{10}} = \frac{1}{\sqrt{10}}\)
• cosec R = \(\frac{1}{\text{sin R}} = \frac{\sqrt{10}}{3}\)
• sec R = \(\frac{1}{\text{cos R}} = \frac{\sqrt{10}}{1} = \sqrt{10}\)

Trigonometric Ratios for Angle P:

Since ∠Q = 90°, ∠P + ∠R = 90°. Angles P and R are complementary. We can use this property, or calculate ratios directly from the sides.

• sin P = \(\frac{\text{Opposite}}{\text{Hypotenuse}} = \frac{QR}{PR} = \frac{k}{k\sqrt{10}} = \frac{1}{\sqrt{10}}\) (Note: sin P = cos R)
• cos P = \(\frac{\text{Adjacent}}{\text{Hypotenuse}} = \frac{PQ}{PR} = \frac{3k}{k\sqrt{10}} = \frac{3}{\sqrt{10}}\) (Note: cos P = sin R)
• sec P = \(\frac{1}{\text{cos P}} = \frac{\sqrt{10}}{3}\) (Note: sec P = cosec R)
• cosec P = \(\frac{1}{\text{sin P}} = \frac{\sqrt{10}}{1} = \sqrt{10}\) (Note: cosec P = sec R)

Let's summarize the relevant ratios needed for the expression:

Evaluating the Given Trigonometric Expression

The expression we need to evaluate is: \(\frac{secP(cosR+sinP)}{cosecR(sinR-cosec P)}\)

Substitute the values we calculated into the expression:

Numerator: $secP(cosR+sinP) = \left(\frac{\sqrt{10}}{3}\right)\left(\frac{1}{\sqrt{10}} + \frac{1}{\sqrt{10}}\right)$

$= \left(\frac{\sqrt{10}}{3}\right)\left(\frac{1+1}{\sqrt{10}}\right)$

$= \left(\frac{\sqrt{10}}{3}\right)\left(\frac{2}{\sqrt{10}}\right)$

The \(\sqrt{10}\) in the numerator and denominator cancels out:

$= \frac{2}{3}$

Denominator: $cosecR(sinR-cosec P) = \left(\frac{\sqrt{10}}{3}\right)\left(\frac{3}{\sqrt{10}} - \sqrt{10}\right)$

To simplify the term inside the second parenthesis, find a common denominator:

$\frac{3}{\sqrt{10}} - \sqrt{10} = \frac{3}{\sqrt{10}} - \frac{\sqrt{10} \times \sqrt{10}}{\sqrt{10}} = \frac{3 - 10}{\sqrt{10}} = \frac{-7}{\sqrt{10}}$

Now substitute this back into the denominator expression:

$cosecR(sinR-cosec P) = \left(\frac{\sqrt{10}}{3}\right)\left(\frac{-7}{\sqrt{10}}\right)$

The \(\sqrt{10}\) in the numerator and denominator cancels out:

$= \frac{-7}{3}$

Now, assemble the full expression:

$\frac{secP(cosR+sinP)}{cosecR(sinR-cosec P)} = \frac{\text{Numerator}}{\text{Denominator}} = \frac{\frac{2}{3}}{\frac{-7}{3}}$

To divide by a fraction, multiply by its reciprocal:

$= \frac{2}{3} \times \frac{3}{-7}$

The 3 in the numerator and denominator cancels out:

$= \frac{2}{-7} = -\frac{2}{7}$

Thus, the value of the given expression is \(-\frac{2}{7}\).

Revision Table: Key Concepts Used

Additional Information on Right Triangle Trigonometry

In any right-angled triangle, the relationships between angles and sides are fundamental to trigonometry. For an acute angle θ:

• sin θ = Opposite side / Hypotenuse
• cos θ = Adjacent side / Hypotenuse
• tan θ = Opposite side / Adjacent side

The reciprocal ratios are:

• cosec θ = 1 / sin θ = Hypotenuse / Opposite side
• sec θ = 1 / cos θ = Hypotenuse / Adjacent side
• cot θ = 1 / tan θ = Adjacent side / Opposite side

These definitions are crucial for solving problems involving right triangles, like the one discussed here, where we used the given cot R to determine the side ratios and then calculated other trigonometric values.