Solving Trigonometry Problems in Right Triangles

This problem involves finding the value of a trigonometric expression in a right-angled triangle ΔPQR, where ∠Q is 90°. We are given the value of tan R and need to use this information to find the values of other trigonometric ratios for angles P and R.

Understanding the Right Triangle ΔPQR

Given that ΔPQR is a right-angled triangle at Q, angles P and R are acute angles and are complementary. This means that the sum of angles P and R is 90° ($\text{P} + \text{R} = 90°$). This property is very useful as it allows us to relate the trigonometric ratios of angle P to those of angle R using complementary angle identities.

Using the Given Information: tan R

We are given that tan R = $\frac{1}{3}$. In a right-angled triangle, the tangent of an angle is defined as the ratio of the length of the side opposite the angle to the length of the side adjacent to the angle.

For angle R in ΔPQR:

• Side opposite to R is PQ.
• Side adjacent to R is QR.

So, tan R = $\frac{\text{PQ}}{\text{QR}} = \frac{1}{3}$.

We can assume that PQ = $k$ and QR = $3k$ for some positive constant $k$.

Finding the Hypotenuse PR

In a right-angled triangle, the square of the hypotenuse is equal to the sum of the squares of the other two sides (Pythagoras theorem). Here, PR is the hypotenuse.

$\text{PR}^2 = \text{PQ}^2 + \text{QR}^2$

$\text{PR}^2 = (k)^2 + (3k)^2$

$\text{PR}^2 = k^2 + 9k^2$

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

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

Now we have the lengths of all three sides in terms of $k$: PQ = $k$, QR = $3k$, and PR = $k\sqrt{10}$.

Calculating Trigonometric Ratios for Angle R

Using the side lengths, we can find the trigonometric ratios for angle R:

• sin R = $\frac{\text{Opposite}}{\text{Hypotenuse}} = \frac{\text{PQ}}{\text{PR}} = \frac{k}{k\sqrt{10}} = \frac{1}{\sqrt{10}}$
• cos R = $\frac{\text{Adjacent}}{\text{Hypotenuse}} = \frac{\text{QR}}{\text{PR}} = \frac{3k}{k\sqrt{10}} = \frac{3}{\sqrt{10}}$
• tan R = $\frac{\text{Opposite}}{\text{Adjacent}} = \frac{\text{PQ}}{\text{QR}} = \frac{k}{3k} = \frac{1}{3}$ (This matches the given information)
• cosec R = $\frac{1}{\text{sin R}} = \frac{1}{1/\sqrt{10}} = \sqrt{10}$
• sec R = $\frac{1}{\text{cos R}} = \frac{1}{3/\sqrt{10}} = \frac{\sqrt{10}}{3}$

Calculating Trigonometric Ratios for Angle P

Since P + R = 90°, we can use the complementary angle identities:

• sin P = cos R = $\frac{3}{\sqrt{10}}$
• cos P = sin R = $\frac{1}{\sqrt{10}}$
• tan P = cot R = $\frac{1}{\text{tan R}} = \frac{1}{1/3} = 3$
• cosec P = sec R = $\frac{\sqrt{10}}{3}$
• sec P = cosec R = $\sqrt{10}$

Evaluating the Expression

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

Let's substitute the values we found:

• sec P = $\sqrt{10}$
• cos R = $\frac{3}{\sqrt{10}}$
• sin P = $\frac{3}{\sqrt{10}}$
• cosec R = $\sqrt{10}$
• sin R = $\frac{1}{\sqrt{10}}$
• cosec P = $\frac{\sqrt{10}}{3}$

Substitute these values into the expression:

Numerator: sec P (cos R + sin P)

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

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

= $\sqrt{10} \left(\frac{6}{\sqrt{10}}\right)$

= $6$

Denominator: cosec R (sin R - cosec P)

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

To simplify the term in the parenthesis, find a common denominator:

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

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

= $\sqrt{10} \left(\frac{-7}{3\sqrt{10}}\right)$

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

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

Now, divide the Numerator by the Denominator:

Value of expression = $\frac{6}{-7/3}$

= $6 \times \left(-\frac{3}{7}\right)$

= $-\frac{18}{7}$

Summary of Ratios

Final Answer Calculation

Substituting values into $\rm \frac{secP(cosR+sinP)}{cosecR(sinR-cosecP)}$:

Numerator: $\sqrt{10} \left(\frac{3}{\sqrt{10}} + \frac{3}{\sqrt{10}}\right) = \sqrt{10} \left(\frac{6}{\sqrt{10}}\right) = 6$

Denominator: $\sqrt{10} \left(\frac{1}{\sqrt{10}} - \frac{\sqrt{10}}{3}\right) = \sqrt{10} \left(\frac{3-10}{3\sqrt{10}}\right) = \sqrt{10} \left(\frac{-7}{3\sqrt{10}}\right) = -\frac{7}{3}$

Expression value = $\frac{6}{-7/3} = 6 \times \left(-\frac{3}{7}\right) = -\frac{18}{7}$

Revision Table: Right Triangle Trigonometry

Review key concepts related to right-angled triangles and trigonometric ratios.

Additional Information: Trigonometric Identities

Understanding trigonometric identities helps simplify expressions and solve problems efficiently. Some fundamental identities include:

• Pythagorean Identities:
  • sin² $\theta$ + cos² $\theta$ = 1
  • 1 + tan² $\theta$ = sec² $\theta$
  • 1 + cot² $\theta$ = cosec² $\theta$
• Quotient Identities:
  • tan $\theta$ = $\frac{\text{sin } \theta}{\text{cos } \theta}$
  • cot $\theta$ = $\frac{\text{cos } \theta}{\text{sin } \theta}$
• Reciprocal Identities: (Already covered in the table)

These identities are derived from the definitions of the trigonometric ratios and the Pythagoras theorem. They are crucial tools for manipulating and simplifying trigonometric expressions in various problems.