Q03 Test 39 English

Test No. 39 – Complete Solutions

156

Prove: \(\cos\theta \cdot \sec\theta = 1\)

🔑 Key Concept – Reciprocal Identity: By definition, secant is the reciprocal of cosine: \[ \sec\theta = \frac{1}{\cos\theta} \qquad (\cos\theta \neq 0) \] Multiplying a quantity by its reciprocal always equals 1.

Step 1 – Start with the Left Hand Side (LHS): \[ \text{LHS} = \cos\theta \cdot \sec\theta \]

Step 2 – Replace \(\sec\theta\) with its reciprocal definition: \[ = \cos\theta \cdot \frac{1}{\cos\theta} \]

Step 3 – Cancel \(\cos\theta\): \[ = \frac{\cos\theta}{\cos\theta} = 1 = \text{RHS} \]

\( \cos\theta \cdot \sec\theta = \cos\theta \cdot \dfrac{1}{\cos\theta} = 1 \qquad \textbf{Proved} \)

157

Find equation: Line passing through \((-8,\,-5)\) with undefined slope.

🔑 Key Concept – Undefined Slope → Vertical Line: An undefined slope means the line is vertical. A vertical line passes through all points with the same x-coordinate: \[ x = c \] where \(c\) is the x-coordinate of the given point.

Given: Point \((-8,\ -5)\), slope is undefined.

Step 1 – Identify the x-coordinate: The x-coordinate of the given point is \(-8\).

Step 2 – Write the equation: Since the line is vertical, every point on it has \(x = -8\): \[ x = -8 \]

\( \boxed{x = -8} \)

158

Find the value of \(\cos 60°\cos 30° - \sin 60°\sin 30°\).

🔑 Key Concept – Cosine Addition Formula: \[ \cos(A + B) = \cos A\cos B - \sin A\sin B \] So \(\cos 60°\cos 30° - \sin 60°\sin 30° = \cos(60° + 30°) = \cos 90°\).
Standard values: \[ \cos 60° = \frac{1}{2},\quad \cos 30° = \frac{\sqrt{3}}{2},\quad \sin 60° = \frac{\sqrt{3}}{2},\quad \sin 30° = \frac{1}{2} \]

Method 1 – Using the cosine addition formula: \[ \cos 60°\cos 30° - \sin 60°\sin 30° = \cos(60° + 30°) = \cos 90° = 0 \]

Method 2 – Direct substitution: \[ = \frac{1}{2} \times \frac{\sqrt{3}}{2} - \frac{\sqrt{3}}{2} \times \frac{1}{2} \] \[ = \frac{\sqrt{3}}{4} - \frac{\sqrt{3}}{4} = 0 \]

\( \boxed{\cos 60°\cos 30° - \sin 60°\sin 30° = 0} \)

159

Find the unknown quantities in the following figures (similar triangles).

🔑 Key Concept – Ratio of Areas of Similar Figures: \[ \frac{A_1}{A_2} = \left(\frac{l_1}{l_2}\right)^2 \] Rearranging to find the unknown area: \[ A_1 = A_2 \times \left(\frac{l_1}{l_2}\right)^2 \]

Given: Triangle 1: \(l_1 = 15\ \text{cm},\quad A_1 = ?\)
Triangle 2: \(l_2 = 12\ \text{cm},\quad A_2 = 96\ \text{cm}^2\)

Step 1 – Write the formula: \[ \frac{A_1}{A_2} = \left(\frac{l_1}{l_2}\right)^2 \]

Step 2 – Substitute known values: \[ \frac{A_1}{96} = \left(\frac{15}{12}\right)^2 \]

Step 3 – Simplify the ratio: \[ \frac{15}{12} = \frac{5}{4} \]

Step 4 – Square the ratio: \[ \left(\frac{5}{4}\right)^2 = \frac{25}{16} \]

Step 5 – Solve for \(A_1\): \[ A_1 = 96 \times \frac{25}{16} = \frac{96 \times 25}{16} = 6 \times 25 = 150\ \text{cm}^2 \]

\( \boxed{A_1 = 150\ \text{cm}^2} \)