Ganit Prakash - Class-X - Trigonometry
Let us work out 23.2 Trigonometric Ratios
📘 Exercise 23.2 Solutions (Q1 - Q11)
Trigonometric RatiosExplanation:
Let the window be at point $A$ and the ground be the horizontal line $BC$. Let $AB$ be the height of the window from the ground.
The ladder is represented by $AC$, so the length of the hypotenuse $AC = 2\sqrt{3}\text{ m.}$
The ladder makes an angle of $60^\circ$ with the ground, so $\angle ACB = 60^\circ$.
We need to find the height $AB$ (Perpendicular).
In the right-angled triangle $\Delta ABC$,
\[\begin{array}{l} \sin(\angle ACB) = \frac{\text{Perpendicular}}{\text{Hypotenuse}} = \frac{AB}{AC} \\ \sin(60^\circ) = \frac{AB}{2\sqrt{3}} \end{array}\]We know that $\sin(60^\circ) = \frac{\sqrt{3}}{2}$. Substituting this value:
\[\begin{array}{l} \frac{\sqrt{3}}{2} = \frac{AB}{2\sqrt{3}} \\ AB = \frac{\sqrt{3} \times 2\sqrt{3}}{2} \\ AB = \frac{2 \times 3}{2} \\ AB = 3\text{ m.} \end{array}\]Therefore, the height of the window from the ground is $3\text{ m.}$
Explanation:
In the right-angled triangle ABC, $\angle B = 90^\circ$.
Given: Perpendicular $AB = 8\sqrt{3}\text{ cm.}$ and Base $BC = 8\text{ cm.}$ (with respect to angle $C$).
To find $\angle ACB$:
\[\begin{array}{l} \tan(\angle ACB) = \frac{\text{Perpendicular}}{\text{Base}} = \frac{AB}{BC} \\ \tan(\angle ACB) = \frac{8\sqrt{3}}{8} \\ \tan(\angle ACB) = \sqrt{3} \end{array}\]Since $\tan(60^\circ) = \sqrt{3}$, we have:
\[\begin{array}{l} \angle ACB = \mathbf{60^\circ} \end{array}\]To find $\angle BAC$:
The sum of angles in a triangle is $180^\circ$. Since $\angle B = 90^\circ$, the other two acute angles must add up to $90^\circ$.
\[\begin{array}{l} \angle BAC + \angle ACB = 90^\circ \\ \angle BAC + 60^\circ = 90^\circ \\ \angle BAC = 90^\circ - 60^\circ \\ \angle BAC = \mathbf{30^\circ} \end{array}\]Alternatively, using trigonometric ratios for $\angle BAC$:
\[\begin{array}{l} \tan(\angle BAC) = \frac{\text{Opposite}}{\text{Adjacent}} = \frac{BC}{AB} \\ \tan(\angle BAC) = \frac{8}{8\sqrt{3}} = \frac{1}{\sqrt{3}} \end{array}\]Since $\tan(30^\circ) = \frac{1}{\sqrt{3}}$, we get $\angle BAC = \mathbf{30^\circ}$.
Explanation:
In the right-angled triangle ABC, $\angle B = 90^\circ$ and Hypotenuse $AC = 20\text{ cm.}$
We are given the reference angle $\angle A = 30^\circ$.
To find the length of side BC (Perpendicular opposite to $\angle A$):
\[\begin{array}{l} \sin(\angle A) = \frac{\text{Perpendicular}}{\text{Hypotenuse}} = \frac{BC}{AC} \\ \sin(30^\circ) = \frac{BC}{20} \end{array}\]We know $\sin(30^\circ) = \frac{1}{2}$:
\[\begin{array}{l} \frac{1}{2} = \frac{BC}{20} \\ BC = \frac{20}{2} = \mathbf{10\text{ cm.}} \end{array}\]To find the length of side AB (Base adjacent to $\angle A$):
\[\begin{array}{l} \cos(\angle A) = \frac{\text{Base}}{\text{Hypotenuse}} = \frac{AB}{AC} \\ \cos(30^\circ) = \frac{AB}{20} \end{array}\]We know $\cos(30^\circ) = \frac{\sqrt{3}}{2}$:
\[\begin{array}{l} \frac{\sqrt{3}}{2} = \frac{AB}{20} \\ AB = \frac{20\sqrt{3}}{2} = \mathbf{10\sqrt{3}\text{ cm.}} \end{array}\]Explanation:
In the right-angled triangle PQR, $\angle Q = 90^\circ$ and Hypotenuse $PR = 3\sqrt{2}\text{ m.}$
We are given the reference angle $\angle R = 45^\circ$.
To find the length of side PQ (Perpendicular opposite to $\angle R$):
\[\begin{array}{l} \sin(\angle R) = \frac{\text{Perpendicular}}{\text{Hypotenuse}} = \frac{PQ}{PR} \\ \sin(45^\circ) = \frac{PQ}{3\sqrt{2}} \end{array}\]We know $\sin(45^\circ) = \frac{1}{\sqrt{2}}$:
\[\begin{array}{l} \frac{1}{\sqrt{2}} = \frac{PQ}{3\sqrt{2}} \\ PQ = \frac{3\sqrt{2}}{\sqrt{2}} = \mathbf{3\text{ m.}} \end{array}\]To find the length of side QR (Base adjacent to $\angle R$):
\[\begin{array}{l} \cos(\angle R) = \frac{\text{Base}}{\text{Hypotenuse}} = \frac{QR}{PR} \\ \cos(45^\circ) = \frac{QR}{3\sqrt{2}} \end{array}\]We know $\cos(45^\circ) = \frac{1}{\sqrt{2}}$:
\[\begin{array}{l} \frac{1}{\sqrt{2}} = \frac{QR}{3\sqrt{2}} \\ QR = \frac{3\sqrt{2}}{\sqrt{2}} = \mathbf{3\text{ m.}} \end{array}\]Note: Since one acute angle is $45^\circ$ in a right triangle, it is an isosceles right triangle, which implies $PQ = QR$.
$\sin^2 45^\circ - \operatorname{cosec}^2 60^\circ + \sec^2 30^\circ$
Explanation:
Substituting the standard trigonometric values:
\[\begin{array}{l} = \left(\frac{1}{\sqrt{2}}\right)^2 - \left(\frac{2}{\sqrt{3}}\right)^2 + \left(\frac{2}{\sqrt{3}}\right)^2 \\ = \frac{1}{2} - \frac{4}{3} + \frac{4}{3} \\ = \mathbf{\frac{1}{2}} \end{array}\]$\sec^2 45^\circ - \cot^2 45^\circ - \sin^2 30^\circ - \sin^2 60^\circ$
Explanation:
Substituting the standard trigonometric values:
\[\begin{array}{l} = (\sqrt{2})^2 - (1)^2 - \left(\frac{1}{2}\right)^2 - \left(\frac{\sqrt{3}}{2}\right)^2 \\ = 2 - 1 - \frac{1}{4} - \frac{3}{4} \\ = 1 - \left(\frac{1}{4} + \frac{3}{4}\right) \\ = 1 - \frac{4}{4} \\ = 1 - 1 = \mathbf{0} \end{array}\]$3\tan^2 45^\circ - \sin^2 60^\circ - \frac{1}{3}\cot^2 30^\circ - \frac{1}{8}\sec^2 45^\circ$
Explanation:
Substituting the standard trigonometric values:
\[\begin{array}{l} = 3(1)^2 - \left(\frac{\sqrt{3}}{2}\right)^2 - \frac{1}{3}(\sqrt{3})^2 - \frac{1}{8}(\sqrt{2})^2 \\ = 3(1) - \frac{3}{4} - \frac{1}{3}(3) - \frac{1}{8}(2) \\ = 3 - \frac{3}{4} - 1 - \frac{1}{4} \\ = (3 - 1) - \left(\frac{3}{4} + \frac{1}{4}\right) \\ = 2 - \frac{4}{4} \\ = 2 - 1 = \mathbf{1} \end{array}\]$\frac{4}{3}\cot^2 30^\circ + 3\sin^2 60^\circ - 2\operatorname{cosec}^2 60^\circ - \frac{3}{4}\tan^2 30^\circ$
Explanation:
Substituting the standard trigonometric values:
\[\begin{array}{l} = \frac{4}{3}(\sqrt{3})^2 + 3\left(\frac{\sqrt{3}}{2}\right)^2 - 2\left(\frac{2}{\sqrt{3}}\right)^2 - \frac{3}{4}\left(\frac{1}{\sqrt{3}}\right)^2 \\ = \frac{4}{3}(3) + 3\left(\frac{3}{4}\right) - 2\left(\frac{4}{3}\right) - \frac{3}{4}\left(\frac{1}{3}\right) \\ = 4 + \frac{9}{4} - \frac{8}{3} - \frac{1}{4} \\ = 4 + \left(\frac{9}{4} - \frac{1}{4}\right) - \frac{8}{3} \\ = 4 + \frac{8}{4} - \frac{8}{3} \\ = 4 + 2 - \frac{8}{3} \\ = 6 - \frac{8}{3} \\ = \frac{18 - 8}{3} = \frac{10}{3} = \mathbf{3\frac{1}{3}} \end{array}\]$\frac{\frac{1}{3}\cos 30^\circ}{\frac{1}{2}\sin 45^\circ} + \frac{\tan 60^\circ}{\cos 30^\circ}$
Explanation:
Substituting the standard trigonometric values:
\[\begin{array}{l} = \frac{\frac{1}{3} \cdot \frac{\sqrt{3}}{2}}{\frac{1}{2} \cdot \frac{1}{\sqrt{2}}} + \frac{\sqrt{3}}{\frac{\sqrt{3}}{2}} \\ = \frac{\frac{\sqrt{3}}{6}}{\frac{1}{2\sqrt{2}}} + \sqrt{3} \times \frac{2}{\sqrt{3}} \\ = \frac{\sqrt{3}}{6} \times \frac{2\sqrt{2}}{1} + 2 \\ = \frac{2\sqrt{6}}{6} + 2 \\ = \mathbf{\frac{\sqrt{6}}{3} + 2} \end{array}\]$\cot^2 30^\circ - 2\cos^2 60^\circ - \frac{3}{4}\sec^2 45^\circ - 4\sin^2 30^\circ$
Explanation:
Substituting the standard trigonometric values:
\[\begin{array}{l} = (\sqrt{3})^2 - 2\left(\frac{1}{2}\right)^2 - \frac{3}{4}(\sqrt{2})^2 - 4\left(\frac{1}{2}\right)^2 \\ = 3 - 2\left(\frac{1}{4}\right) - \frac{3}{4}(2) - 4\left(\frac{1}{4}\right) \\ = 3 - \frac{1}{2} - \frac{3}{2} - 1 \\ = 3 - \left(\frac{1}{2} + \frac{3}{2}\right) - 1 \\ = 3 - \frac{4}{2} - 1 \\ = 3 - 2 - 1 = \mathbf{0} \end{array}\]$\sec^2 60^\circ - \cot^2 30^\circ - \frac{2\tan 30^\circ \operatorname{cosec} 60^\circ}{1 + \tan^2 30^\circ}$
Explanation:
Substituting the standard trigonometric values:
\[\begin{array}{l} = (2)^2 - (\sqrt{3})^2 - \frac{2\left(\frac{1}{\sqrt{3}}\right)\left(\frac{2}{\sqrt{3}}\right)}{1 + \left(\frac{1}{\sqrt{3}}\right)^2} \\ = 4 - 3 - \frac{\frac{4}{3}}{1 + \frac{1}{3}} \\ = 1 - \frac{\frac{4}{3}}{\frac{4}{3}} \\ = 1 - 1 = \mathbf{0} \end{array}\]$\frac{\tan 60^\circ - \tan 30^\circ}{1 + \tan 60^\circ \tan 30^\circ} + \cos 60^\circ \cos 30^\circ + \sin 60^\circ \sin 30^\circ$
Explanation:
Substituting the standard trigonometric values:
\[\begin{array}{l} = \frac{\sqrt{3} - \frac{1}{\sqrt{3}}}{1 + (\sqrt{3})\left(\frac{1}{\sqrt{3}}\right)} + \left(\frac{1}{2}\right)\left(\frac{\sqrt{3}}{2}\right) + \left(\frac{\sqrt{3}}{2}\right)\left(\frac{1}{2}\right) \\ = \frac{\frac{3 - 1}{\sqrt{3}}}{1 + 1} + \frac{\sqrt{3}}{4} + \frac{\sqrt{3}}{4} \\ = \frac{\frac{2}{\sqrt{3}}}{2} + \frac{2\sqrt{3}}{4} \\ = \frac{1}{\sqrt{3}} + \frac{\sqrt{3}}{2} \\ = \frac{2 + 3}{2\sqrt{3}} \\ = \frac{5}{2\sqrt{3}} = \mathbf{\frac{5\sqrt{3}}{6}} \end{array}\]$\frac{1 - \sin^2 30^\circ}{1 + \sin^2 30^\circ} \times \frac{\cos^2 60^\circ + \cos^2 30^\circ}{\operatorname{cosec}^2 90^\circ - \cot^2 90^\circ} \div (\sin 60^\circ \tan 30^\circ)$
Explanation:
Substituting the standard trigonometric values:
\[\begin{array}{l} = \frac{1 - \left(\frac{1}{2}\right)^2}{1 + \left(\frac{1}{2}\right)^2} \times \frac{\left(\frac{1}{2}\right)^2 + \left(\frac{\sqrt{3}}{2}\right)^2}{(1)^2 - (0)^2} \div \left( \frac{\sqrt{3}}{2} \times \frac{1}{\sqrt{3}} \right) \\ = \frac{1 - \frac{1}{4}}{1 + \frac{1}{4}} \times \frac{\frac{1}{4} + \frac{3}{4}}{1 - 0} \div \left( \frac{1}{2} \right) \\ = \frac{\frac{3}{4}}{\frac{5}{4}} \times \frac{\frac{4}{4}}{1} \times 2 \\ = \frac{3}{5} \times 1 \times 2 \\ = \frac{6}{5} = \mathbf{1\frac{1}{5}} \end{array}\]📘 Exercise 23.2 Solutions (Q6)
Prove the followingProof:
Taking the Left Hand Side (L.H.S.):
\[\begin{array}{l} \text{L.H.S.} = \sin^2 45^\circ + \cos^2 45^\circ \\ = \left(\frac{1}{\sqrt{2}}\right)^2 + \left(\frac{1}{\sqrt{2}}\right)^2 \\ = \frac{1}{2} + \frac{1}{2} \\ = 1 = \text{R.H.S.} \end{array}\][Proved]
Proof:
\[\begin{array}{l} \text{L.H.S.} = \cos 60^\circ = \frac{1}{2} \\ \\ \text{R.H.S.} = \cos^2 30^\circ - \sin^2 30^\circ \\ = \left(\frac{\sqrt{3}}{2}\right)^2 - \left(\frac{1}{2}\right)^2 \\ = \frac{3}{4} - \frac{1}{4} \\ = \frac{2}{4} = \frac{1}{2} \end{array}\]Since L.H.S. = R.H.S., the given statement is [Proved].
Proof:
\[\begin{array}{l} \text{L.H.S.} = \frac{2\tan 30^\circ}{1-\tan^2 30^\circ} \\ = \frac{2\left(\frac{1}{\sqrt{3}}\right)}{1 - \left(\frac{1}{\sqrt{3}}\right)^2} \\ = \frac{\frac{2}{\sqrt{3}}}{1 - \frac{1}{3}} \\ = \frac{\frac{2}{\sqrt{3}}}{\frac{2}{3}} \\ = \frac{2}{\sqrt{3}} \times \frac{3}{2} \\ = \frac{3}{\sqrt{3}} = \sqrt{3} = \text{R.H.S.} \end{array}\][Proved]
Proof:
\[\begin{array}{l} \text{L.H.S.} = \sqrt{\frac{1+\cos 30^\circ}{1-\cos 30^\circ}} \\ = \sqrt{\frac{1 + \frac{\sqrt{3}}{2}}{1 - \frac{\sqrt{3}}{2}}} \\ = \sqrt{\frac{\frac{2+\sqrt{3}}{2}}{\frac{2-\sqrt{3}}{2}}} \\ = \sqrt{\frac{2+\sqrt{3}}{2-\sqrt{3}}} \end{array}\]Multiplying the numerator and denominator by the conjugate $(2+\sqrt{3})$:
\[\begin{array}{l} = \sqrt{\frac{(2+\sqrt{3})(2+\sqrt{3})}{(2-\sqrt{3})(2+\sqrt{3})}} \\ = \sqrt{\frac{(2+\sqrt{3})^2}{4 - 3}} \\ = \sqrt{(2+\sqrt{3})^2} = 2 + \sqrt{3} \end{array}\]\[\begin{array}{l} \text{R.H.S.} = \sec 60^\circ + \tan 60^\circ \\ = 2 + \sqrt{3} \end{array}\]
Since L.H.S. = R.H.S., the given statement is [Proved].
Proof:
\[\begin{array}{l} \text{L.H.S.} = \frac{2\tan^2 30^\circ}{1-\tan^2 30^\circ} + \sec^2 45^\circ - \cot^2 45^\circ \\ = \frac{2\left(\frac{1}{\sqrt{3}}\right)^2}{1 - \left(\frac{1}{\sqrt{3}}\right)^2} + (\sqrt{2})^2 - (1)^2 \\ = \frac{2\left(\frac{1}{3}\right)}{1 - \frac{1}{3}} + 2 - 1 \\ = \frac{\frac{2}{3}}{\frac{2}{3}} + 1 \\ = 1 + 1 = 2 \end{array}\]\[\begin{array}{l} \text{R.H.S.} = \sec 60^\circ = 2 \end{array}\]
Since L.H.S. = R.H.S., the given statement is [Proved].
Proof:
Note: The angles are in radians. $\frac{\pi}{4} = 45^\circ$, $\frac{\pi}{3} = 60^\circ$, and $\frac{\pi}{6} = 30^\circ$.
\[\begin{array}{l} \text{L.H.S.} = \tan^2 45^\circ \cdot \sin 60^\circ \cdot \tan 30^\circ \cdot \tan^2 60^\circ \\ = (1)^2 \cdot \left(\frac{\sqrt{3}}{2}\right) \cdot \left(\frac{1}{\sqrt{3}}\right) \cdot (\sqrt{3})^2 \\ = 1 \cdot \frac{\sqrt{3}}{2} \cdot \frac{1}{\sqrt{3}} \cdot 3 \\ = \frac{3}{2} \\ = 1\frac{1}{2} = \text{R.H.S.} \end{array}\][Proved]
Proof:
Note: $\frac{\pi}{3} = 60^\circ$, $\frac{\pi}{6} = 30^\circ$, $\frac{\pi}{2} = 90^\circ$, $\frac{\pi}{4} = 45^\circ$.
\[\begin{array}{l} \text{L.H.S.} = \sin 60^\circ \tan 30^\circ + \sin 90^\circ \cos 60^\circ \\ = \left(\frac{\sqrt{3}}{2}\right) \cdot \left(\frac{1}{\sqrt{3}}\right) + (1) \cdot \left(\frac{1}{2}\right) \\ = \frac{1}{2} + \frac{1}{2} \\ = 1 \end{array}\]\[\begin{array}{l} \text{R.H.S.} = 2\sin^2 45^\circ \\ = 2 \cdot \left(\frac{1}{\sqrt{2}}\right)^2 \\ = 2 \cdot \frac{1}{2} = 1 \end{array}\]
Since L.H.S. = R.H.S., the given statement is [Proved].
📘 Exercise 23.2 Solutions (Q7)
Determine the value of xExplanation:
Substituting the standard trigonometric values:
\[\begin{array}{l} x \cdot \left(\frac{1}{\sqrt{2}}\right) \cdot \left(\frac{1}{\sqrt{2}}\right) \cdot (\sqrt{3}) = (1)^2 - \frac{1}{2} \\ x \cdot \frac{1}{2} \cdot \sqrt{3} = 1 - \frac{1}{2} \\ x \cdot \frac{\sqrt{3}}{2} = \frac{1}{2} \end{array}\]Solving for $x$:
\[\begin{array}{l} x = \frac{1}{2} \times \frac{2}{\sqrt{3}} \\ x = \frac{1}{\sqrt{3}} = \mathbf{\frac{\sqrt{3}}{3}} \end{array}\]Explanation:
Substituting the standard trigonometric values:
\[\begin{array}{l} x \cdot \left(\frac{\sqrt{3}}{2}\right) \cdot \left(\frac{\sqrt{3}}{2}\right)^2 = \frac{(1)^2 \cdot 2}{\frac{2}{\sqrt{3}}} \\ x \cdot \frac{\sqrt{3}}{2} \cdot \frac{3}{4} = \frac{2}{\frac{2}{\sqrt{3}}} \\ x \cdot \frac{3\sqrt{3}}{8} = 2 \times \frac{\sqrt{3}}{2} \\ x \cdot \frac{3\sqrt{3}}{8} = \sqrt{3} \end{array}\]Solving for $x$:
\[\begin{array}{l} x = \frac{\sqrt{3} \cdot 8}{3\sqrt{3}} \\ x = \mathbf{\frac{8}{3}} \text{ or } \mathbf{2\frac{2}{3}} \end{array}\]Explanation:
Substituting the standard trigonometric values:
\[\begin{array}{l} x^2 = \left(\frac{1}{2}\right)^2 + 4(1)^2 - (2)^2 \\ x^2 = \frac{1}{4} + 4(1) - 4 \\ x^2 = \frac{1}{4} + 4 - 4 \\ x^2 = \frac{1}{4} \end{array}\]Taking the square root on both sides:
\[\begin{array}{l} x = \pm\sqrt{\frac{1}{4}} \\ x = \mathbf{\pm\frac{1}{2}} \end{array}\]Explanation:
Substitute the standard trigonometric values into the given equations.
Equation 1:
\[\begin{array}{l} x \tan 30^\circ + y \cot 60^\circ = 0 \\ x \left(\frac{1}{\sqrt{3}}\right) + y \left(\frac{1}{\sqrt{3}}\right) = 0 \\ \frac{x + y}{\sqrt{3}} = 0 \\ x + y = 0 \\ y = -x \quad \text{--- (i)} \end{array}\]Equation 2:
\[\begin{array}{l} 2x - y \tan 45^\circ = 1 \\ 2x - y(1) = 1 \\ 2x - y = 1 \quad \text{--- (ii)} \end{array}\]Substitute the value of $y$ from equation (i) into equation (ii):
\[\begin{array}{l} 2x - (-x) = 1 \\ 2x + x = 1 \\ 3x = 1 \\ x = \mathbf{\frac{1}{3}} \end{array}\]Now, find $y$ using equation (i):
\[\begin{array}{l} y = -x = \mathbf{-\frac{1}{3}} \end{array}\]Answer: The values are $x = \frac{1}{3}$ and $y = -\frac{1}{3}$.
(i) $\sin(A+B) = \sin A \cos B + \cos A \sin B$
(ii) $\cos(A+B) = \cos A \cos B - \sin A \sin B$
Proof for (i):
Given $A = 45^\circ$ and $B = 45^\circ$.
\[\begin{array}{l} \text{L.H.S.} = \sin(A+B) = \sin(45^\circ + 45^\circ) = \sin 90^\circ = 1 \end{array}\] \[\begin{array}{l} \text{R.H.S.} = \sin 45^\circ \cos 45^\circ + \cos 45^\circ \sin 45^\circ \\ = \left(\frac{1}{\sqrt{2}}\right) \left(\frac{1}{\sqrt{2}}\right) + \left(\frac{1}{\sqrt{2}}\right) \left(\frac{1}{\sqrt{2}}\right) \\ = \frac{1}{2} + \frac{1}{2} = 1 \end{array}\]Since L.H.S. = R.H.S., statement (i) is [Justified].
Proof for (ii):
\[\begin{array}{l} \text{L.H.S.} = \cos(A+B) = \cos(45^\circ + 45^\circ) = \cos 90^\circ = 0 \end{array}\] \[\begin{array}{l} \text{R.H.S.} = \cos 45^\circ \cos 45^\circ - \sin 45^\circ \sin 45^\circ \\ = \left(\frac{1}{\sqrt{2}}\right) \left(\frac{1}{\sqrt{2}}\right) - \left(\frac{1}{\sqrt{2}}\right) \left(\frac{1}{\sqrt{2}}\right) \\ = \frac{1}{2} - \frac{1}{2} = 0 \end{array}\]Since L.H.S. = R.H.S., statement (ii) is [Justified].
Proof:
In an equilateral triangle ABC, all internal angles are $60^\circ$.
Therefore, $\angle A = \angle BAD = 60^\circ$.
In an equilateral triangle, the median also acts as the angle bisector. So, the median BD bisects $\angle B$.
\[\begin{array}{l} \angle ABD = \frac{\angle ABC}{2} = \frac{60^\circ}{2} = 30^\circ \end{array}\]Now, let's evaluate both sides of the equation:
\[\begin{array}{l} \text{L.H.S.} = \tan \angle ABD = \tan 30^\circ = \frac{1}{\sqrt{3}} \\ \text{R.H.S.} = \cot \angle BAD = \cot 60^\circ = \frac{1}{\sqrt{3}} \end{array}\]Since L.H.S. = R.H.S., the statement is [Proved].
Let us prove that, $\frac{\sec \angle ACD}{\sin \angle CAD} = \operatorname{cosec}^2 \angle CAD$
Proof:
In the isosceles right-angled triangle ABC, where $\angle A = 90^\circ$ and $AB = AC$, the base angles are equal.
\[\begin{array}{l} \angle ABC = \angle ACB = \frac{180^\circ - 90^\circ}{2} = 45^\circ \end{array}\]Therefore, $\angle ACD = \angle ACB = 45^\circ$.
AD is the angle bisector of $\angle BAC$ ($90^\circ$).
\[\begin{array}{l} \angle CAD = \frac{90^\circ}{2} = 45^\circ \end{array}\]Now, let's evaluate both sides of the given equation:
\[\begin{array}{l} \text{L.H.S.} = \frac{\sec \angle ACD}{\sin \angle CAD} \\ = \frac{\sec 45^\circ}{\sin 45^\circ} \\ = \frac{\sqrt{2}}{\frac{1}{\sqrt{2}}} \\ = \sqrt{2} \times \sqrt{2} = 2 \end{array}\]\[\begin{array}{l} \text{R.H.S.} = \operatorname{cosec}^2 \angle CAD \\ = \operatorname{cosec}^2 45^\circ \\ = (\sqrt{2})^2 = 2 \end{array}\]
Since L.H.S. = R.H.S., the statement is [Proved].
Explanation:
We are given a quadratic equation in terms of $\cos\theta$:
\[\begin{array}{l} 2\cos^2\theta - 3\cos\theta + 1 = 0 \end{array}\]Let $\cos\theta = y$. The equation becomes:
\[\begin{array}{l} 2y^2 - 3y + 1 = 0 \end{array}\]We can solve this by factoring the middle term:
\[\begin{array}{l} 2y^2 - 2y - y + 1 = 0 \\ 2y(y - 1) - 1(y - 1) = 0 \\ (2y - 1)(y - 1) = 0 \end{array}\]This gives two possible values for $y$:
\[\begin{array}{l} 2y - 1 = 0 \implies y = \frac{1}{2} \\ y - 1 = 0 \implies y = 1 \end{array}\]Now substitute back $\cos\theta = y$ to find $\theta$ (knowing $0^\circ \le \theta \le 90^\circ$):
Case 1: $\cos\theta = 1$
\[\begin{array}{l} \theta = 0^\circ \quad (\text{since } \cos 0^\circ = 1) \end{array}\]Case 2: $\cos\theta = \frac{1}{2}$
\[\begin{array}{l} \theta = 60^\circ \quad (\text{since } \cos 60^\circ = \frac{1}{2}) \end{array}\]Both values lie within the required range $0^\circ \le \theta \le 90^\circ$.
Answer: The values of $\theta$ are $0^\circ$ and $60^\circ$.
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