Q03 Test 40 English

Test No. 40 – Complete Solutions

160

Prove: \(\tan\theta \cdot \cot\theta = 1\)

🔑 Key Concept – Reciprocal Identity: By definition, cotangent is the reciprocal of tangent: \[ \cot\theta = \frac{1}{\tan\theta} \qquad (\tan\theta \neq 0) \] Also expressible as: \[ \tan\theta = \frac{\sin\theta}{\cos\theta} \qquad \cot\theta = \frac{\cos\theta}{\sin\theta} \]

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

Step 2 – Replace with sin/cos definitions: \[ = \frac{\sin\theta}{\cos\theta} \times \frac{\cos\theta}{\sin\theta} \]

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

\( \tan\theta \cdot \cot\theta = \dfrac{\sin\theta}{\cos\theta} \times \dfrac{\cos\theta}{\sin\theta} = 1 \qquad \textbf{Proved} \)

161

Find equation of line passing through \((-1,\,9)\) and \((-5,\,-3)\).

🔑 Key Concept – Equation of Line Through Two Points: First find slope: \[ m = \frac{y_2 - y_1}{x_2 - x_1} \] Then apply point-slope form: \[ y - y_1 = m(x - x_1) \]

Given: \((x_1, y_1) = (-1,\ 9)\),  \((x_2, y_2) = (-5,\ -3)\)

Step 1 – Find the slope: \[ m = \frac{-3 - 9}{-5 - (-1)} = \frac{-12}{-5 + 1} = \frac{-12}{-4} = 3 \]

Step 2 – Apply point-slope form using \((-1,\ 9)\): \[ y - 9 = 3\bigl(x - (-1)\bigr) \] \[ y - 9 = 3(x + 1) \]

Step 3 – Expand and rearrange: \[ y - 9 = 3x + 3 \] \[ y = 3x + 3 + 9 \] \[ y = 3x + 12 \] \[ 3x - y + 12 = 0 \]

\( \boxed{3x - y + 12 = 0} \)

162

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

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

Method 1 – Using sine subtraction formula: \[ \sin 60°\cos 30° - \cos 60°\sin 30° = \sin(60° - 30°) = \sin 30° = \frac{1}{2} \]

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

\( \boxed{\sin 60°\cos 30° - \cos 60°\sin 30° = \dfrac{1}{2}} \)

163

Find the unknown volume \(V_2\) for similar cones with radii 2.5 cm and 4 cm, given \(V_1 = 200\ \text{cm}^3\).

🔑 Key Concept – Volumes of Similar Solids: For two similar 3D solids with corresponding lengths \(r_1\) and \(r_2\): \[ \frac{V_1}{V_2} = \left(\frac{r_1}{r_2}\right)^3 \] Volumes scale as the cube of the ratio of corresponding lengths, because volume is a three-dimensional measure.

Given: Cone 1: radius \(r_1 = 2.5\ \text{cm}\),  \(V_1 = 200\ \text{cm}^3\)
Cone 2: radius \(r_2 = 4\ \text{cm}\),  \(V_2 = ?\)

Step 1 – Write the formula: \[ \frac{V_1}{V_2} = \left(\frac{r_1}{r_2}\right)^3 \]

Step 2 – Substitute known values: \[ \frac{200}{V_2} = \left(\frac{2.5}{4}\right)^3 \]

Step 3 – Simplify the ratio: \[ \frac{2.5}{4} = \frac{25}{40} = \frac{5}{8} \]

Step 4 – Cube the ratio: \[ \left(\frac{5}{8}\right)^3 = \frac{125}{512} \]

Step 5 – Solve for \(V_2\): \[ \frac{200}{V_2} = \frac{125}{512} \] \[ V_2 = \frac{200 \times 512}{125} = \frac{102400}{125} = 819.2\ \text{cm}^3 \]

\( \boxed{V_2 = 819.2\ \text{cm}^3} \)