Ganit Prakash - Class-X - Trigonometry
Let us work out 23.3 Trigonometric Ratios
📘 Exercise 23.3 Solutions (Q1 - Q2)
Trigonometric ProblemsExplanation:
Given \( \sin\theta = \frac{4}{5} \).
We know that \( \operatorname{cosec}\theta = \frac{1}{\sin\theta} \), therefore:
\[\begin{array}{l} \operatorname{cosec}\theta = \frac{5}{4} \end{array}\]To find \( \cot\theta \), we first need \( \cos\theta \). Using the identity \( \sin^2\theta + \cos^2\theta = 1 \):
\[\begin{array}{l} \cos\theta = \sqrt{1 - \sin^2\theta} \\ = \sqrt{1 - \left(\frac{4}{5}\right)^2} \\ = \sqrt{1 - \frac{16}{25}} \\ = \sqrt{\frac{9}{25}} = \frac{3}{5} \end{array}\]Now, calculate \( \cot\theta \):
\[\begin{array}{l} \cot\theta = \frac{\cos\theta}{\sin\theta} = \frac{\frac{3}{5}}{\frac{4}{5}} = \frac{3}{4} \end{array}\]Substitute these values into the given expression:
\[\begin{array}{l} \frac{\operatorname{cosec}\theta}{1+\cot\theta} = \frac{\frac{5}{4}}{1 + \frac{3}{4}} \\ = \frac{\frac{5}{4}}{\frac{4+3}{4}} \\ = \frac{\frac{5}{4}}{\frac{7}{4}} \\ = \frac{5}{4} \times \frac{4}{7} = \mathbf{\frac{5}{7}} \end{array}\]Proof:
Given \( \tan\theta = \frac{3}{4} \). We know \( \tan\theta = \frac{\text{Perpendicular}}{\text{Base}} \).
Let Perpendicular \( = 3k \) and Base \( = 4k \) (where \( k > 0 \)).
By Pythagoras theorem, Hypotenuse \( = \sqrt{(3k)^2 + (4k)^2} = \sqrt{9k^2 + 16k^2} = \sqrt{25k^2} = 5k \).
Now, find \( \sin\theta \):
\[\begin{array}{l} \sin\theta = \frac{\text{Perpendicular}}{\text{Hypotenuse}} = \frac{3k}{5k} = \frac{3}{5} \end{array}\]Evaluate the Left Hand Side (L.H.S.):
\[\begin{array}{l} \text{L.H.S.} = \sqrt{\frac{1-\sin\theta}{1+\sin\theta}} \\ = \sqrt{\frac{1 - \frac{3}{5}}{1 + \frac{3}{5}}} \\ = \sqrt{\frac{\frac{2}{5}}{\frac{8}{5}}} \\ = \sqrt{\frac{2}{8}} \\ = \sqrt{\frac{1}{4}} = \frac{1}{2} \end{array}\]Since L.H.S. = R.H.S., the relation is [Proved].
Explanation:
Given \( \tan\theta = 1 \). This means \( \theta = 45^\circ \) (for acute angle).
Therefore, \( \sin\theta = \frac{1}{\sqrt{2}} \) and \( \cos\theta = \frac{1}{\sqrt{2}} \).
Substitute these values into the given expression:
Numerator:
\[\begin{array}{l} 8\sin\theta + 5\cos\theta = 8\left(\frac{1}{\sqrt{2}}\right) + 5\left(\frac{1}{\sqrt{2}}\right) = \frac{13}{\sqrt{2}} \end{array}\]Denominator:
\[\begin{array}{l} \sin^3\theta - 2\cos^3\theta + 7\cos\theta \\ = \left(\frac{1}{\sqrt{2}}\right)^3 - 2\left(\frac{1}{\sqrt{2}}\right)^3 + 7\left(\frac{1}{\sqrt{2}}\right) \\ = \frac{1}{2\sqrt{2}} - \frac{2}{2\sqrt{2}} + \frac{7}{\sqrt{2}} \\ = -\frac{1}{2\sqrt{2}} + \frac{14}{2\sqrt{2}} \\ = \frac{13}{2\sqrt{2}} \end{array}\]Now, divide the Numerator by the Denominator:
\[\begin{array}{l} \text{Value} = \frac{\frac{13}{\sqrt{2}}}{\frac{13}{2\sqrt{2}}} \\ = \frac{13}{\sqrt{2}} \times \frac{2\sqrt{2}}{13} \\ = \mathbf{2} \end{array}\]Explanation:
To express \( \operatorname{cosec}\theta \) in terms of \( \sin\theta \), we use the reciprocal identity:
\[\begin{array}{l} \operatorname{cosec}\theta = \mathbf{\frac{1}{\sin\theta}} \end{array}\]To express \( \tan\theta \) in terms of \( \sin\theta \), we start with the quotient identity:
\[\begin{array}{l} \tan\theta = \frac{\sin\theta}{\cos\theta} \end{array}\]Using the Pythagorean identity \( \sin^2\theta + \cos^2\theta = 1 \), we can write \( \cos\theta = \sqrt{1 - \sin^2\theta} \) (considering positive acute angles).
Substitute this into the expression for \( \tan\theta \):
\[\begin{array}{l} \tan\theta = \mathbf{\frac{\sin\theta}{\sqrt{1 - \sin^2\theta}}} \end{array}\]Explanation:
First, we express \( \sin\theta \) in terms of \( \cos\theta \) using the identity \( \sin^2\theta + \cos^2\theta = 1 \):
\[\begin{array}{l} \sin\theta = \sqrt{1 - \cos^2\theta} \end{array}\]To express \( \operatorname{cosec}\theta \) in terms of \( \cos\theta \):
\[\begin{array}{l} \operatorname{cosec}\theta = \frac{1}{\sin\theta} \\ \operatorname{cosec}\theta = \mathbf{\frac{1}{\sqrt{1 - \cos^2\theta}}} \end{array}\]To express \( \tan\theta \) in terms of \( \cos\theta \):
\[\begin{array}{l} \tan\theta = \frac{\sin\theta}{\cos\theta} \\ \tan\theta = \mathbf{\frac{\sqrt{1 - \cos^2\theta}}{\cos\theta}} \end{array}\]Explanation:
We know the standard trigonometric identity:
\[\begin{array}{l} \sec^2\theta - \tan^2\theta = 1 \end{array}\]Using the algebraic formula \( a^2 - b^2 = (a+b)(a-b) \):
\[\begin{array}{l} (\sec\theta + \tan\theta)(\sec\theta - \tan\theta) = 1 \end{array}\]Substitute the given value \( \sec\theta + \tan\theta = 2 \):
\[\begin{array}{l} 2 \times (\sec\theta - \tan\theta) = 1 \\ \sec\theta - \tan\theta = \mathbf{\frac{1}{2}} \end{array}\]Explanation:
We know the standard trigonometric identity:
\[\begin{array}{l} \operatorname{cosec}^2\theta - \cot^2\theta = 1 \\ (\operatorname{cosec}\theta - \cot\theta)(\operatorname{cosec}\theta + \cot\theta) = 1 \end{array}\]Substitute the given value \( \operatorname{cosec}\theta - \cot\theta = \sqrt{2} - 1 \):
\[\begin{array}{l} (\sqrt{2} - 1)(\operatorname{cosec}\theta + \cot\theta) = 1 \\ \operatorname{cosec}\theta + \cot\theta = \frac{1}{\sqrt{2} - 1} \end{array}\]Rationalize the denominator by multiplying the numerator and denominator by the conjugate \( (\sqrt{2} + 1) \):
\[\begin{array}{l} = \frac{1 \times (\sqrt{2} + 1)}{(\sqrt{2} - 1)(\sqrt{2} + 1)} \\ = \frac{\sqrt{2} + 1}{(\sqrt{2})^2 - (1)^2} \\ = \frac{\sqrt{2} + 1}{2 - 1} = \mathbf{\sqrt{2} + 1} \end{array}\]Explanation:
Given \( \sin\theta + \cos\theta = 1 \). Square both sides of the equation:
\[\begin{array}{l} (\sin\theta + \cos\theta)^2 = (1)^2 \\ \sin^2\theta + \cos^2\theta + 2\sin\theta\cos\theta = 1 \end{array}\]Since \( \sin^2\theta + \cos^2\theta = 1 \):
\[\begin{array}{l} 1 + 2\sin\theta\cos\theta = 1 \\ 2\sin\theta\cos\theta = 1 - 1 \\ 2\sin\theta\cos\theta = 0 \\ \sin\theta\cos\theta = \mathbf{0} \end{array}\]Explanation:
We use the algebraic identity \( (a - b)^2 = (a + b)^2 - 4ab \):
\[\begin{array}{l} (\tan\theta - \cot\theta)^2 = (\tan\theta + \cot\theta)^2 - 4\tan\theta\cot\theta \end{array}\]We know that \( \tan\theta \cdot \cot\theta = 1 \). Substitute the given value \( \tan\theta + \cot\theta = 2 \):
\[\begin{array}{l} (\tan\theta - \cot\theta)^2 = (2)^2 - 4(1) \\ (\tan\theta - \cot\theta)^2 = 4 - 4 = 0 \\ \tan\theta - \cot\theta = \mathbf{0} \end{array}\]Explanation:
We use the algebraic identity \( (a + b)^2 + (a - b)^2 = 2(a^2 + b^2) \):
\[\begin{array}{l} (\sin\theta + \cos\theta)^2 + (\sin\theta - \cos\theta)^2 = 2(\sin^2\theta + \cos^2\theta) \end{array}\]Since \( \sin^2\theta + \cos^2\theta = 1 \):
\[\begin{array}{l} (\sin\theta + \cos\theta)^2 + \left(\frac{7}{13}\right)^2 = 2(1) \\ (\sin\theta + \cos\theta)^2 + \frac{49}{169} = 2 \\ (\sin\theta + \cos\theta)^2 = 2 - \frac{49}{169} \\ (\sin\theta + \cos\theta)^2 = \frac{338 - 49}{169} = \frac{289}{169} \end{array}\]Taking the square root (since \( \theta \) is a positive acute angle, \( \sin\theta + \cos\theta \) is positive):
\[\begin{array}{l} \sin\theta + \cos\theta = \mathbf{\frac{17}{13}} \end{array}\]Explanation:
We know the expansion:
\[\begin{array}{l} (\sin\theta + \cos\theta)^2 = \sin^2\theta + \cos^2\theta + 2\sin\theta\cos\theta \end{array}\]Since \( \sin^2\theta + \cos^2\theta = 1 \) and \( \sin\theta\cos\theta = \frac{1}{2} \):
\[\begin{array}{l} (\sin\theta + \cos\theta)^2 = 1 + 2\left(\frac{1}{2}\right) \\ (\sin\theta + \cos\theta)^2 = 1 + 1 = 2 \end{array}\]Taking the square root (assuming positive value for acute angle):
\[\begin{array}{l} \sin\theta + \cos\theta = \mathbf{\sqrt{2}} \end{array}\]Explanation:
We know \( \sec^2\theta - \tan^2\theta = 1 \).
\[\begin{array}{l} (\sec\theta - \tan\theta)(\sec\theta + \tan\theta) = 1 \\ \left(\frac{1}{\sqrt{3}}\right)(\sec\theta + \tan\theta) = 1 \\ \sec\theta + \tan\theta = \sqrt{3} \quad \text{--- (1)} \end{array}\]We are given:
\[\begin{array}{l} \sec\theta - \tan\theta = \frac{1}{\sqrt{3}} \quad \text{--- (2)} \end{array}\]Adding equations (1) and (2):
\[\begin{array}{l} 2\sec\theta = \sqrt{3} + \frac{1}{\sqrt{3}} = \frac{3 + 1}{\sqrt{3}} = \frac{4}{\sqrt{3}} \\ \sec\theta = \mathbf{\frac{2}{\sqrt{3}}} \end{array}\]Subtracting equation (2) from (1):
\[\begin{array}{l} 2\tan\theta = \sqrt{3} - \frac{1}{\sqrt{3}} = \frac{3 - 1}{\sqrt{3}} = \frac{2}{\sqrt{3}} \\ \tan\theta = \mathbf{\frac{1}{\sqrt{3}}} \end{array}\]Explanation:
We know \( \operatorname{cosec}^2\theta - \cot^2\theta = 1 \).
\[\begin{array}{l} (\operatorname{cosec}\theta + \cot\theta)(\operatorname{cosec}\theta - \cot\theta) = 1 \\ \sqrt{3}(\operatorname{cosec}\theta - \cot\theta) = 1 \\ \operatorname{cosec}\theta - \cot\theta = \frac{1}{\sqrt{3}} \quad \text{--- (1)} \end{array}\]We are given:
\[\begin{array}{l} \operatorname{cosec}\theta + \cot\theta = \sqrt{3} \quad \text{--- (2)} \end{array}\]Adding equations (1) and (2):
\[\begin{array}{l} 2\operatorname{cosec}\theta = \sqrt{3} + \frac{1}{\sqrt{3}} = \frac{3+1}{\sqrt{3}} = \frac{4}{\sqrt{3}} \\ \operatorname{cosec}\theta = \mathbf{\frac{2}{\sqrt{3}}} \end{array}\]Subtracting equation (1) from (2):
\[\begin{array}{l} 2\cot\theta = \sqrt{3} - \frac{1}{\sqrt{3}} = \frac{3-1}{\sqrt{3}} = \frac{2}{\sqrt{3}} \\ \cot\theta = \mathbf{\frac{1}{\sqrt{3}}} \end{array}\]Explanation:
Given equation:
\[\begin{array}{l} \frac{\sin\theta + \cos\theta}{\sin\theta - \cos\theta} = \frac{7}{1} \end{array}\]Applying the componendo and dividendo rule \( \left(\frac{a+b}{a-b} = \frac{c+d}{c-d}\right) \):
\[\begin{array}{l} \frac{(\sin\theta + \cos\theta) + (\sin\theta - \cos\theta)}{(\sin\theta + \cos\theta) - (\sin\theta - \cos\theta)} = \frac{7 + 1}{7 - 1} \\ \frac{2\sin\theta}{2\cos\theta} = \frac{8}{6} \\ \frac{\sin\theta}{\cos\theta} = \frac{4}{3} \\ \tan\theta = \mathbf{\frac{4}{3}} \end{array}\]Explanation:
Cross-multiplying the given equation:
\[\begin{array}{l} 2(\operatorname{cosec}\theta + \sin\theta) = 5(\operatorname{cosec}\theta - \sin\theta) \\ 2\operatorname{cosec}\theta + 2\sin\theta = 5\operatorname{cosec}\theta - 5\sin\theta \\ 2\sin\theta + 5\sin\theta = 5\operatorname{cosec}\theta - 2\operatorname{cosec}\theta \\ 7\sin\theta = 3\operatorname{cosec}\theta \end{array}\]Since \( \operatorname{cosec}\theta = \frac{1}{\sin\theta} \):
\[\begin{array}{l} 7\sin\theta = \frac{3}{\sin\theta} \\ 7\sin^2\theta = 3 \\ \sin^2\theta = \frac{3}{7} \end{array}\]Taking the positive square root (since \( \theta \) is an acute angle):
\[\begin{array}{l} \sin\theta = \mathbf{\sqrt{\frac{3}{7}}} \end{array}\]Explanation:
We use the algebraic identity \( (a - b)^2 = (a + b)^2 - 4ab \):
\[\begin{array}{l} (\sec\theta - \cos\theta)^2 = (\sec\theta + \cos\theta)^2 - 4\sec\theta\cos\theta \end{array}\]Since \( \sec\theta \cdot \cos\theta = 1 \) and \( \sec\theta + \cos\theta = \frac{5}{2} \):
\[\begin{array}{l} (\sec\theta - \cos\theta)^2 = \left(\frac{5}{2}\right)^2 - 4(1) \\ (\sec\theta - \cos\theta)^2 = \frac{25}{4} - 4 \\ (\sec\theta - \cos\theta)^2 = \frac{25 - 16}{4} = \frac{9}{4} \end{array}\]Since for a positive acute angle (\( \theta \neq 0^\circ \)), \( \sec\theta \ge 1 \) and \( \cos\theta \le 1 \), so \( \sec\theta - \cos\theta \) will be positive:
\[\begin{array}{l} \sec\theta - \cos\theta = \mathbf{\frac{3}{2}} \end{array}\]Explanation:
Given equation:
\[\begin{array}{l} 5\sin^2\theta + 4\cos^2\theta = \frac{9}{2} \end{array}\]We can rewrite \( 5\sin^2\theta \) as \( \sin^2\theta + 4\sin^2\theta \):
\[\begin{array}{l} \sin^2\theta + 4\sin^2\theta + 4\cos^2\theta = \frac{9}{2} \\ \sin^2\theta + 4(\sin^2\theta + \cos^2\theta) = \frac{9}{2} \end{array}\]Since \( \sin^2\theta + \cos^2\theta = 1 \):
\[\begin{array}{l} \sin^2\theta + 4(1) = \frac{9}{2} \\ \sin^2\theta = \frac{9}{2} - 4 \\ \sin^2\theta = \frac{9 - 8}{2} = \frac{1}{2} \end{array}\]Now, substitute \( \sin^2\theta = \frac{1}{2} \) to find \( \cos^2\theta \):
\[\begin{array}{l} \cos^2\theta = 1 - \sin^2\theta = 1 - \frac{1}{2} = \frac{1}{2} \end{array}\]Therefore, we calculate \( \tan^2\theta \):
\[\begin{array}{l} \tan^2\theta = \frac{\sin^2\theta}{\cos^2\theta} = \frac{\frac{1}{2}}{\frac{1}{2}} = 1 \end{array}\]Since \( \theta \) is an acute angle, \( \tan\theta \) is positive:
\[\begin{array}{l} \tan\theta = \mathbf{1} \end{array}\]Explanation:
We know \( \tan\theta \cdot \cot\theta = 1 \).
For \( (\tan\theta + \cot\theta) \):
\[\begin{array}{l} (\tan\theta + \cot\theta)^2 = \tan^2\theta + \cot^2\theta + 2\tan\theta\cot\theta \\ = \frac{10}{3} + 2(1) \\ = \frac{10 + 6}{3} = \frac{16}{3} \end{array}\] \[\begin{array}{l} \tan\theta + \cot\theta = \mathbf{\frac{4}{\sqrt{3}}} \quad \text{--- (1)} \end{array}\]For \( (\tan\theta - \cot\theta) \):
\[\begin{array}{l} (\tan\theta - \cot\theta)^2 = \tan^2\theta + \cot^2\theta - 2\tan\theta\cot\theta \\ = \frac{10}{3} - 2(1) \\ = \frac{10 - 6}{3} = \frac{4}{3} \end{array}\] \[\begin{array}{l} \tan\theta - \cot\theta = \mathbf{\pm \frac{2}{\sqrt{3}}} \quad \text{--- (2)} \end{array}\]To find \( \tan\theta \):
Case 1: If \( \tan\theta - \cot\theta = \frac{2}{\sqrt{3}} \), adding this to (1):
\[\begin{array}{l} 2\tan\theta = \frac{4}{\sqrt{3}} + \frac{2}{\sqrt{3}} = \frac{6}{\sqrt{3}} = 2\sqrt{3} \\ \tan\theta = \mathbf{\sqrt{3}} \end{array}\]Case 2: If \( \tan\theta - \cot\theta = -\frac{2}{\sqrt{3}} \), adding this to (1):
\[\begin{array}{l} 2\tan\theta = \frac{4}{\sqrt{3}} - \frac{2}{\sqrt{3}} = \frac{2}{\sqrt{3}} \\ \tan\theta = \mathbf{\frac{1}{\sqrt{3}}} \end{array}\]Explanation:
We know the algebraic identity \( a^2 - b^2 = (a+b)(a-b) \). Applying this to the given expression:
\[\begin{array}{l} \sec^4\theta - \tan^4\theta = (\sec^2\theta)^2 - (\tan^2\theta)^2 \\ = (\sec^2\theta + \tan^2\theta)(\sec^2\theta - \tan^2\theta) \end{array}\]We already know the standard trigonometric identity \( \sec^2\theta - \tan^2\theta = 1 \). And we are given \( \sec^2\theta + \tan^2\theta = \frac{13}{12} \).
Substitute these values into the factored expression:
\[\begin{array}{l} \sec^4\theta - \tan^4\theta = \left(\frac{13}{12}\right) \times (1) \\ = \mathbf{\frac{13}{12}} \end{array}\]Explanation:
In the right-angled triangle PQR, hypotenuse is $PR = \sqrt{5}$.
We are given $PQ - RQ = 1 \implies PQ = RQ + 1$.
Applying Pythagoras theorem:
\[\begin{array}{l} PQ^2 + RQ^2 = PR^2 \\ (RQ + 1)^2 + RQ^2 = (\sqrt{5})^2 \\ RQ^2 + 2RQ + 1 + RQ^2 = 5 \\ 2RQ^2 + 2RQ - 4 = 0 \end{array}\]Dividing by 2:
\[\begin{array}{l} RQ^2 + RQ - 2 = 0 \\ (RQ + 2)(RQ - 1) = 0 \end{array}\]Since side length cannot be negative, $RQ = 1$ unit.
So, $PQ = RQ + 1 = 1 + 1 = 2$ units.
Now, finding the required trigonometric ratios:
\[\begin{array}{l} \cos P = \frac{\text{Base adjacent to } P}{\text{Hypotenuse}} = \frac{PQ}{PR} = \frac{2}{\sqrt{5}} \\ \cos R = \frac{\text{Base adjacent to } R}{\text{Hypotenuse}} = \frac{RQ}{PR} = \frac{1}{\sqrt{5}} \end{array}\]Therefore:
\[\begin{array}{l} \cos P - \cos R = \frac{2}{\sqrt{5}} - \frac{1}{\sqrt{5}} = \mathbf{\frac{1}{\sqrt{5}}} \end{array}\]Explanation:
In the right-angled triangle XYZ, hypotenuse is $XZ$ and base/perpendicular is $XY = 2\sqrt{3}$.
We are given $XZ - YZ = 2 \implies XZ = YZ + 2$.
Applying Pythagoras theorem:
\[\begin{array}{l} XY^2 + YZ^2 = XZ^2 \\ (2\sqrt{3})^2 + YZ^2 = (YZ + 2)^2 \\ 12 + YZ^2 = YZ^2 + 4YZ + 4 \end{array}\]Subtracting $YZ^2$ from both sides:
\[\begin{array}{l} 12 = 4YZ + 4 \\ 4YZ = 8 \\ YZ = 2 \text{ units} \end{array}\]So, $XZ = YZ + 2 = 2 + 2 = 4$ units.
Now, finding the required trigonometric ratios for angle $X$:
\[\begin{array}{l} \sec X = \frac{\text{Hypotenuse}}{\text{Base adjacent to } X} = \frac{XZ}{XY} = \frac{4}{2\sqrt{3}} = \frac{2}{\sqrt{3}} \\ \tan X = \frac{\text{Perpendicular opposite to } X}{\text{Base adjacent to } X} = \frac{YZ}{XY} = \frac{2}{2\sqrt{3}} = \frac{1}{\sqrt{3}} \end{array}\]Therefore:
\[\begin{array}{l} \sec X - \tan X = \frac{2}{\sqrt{3}} - \frac{1}{\sqrt{3}} = \mathbf{\frac{1}{\sqrt{3}}} \end{array}\]Explanation:
From the given equations, we can express $\sin\theta$ and $\cos\theta$ in terms of $x$ and $y$:
\[\begin{array}{l} \sin\theta = \frac{x}{2} \\ \cos\theta = \frac{y}{3} \end{array}\]We know the fundamental trigonometric identity relating sine and cosine:
\[\begin{array}{l} \sin^2\theta + \cos^2\theta = 1 \end{array}\]Substituting the values into the identity:
\[\begin{array}{l} \left(\frac{x}{2}\right)^2 + \left(\frac{y}{3}\right)^2 = 1 \\ \mathbf{\frac{x^2}{4} + \frac{y^2}{9} = 1} \end{array}\]Explanation:
From the given equations, we can express $\sec\theta$ and $\tan\theta$ in terms of $x$ and $y$:
\[\begin{array}{l} \sec\theta = \frac{5x}{3} \\ \tan\theta = \frac{y}{3} \end{array}\]We know the fundamental trigonometric identity relating secant and tangent:
\[\begin{array}{l} \sec^2\theta - \tan^2\theta = 1 \end{array}\]Substituting the values into the identity:
\[\begin{array}{l} \left(\frac{5x}{3}\right)^2 - \left(\frac{y}{3}\right)^2 = 1 \\ \frac{25x^2}{9} - \frac{y^2}{9} = 1 \\ \mathbf{25x^2 - y^2 = 9} \end{array}\]Proof:
Given $\sin\alpha = \frac{\text{Perpendicular}}{\text{Hypotenuse}} = \frac{5}{13}$.
Let Perpendicular = $5k$ and Hypotenuse = $13k$ ($k > 0$).
By Pythagoras theorem:
\[\begin{array}{l} \text{Base} = \sqrt{\text{Hypotenuse}^2 - \text{Perpendicular}^2} \\ = \sqrt{(13k)^2 - (5k)^2} \\ = \sqrt{169k^2 - 25k^2} \\ = \sqrt{144k^2} = 12k \end{array}\]Now, calculate $\tan\alpha$ and $\sec\alpha$:
\[\begin{array}{l} \tan\alpha = \frac{\text{Perpendicular}}{\text{Base}} = \frac{5k}{12k} = \frac{5}{12} \\ \sec\alpha = \frac{\text{Hypotenuse}}{\text{Base}} = \frac{13k}{12k} = \frac{13}{12} \end{array}\]Now evaluate the Left Hand Side (L.H.S.):
\[\begin{array}{l} \text{L.H.S.} = \tan\alpha + \sec\alpha \\ = \frac{5}{12} + \frac{13}{12} \\ = \frac{18}{12} \\ = \frac{3}{2} = 1.5 = \text{R.H.S.} \end{array}\][Proved]
Explanation:
Given $\tan A = \frac{\text{Perpendicular}}{\text{Base}} = \frac{n}{m}$.
Let Perpendicular = $nk$ and Base = $mk$ ($k > 0$).
By Pythagoras theorem, Hypotenuse is:
\[\begin{array}{l} \text{Hypotenuse} = \sqrt{\text{Base}^2 + \text{Perpendicular}^2} \\ = \sqrt{(mk)^2 + (nk)^2} \\ = \sqrt{m^2k^2 + n^2k^2} \\ = k\sqrt{m^2 + n^2} \end{array}\]Now we determine the required values:
\[\begin{array}{l} \sin A = \frac{\text{Perpendicular}}{\text{Hypotenuse}} = \frac{nk}{k\sqrt{m^2 + n^2}} = \mathbf{\frac{n}{\sqrt{m^2 + n^2}}} \\ \sec A = \frac{\text{Hypotenuse}}{\text{Base}} = \frac{k\sqrt{m^2 + n^2}}{mk} = \mathbf{\frac{\sqrt{m^2 + n^2}}{m}} \end{array}\]Proof:
Given $\cos\theta = \frac{\text{Base}}{\text{Hypotenuse}} = \frac{x}{\sqrt{x^2+y^2}}$.
Let Base = $xk$ and Hypotenuse = $k\sqrt{x^2+y^2}$ ($k > 0$).
By Pythagoras theorem, Perpendicular is:
\[\begin{array}{l} \text{Perpendicular} = \sqrt{\text{Hypotenuse}^2 - \text{Base}^2} \\ = \sqrt{(k\sqrt{x^2+y^2})^2 - (xk)^2} \\ = \sqrt{k^2(x^2+y^2) - k^2x^2} \\ = \sqrt{k^2x^2 + k^2y^2 - k^2x^2} \\ = \sqrt{k^2y^2} = yk \end{array}\]Therefore, $\sin\theta = \frac{\text{Perpendicular}}{\text{Hypotenuse}} = \frac{yk}{k\sqrt{x^2+y^2}} = \frac{y}{\sqrt{x^2+y^2}}$.
Now evaluate L.H.S. and R.H.S.:
\[\begin{array}{l} \text{L.H.S.} = x \sin\theta = x \left(\frac{y}{\sqrt{x^2+y^2}}\right) = \frac{xy}{\sqrt{x^2+y^2}} \\ \text{R.H.S.} = y \cos\theta = y \left(\frac{x}{\sqrt{x^2+y^2}}\right) = \frac{xy}{\sqrt{x^2+y^2}} \end{array}\]Since L.H.S. = R.H.S., it is [Proved].
Proof:
Given $\sin\alpha = \frac{\text{Perpendicular}}{\text{Hypotenuse}} = \frac{a^2-b^2}{a^2+b^2}$.
Let Perpendicular = $(a^2-b^2)k$ and Hypotenuse = $(a^2+b^2)k$ ($k > 0$).
By Pythagoras theorem, Base is:
\[\begin{array}{l} \text{Base} = \sqrt{\text{Hypotenuse}^2 - \text{Perpendicular}^2} \\ = \sqrt{((a^2+b^2)k)^2 - ((a^2-b^2)k)^2} \\ = k\sqrt{(a^2+b^2)^2 - (a^2-b^2)^2} \end{array}\]Using the algebraic identity $(x+y)^2 - (x-y)^2 = 4xy$:
\[\begin{array}{l} \text{Base} = k\sqrt{4(a^2)(b^2)} = k\sqrt{4a^2b^2} = 2abk \end{array}\]Now determine $\cot\alpha$:
\[\begin{array}{l} \cot\alpha = \frac{\text{Base}}{\text{Perpendicular}} \\ = \frac{2abk}{(a^2-b^2)k} \\ = \frac{2ab}{a^2-b^2} \end{array}\]This matches the given expression, hence [Proved].
Proof:
Let $\frac{\sin\theta}{x} = \frac{\cos\theta}{y} = k$.
Then, $\sin\theta = kx$ and $\cos\theta = ky$.
We know the identity $\sin^2\theta + \cos^2\theta = 1$. Substituting the values:
\[\begin{array}{l} (kx)^2 + (ky)^2 = 1 \\ k^2x^2 + k^2y^2 = 1 \\ k^2(x^2 + y^2) = 1 \\ k^2 = \frac{1}{x^2 + y^2} \\ k = \frac{1}{\sqrt{x^2 + y^2}} \end{array}\]Therefore, we get:
\[\begin{array}{l} \sin\theta = \frac{x}{\sqrt{x^2 + y^2}} \\ \cos\theta = \frac{y}{\sqrt{x^2 + y^2}} \end{array}\]Now evaluate the expression:
\[\begin{array}{l} \sin\theta - \cos\theta = \frac{x}{\sqrt{x^2 + y^2}} - \frac{y}{\sqrt{x^2 + y^2}} \\ = \frac{x - y}{\sqrt{x^2 + y^2}} \end{array}\][Proved]
Proof:
From the given equation, we have $\cos A = \frac{4x}{1+4x^2}$.
This means Base = $4x$ and Hypotenuse = $1+4x^2$.
By Pythagoras theorem, Perpendicular is:
\[\begin{array}{l} \text{Perpendicular} = \sqrt{\text{Hypotenuse}^2 - \text{Base}^2} \\ = \sqrt{(1+4x^2)^2 - (4x)^2} \\ = \sqrt{(1 + 8x^2 + 16x^4) - 16x^2} \\ = \sqrt{1 - 8x^2 + 16x^4} \\ = \sqrt{(1-4x^2)^2} \\ = 1-4x^2 \end{array}\]Now find $\operatorname{cosec} A$ and $\cot A$:
\[\begin{array}{l} \operatorname{cosec} A = \frac{\text{Hypotenuse}}{\text{Perpendicular}} = \frac{1+4x^2}{1-4x^2} \\ \cot A = \frac{\text{Base}}{\text{Perpendicular}} = \frac{4x}{1-4x^2} \end{array}\]Adding these gives the L.H.S.:
\[\begin{array}{l} \text{L.H.S.} = \operatorname{cosec} A + \cot A \\ = \frac{1+4x^2}{1-4x^2} + \frac{4x}{1-4x^2} \\ = \frac{1 + 4x + 4x^2}{1 - 4x^2} \end{array}\]Factoring the numerator as a perfect square and the denominator as a difference of squares:
\[\begin{array}{l} = \frac{(1+2x)^2}{(1-2x)(1+2x)} \\ = \frac{1+2x}{1-2x} = \text{R.H.S.} \end{array}\][Proved]
Proof:
Given relations:
\[\begin{array}{l} x = a\sin\theta \implies \frac{a}{x} = \frac{1}{\sin\theta} \implies \frac{a}{x} = \operatorname{cosec}\theta \\ y = b\tan\theta \implies \frac{b}{y} = \frac{1}{\tan\theta} \implies \frac{b}{y} = \cot\theta \end{array}\]Now, taking the Left Hand Side (L.H.S.) of the expression to be proved:
\[\begin{array}{l} \text{L.H.S.} = \frac{a^2}{x^2} - \frac{b^2}{y^2} \\ = \left(\frac{a}{x}\right)^2 - \left(\frac{b}{y}\right)^2 \end{array}\]Substituting the values derived above:
\[\begin{array}{l} = (\operatorname{cosec}\theta)^2 - (\cot\theta)^2 \\ = \operatorname{cosec}^2\theta - \cot^2\theta \end{array}\]We know the standard trigonometric identity \( \operatorname{cosec}^2\theta - \cot^2\theta = 1 \).
\[\begin{array}{l} = 1 = \text{R.H.S.} \end{array}\]Therefore, \( \frac{a^2}{x^2} - \frac{b^2}{y^2} = 1 \). [Proved]
Proof:
Given equation:
\[\begin{array}{l} \sin\theta + \sin^2\theta = 1 \\ \sin\theta = 1 - \sin^2\theta \end{array}\]We know the identity \( \sin^2\theta + \cos^2\theta = 1 \), which means \( 1 - \sin^2\theta = \cos^2\theta \). Therefore:
\[\begin{array}{l} \sin\theta = \cos^2\theta \quad \text{--- (1)} \end{array}\]Now, taking the Left Hand Side (L.H.S.) of the expression to be proved:
\[\begin{array}{l} \text{L.H.S.} = \cos^2\theta + \cos^4\theta \\ = \cos^2\theta + (\cos^2\theta)^2 \end{array}\]Substituting \( \cos^2\theta = \sin\theta \) from equation (1):
\[\begin{array}{l} = \sin\theta + (\sin\theta)^2 \\ = \sin\theta + \sin^2\theta \end{array}\]But it is given in the problem that \( \sin\theta + \sin^2\theta = 1 \).
\[\begin{array}{l} = 1 = \text{R.H.S.} \end{array}\]Therefore, \( \cos^2\theta + \cos^4\theta = 1 \). [Proved]
📘 9. Very short Answer : (V.S.A.)
(A) M.C.Q.Explanation:
From the given equations, we find \( x \) and \( \frac{1}{x} \):
\[\begin{array}{l} 3x = \operatorname{cosec}\alpha \implies x = \frac{\operatorname{cosec}\alpha}{3} \implies x^2 = \frac{\operatorname{cosec}^2\alpha}{9} \\ \frac{3}{x} = \cot\alpha \implies \frac{1}{x} = \frac{\cot\alpha}{3} \implies \frac{1}{x^2} = \frac{\cot^2\alpha}{9} \end{array}\]Now, substitute these into the given expression:
\[\begin{array}{l} 3\left(x^2 - \frac{1}{x^2}\right) = 3\left(\frac{\operatorname{cosec}^2\alpha}{9} - \frac{\cot^2\alpha}{9}\right) \\ = 3 \left( \frac{\operatorname{cosec}^2\alpha - \cot^2\alpha}{9} \right) \end{array}\]Since \( \operatorname{cosec}^2\alpha - \cot^2\alpha = 1 \):
\[\begin{array}{l} = 3 \left( \frac{1}{9} \right) = \frac{3}{9} = \frac{1}{3} \end{array}\]Correct Option: (c) \( \frac{1}{3} \)
Explanation:
From the given equations, we find \( x \) and \( \frac{1}{x} \):
\[\begin{array}{l} 2x = \sec A \implies x = \frac{\sec A}{2} \implies x^2 = \frac{\sec^2 A}{4} \\ \frac{2}{x} = \tan A \implies \frac{1}{x} = \frac{\tan A}{2} \implies \frac{1}{x^2} = \frac{\tan^2 A}{4} \end{array}\]Now, substitute these into the given expression:
\[\begin{array}{l} 2\left(x^2 - \frac{1}{x^2}\right) = 2\left(\frac{\sec^2 A}{4} - \frac{\tan^2 A}{4}\right) \\ = 2 \left( \frac{\sec^2 A - \tan^2 A}{4} \right) \end{array}\]Since \( \sec^2 A - \tan^2 A = 1 \):
\[\begin{array}{l} = 2 \left( \frac{1}{4} \right) = \frac{2}{4} = \frac{1}{2} \end{array}\]Correct Option: (a) \( \frac{1}{2} \)
Explanation:
Given equation:
\[\begin{array}{l} \tan\alpha + \cot\alpha = 2 \end{array}\]We know \( \cot\alpha = \frac{1}{\tan\alpha} \). Substituting this:
\[\begin{array}{l} \tan\alpha + \frac{1}{\tan\alpha} = 2 \\ \frac{\tan^2\alpha + 1}{\tan\alpha} = 2 \\ \tan^2\alpha + 1 = 2\tan\alpha \\ \tan^2\alpha - 2\tan\alpha + 1 = 0 \\ (\tan\alpha - 1)^2 = 0 \\ \tan\alpha = 1 \end{array}\]If \( \tan\alpha = 1 \), then \( \cot\alpha = \frac{1}{1} = 1 \).
Now, evaluate the required expression:
\[\begin{array}{l} \tan^{13}\alpha + \cot^{13}\alpha = (1)^{13} + (1)^{13} \\ = 1 + 1 = 2 \end{array}\]Correct Option: (c) 2
Explanation:
Given equation:
\[\begin{array}{l} \sin\theta - \cos\theta = 0 \\ \sin\theta = \cos\theta \\ \frac{\sin\theta}{\cos\theta} = 1 \\ \tan\theta = 1 \end{array}\]Since \( 0^\circ \le \theta \le 90^\circ \), the angle is \( \theta = 45^\circ \).
Now, substitute \( \theta = 45^\circ \) into the second equation:
\[\begin{array}{l} x = \sec\theta + \operatorname{cosec}\theta \\ x = \sec 45^\circ + \operatorname{cosec} 45^\circ \\ x = \sqrt{2} + \sqrt{2} \\ x = 2\sqrt{2} \end{array}\]Correct Option: (d) \( 2\sqrt{2} \)
Explanation:
Given equation:
\[\begin{array}{l} 2\cos 3\theta = 1 \\ \cos 3\theta = \frac{1}{2} \end{array}\]We know from the standard trigonometric table that \( \cos 60^\circ = \frac{1}{2} \). Therefore:
\[\begin{array}{l} \cos 3\theta = \cos 60^\circ \\ 3\theta = 60^\circ \\ \theta = \frac{60^\circ}{3} \\ \theta = 20^\circ \end{array}\]Correct Option: (c) 20°
📘 9. Very short Answer : (V.S.A.)
(B) True or FalseIf $0^\circ \le \alpha < 90^\circ$, then the least value of $(\sec^2\alpha + \cos^2\alpha)$ is 2.
Explanation:
We can write the given expression using the algebraic identity $a^2 + b^2 = (a-b)^2 + 2ab$:
\[\begin{array}{l} \sec^2\alpha + \cos^2\alpha = (\sec\alpha - \cos\alpha)^2 + 2\sec\alpha\cos\alpha \\ = (\sec\alpha - \cos\alpha)^2 + 2(1) \\ = (\sec\alpha - \cos\alpha)^2 + 2 \end{array}\]For any real number, the square is always greater than or equal to zero. Thus, the minimum value of $(\sec\alpha - \cos\alpha)^2$ is 0 (which occurs when $\sec\alpha = \cos\alpha$, i.e., $\alpha = 0^\circ$).
Therefore, the minimum or least value of the entire expression is $0 + 2 = 2$.
Answer: True
The value of $(\cos 0^\circ \times \cos 1^\circ \times \cos 2^\circ \times \cos 3^\circ \times \dots \times \cos 90^\circ)$ is 1.
Explanation:
The given expression is a product of cosine values from $0^\circ$ to $90^\circ$.
One of the terms in this multiplication series is $\cos 90^\circ$.
We know from the standard trigonometric table that $\cos 90^\circ = 0$.
Since any number multiplied by zero is zero, the entire product becomes 0.
\[\begin{array}{l} \cos 0^\circ \times \cos 1^\circ \times \dots \times \mathbf{\cos 90^\circ} \\ = \cos 0^\circ \times \cos 1^\circ \times \dots \times \mathbf{0} \\ = 0 \end{array}\]Answer: False
📘 9. Very short Answer : (V.S.A.)
(C) Fill in the blanksExplanation:
We apply the basic trigonometric identities:
1. $\frac{1}{\sec^2\theta} = \cos^2\theta$
2. $1 + \cot^2\theta = \operatorname{cosec}^2\theta$, so $\frac{1}{1+\cot^2\theta} = \frac{1}{\operatorname{cosec}^2\theta} = \sin^2\theta$
Substituting these into the expression:
\[\begin{array}{l} \frac{4}{\sec^2\theta} + \frac{1}{1+\cot^2\theta} + 3\sin^2\theta \\ = 4\cos^2\theta + \frac{1}{\operatorname{cosec}^2\theta} + 3\sin^2\theta \\ = 4\cos^2\theta + \sin^2\theta + 3\sin^2\theta \\ = 4\cos^2\theta + 4\sin^2\theta \\ = 4(\cos^2\theta + \sin^2\theta) \end{array}\]Since $\cos^2\theta + \sin^2\theta = 1$:
\[\begin{array}{l} = 4 \times 1 = 4 \end{array}\]Answer: 4
Explanation:
We know that $\sin 30^\circ = \frac{1}{2}$.
Therefore, we can write the equation as:
\[\begin{array}{l} \sin(\theta - 30^\circ) = \sin 30^\circ \end{array}\]Comparing both sides (assuming $\theta$ is an acute angle):
\[\begin{array}{l} \theta - 30^\circ = 30^\circ \\ \theta = 30^\circ + 30^\circ \\ \theta = 60^\circ \end{array}\]Now, we need to find the value of $\cos\theta$:
\[\begin{array}{l} \cos\theta = \cos 60^\circ = \frac{1}{2} \end{array}\]Answer: $\frac{1}{2}$
Explanation:
We can factor the given expression using the algebraic identity $a^2 - b^2 = (a-b)(a+b)$:
\[\begin{array}{l} \cos^4\theta - \sin^4\theta = (\cos^2\theta)^2 - (\sin^2\theta)^2 \\ = (\cos^2\theta - \sin^2\theta)(\cos^2\theta + \sin^2\theta) \end{array}\]We know the standard trigonometric identity $\cos^2\theta + \sin^2\theta = 1$.
We are given $\cos^2\theta - \sin^2\theta = \frac{1}{2}$.
Substitute these values into the factored expression:
\[\begin{array}{l} = \left(\frac{1}{2}\right) \times (1) \\ = \frac{1}{2} \end{array}\]Answer: $\frac{1}{2}$
📘 10. Short answer type questions : (S.A.)
Exercise 23.3Explanation:
Given the two equations:
\[\begin{array}{l} r\cos\theta = 2\sqrt{3} \quad \text{--- (1)} \\ r\sin\theta = 2 \quad \text{--- (2)} \end{array}\]Step 1: Find the value of \( r \)
Square both sides of equations (1) and (2), then add them together:
\[\begin{array}{l} (r\cos\theta)^2 + (r\sin\theta)^2 = (2\sqrt{3})^2 + 2^2 \\ r^2\cos^2\theta + r^2\sin^2\theta = 12 + 4 \\ r^2(\cos^2\theta + \sin^2\theta) = 16 \end{array}\]We know the identity \( \cos^2\theta + \sin^2\theta = 1 \):
\[\begin{array}{l} r^2(1) = 16 \\ r = \sqrt{16} = 4 \quad [\text{Since } r \text{ is a positive magnitude}] \end{array}\]Step 2: Find the value of \( \theta \)
Divide equation (2) by equation (1):
\[\begin{array}{l} \frac{r\sin\theta}{r\cos\theta} = \frac{2}{2\sqrt{3}} \\ \tan\theta = \frac{1}{\sqrt{3}} \end{array}\]Since \( 0^\circ < \theta < 90^\circ \) and \( \tan 30^\circ = \frac{1}{\sqrt{3}} \):
\[\begin{array}{l} \theta = 30^\circ \end{array}\]Answer: The values are \( r = 4 \) and \( \theta = 30^\circ \).
Explanation:
We are given that \( \sin A + \sin B = 2 \).
For any angle \( \theta \) between \( 0^\circ \) and \( 90^\circ \), the maximum possible value of \( \sin\theta \) is \( 1 \).
Therefore, \( \sin A \le 1 \) and \( \sin B \le 1 \).
The sum of two quantities, each of which cannot exceed 1, can only be equal to 2 if both quantities are exactly equal to 1.
\[\begin{array}{l} \sin A = 1 \implies A = 90^\circ \\ \sin B = 1 \implies B = 90^\circ \end{array}\]Now, we substitute these angle values into the expression we need to find:
\[\begin{array}{l} \cos A + \cos B = \cos 90^\circ + \cos 90^\circ \\ = 0 + 0 \\ = \mathbf{0} \end{array}\]Explanation:
We can rewrite the given expression using the algebraic identity \( a^2 + b^2 = (a-b)^2 + 2ab \):
\[\begin{array}{l} 9\tan^2\theta + 4\cot^2\theta = (3\tan\theta)^2 + (2\cot\theta)^2 \\ = (3\tan\theta - 2\cot\theta)^2 + 2(3\tan\theta)(2\cot\theta) \\ = (3\tan\theta - 2\cot\theta)^2 + 12(\tan\theta \cdot \cot\theta) \end{array}\]We know the trigonometric identity \( \tan\theta \cdot \cot\theta = 1 \). Substituting this in:
\[\begin{array}{l} = (3\tan\theta - 2\cot\theta)^2 + 12(1) \\ = (3\tan\theta - 2\cot\theta)^2 + 12 \end{array}\]For any real number, its square is always non-negative. Therefore, the minimum possible value of \( (3\tan\theta - 2\cot\theta)^2 \) is \( 0 \).
So, the least value of the entire expression occurs when the squared term is zero:
\[\begin{array}{l} \text{Least value} = 0 + 12 = \mathbf{12} \end{array}\]Explanation:
Let's rewrite the given expression by recognizing it as a sum of cubes:
\[\begin{array}{l} \sin^6\alpha + \cos^6\alpha + 3\sin^2\alpha\cos^2\alpha \\ = (\sin^2\alpha)^3 + (\cos^2\alpha)^3 + 3\sin^2\alpha\cos^2\alpha \cdot (1) \end{array}\]We know the fundamental trigonometric identity \( \sin^2\alpha + \cos^2\alpha = 1 \). We substitute this identity in place of the '\( 1 \)' in our expression:
\[\begin{array}{l} = (\sin^2\alpha)^3 + (\cos^2\alpha)^3 + 3\sin^2\alpha\cos^2\alpha(\sin^2\alpha + \cos^2\alpha) \end{array}\]Now, observe that this expression exactly matches the algebraic expansion formula: \( a^3 + b^3 + 3ab(a+b) = (a+b)^3 \), where \( a = \sin^2\alpha \) and \( b = \cos^2\alpha \).
\[\begin{array}{l} = (\sin^2\alpha + \cos^2\alpha)^3 \end{array}\]Applying the identity \( \sin^2\alpha + \cos^2\alpha = 1 \) one final time:
\[\begin{array}{l} = (1)^3 = \mathbf{1} \end{array}\]Explanation:
Given the equation:
\[\begin{array}{l} \operatorname{cosec}^2\theta = 2\cot\theta \end{array}\]We use the standard trigonometric identity relating cosecant and cotangent: \( \operatorname{cosec}^2\theta = 1 + \cot^2\theta \).
Substitute this identity into the equation:
\[\begin{array}{l} 1 + \cot^2\theta = 2\cot\theta \end{array}\]Rearrange the terms to form a quadratic equation in terms of \( \cot\theta \):
\[\begin{array}{l} \cot^2\theta - 2\cot\theta + 1 = 0 \end{array}\]This is a perfect square trinomial, \( a^2 - 2ab + b^2 = (a-b)^2 \):
\[\begin{array}{l} (\cot\theta - 1)^2 = 0 \\ \cot\theta - 1 = 0 \\ \cot\theta = 1 \end{array}\]Since we are given that \( \theta \) is an acute angle (\( 0^\circ < \theta < 90^\circ \)) and we know from the trigonometric table that \( \cot 45^\circ = 1 \):
\[\begin{array}{l} \theta = \mathbf{45^\circ} \end{array}\]
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