Ganit Prakash - Class-X - Trigonometry
Let us work out 24 Trigonometric Ratios of Complementary Angles
📘 Exercise 24 Solutions (Q1 - Q13)
Chapter 24Explanation:
We use the complementary angle identities: \( \cos(90^\circ - \theta) = \sin\theta \), \( \sec(90^\circ - \theta) = \operatorname{cosec}\theta \), and \( \cot(90^\circ - \theta) = \tan\theta \).
(i)
\[\begin{array}{l} \frac{\sin 38^\circ}{\cos 52^\circ} \\ = \frac{\sin 38^\circ}{\cos(90^\circ - 38^\circ)} \\ = \frac{\sin 38^\circ}{\sin 38^\circ} = \mathbf{1} \end{array}\](ii)
\[\begin{array}{l} \frac{\operatorname{cosec} 79^\circ}{\sec 11^\circ} \\ = \frac{\operatorname{cosec} 79^\circ}{\sec(90^\circ - 79^\circ)} \\ = \frac{\operatorname{cosec} 79^\circ}{\operatorname{cosec} 79^\circ} = \mathbf{1} \end{array}\](iii)
\[\begin{array}{l} \frac{\tan 27^\circ}{\cot 63^\circ} \\ = \frac{\tan 27^\circ}{\cot(90^\circ - 27^\circ)} \\ = \frac{\tan 27^\circ}{\tan 27^\circ} = \mathbf{1} \end{array}\](i) \( \sin 66^\circ - \cos 24^\circ = 0 \)
(ii) \( \cos^2 57^\circ + \cos^2 33^\circ = 1 \)
(iii) \( \cos^2 75^\circ - \sin^2 15^\circ = 0 \)
(iv) \( \operatorname{cosec}^2 48^\circ - \tan^2 42^\circ = 1 \)
(v) \( \sec 70^\circ \sin 20^\circ + \cos 20^\circ \operatorname{cosec} 70^\circ = 2 \)
Proof:
(i)
\[\begin{array}{l} \text{L.H.S.} = \sin 66^\circ - \cos 24^\circ \\ = \sin 66^\circ - \cos(90^\circ - 66^\circ) \\ = \sin 66^\circ - \sin 66^\circ \\ = 0 = \text{R.H.S.} \quad \textbf{[Proved]} \end{array}\](ii)
\[\begin{array}{l} \text{L.H.S.} = \cos^2 57^\circ + \cos^2 33^\circ \\ = \cos^2 57^\circ + \cos^2(90^\circ - 57^\circ) \\ = \cos^2 57^\circ + \sin^2 57^\circ \\ = 1 = \text{R.H.S.} \quad \textbf{[Proved]} \end{array}\](iii)
\[\begin{array}{l} \text{L.H.S.} = \cos^2 75^\circ - \sin^2 15^\circ \\ = \cos^2 75^\circ - \sin^2(90^\circ - 75^\circ) \\ = \cos^2 75^\circ - \cos^2 75^\circ \\ = 0 = \text{R.H.S.} \quad \textbf{[Proved]} \end{array}\](iv)
\[\begin{array}{l} \text{L.H.S.} = \operatorname{cosec}^2 48^\circ - \tan^2 42^\circ \\ = \operatorname{cosec}^2 48^\circ - \tan^2(90^\circ - 48^\circ) \\ = \operatorname{cosec}^2 48^\circ - \cot^2 48^\circ \\ = 1 = \text{R.H.S.} \quad \textbf{[Proved]} \end{array}\](v)
\[\begin{array}{l} \text{L.H.S.} = \sec 70^\circ \sin 20^\circ + \cos 20^\circ \operatorname{cosec} 70^\circ \\ = \sec 70^\circ \sin(90^\circ - 70^\circ) + \cos(90^\circ - 70^\circ) \operatorname{cosec} 70^\circ \\ = \sec 70^\circ \cos 70^\circ + \sin 70^\circ \operatorname{cosec} 70^\circ \\ = \left( \frac{1}{\cos 70^\circ} \times \cos 70^\circ \right) + \left( \sin 70^\circ \times \frac{1}{\sin 70^\circ} \right) \\ = 1 + 1 \\ = 2 = \text{R.H.S.} \quad \textbf{[Proved]} \end{array}\](i) \( \sin^2 \alpha + \sin^2 \beta = 1 \)
(ii) \( \cot \beta + \cos \beta = \frac{\cos \beta}{\cos \alpha} (1 + \sin \beta) \)
(iii) \( \frac{\sec \alpha}{\cos \alpha} - \cot^2 \beta = 1 \)
Proof:
Since \( \alpha \) and \( \beta \) are complementary angles, \( \alpha + \beta = 90^\circ \), which means \( \beta = 90^\circ - \alpha \) and \( \alpha = 90^\circ - \beta \).
(i)
\[\begin{array}{l} \text{L.H.S.} = \sin^2 \alpha + \sin^2 \beta \\ = \sin^2 \alpha + \sin^2(90^\circ - \alpha) \\ = \sin^2 \alpha + \cos^2 \alpha \\ = 1 = \text{R.H.S.} \quad \textbf{[Proved]} \end{array}\](ii)
\[\begin{array}{l} \text{L.H.S.} = \cot \beta + \cos \beta \\ = \frac{\cos \beta}{\sin \beta} + \cos \beta \\ = \cos \beta \left( \frac{1}{\sin \beta} + 1 \right) \\ = \cos \beta \left( \frac{1 + \sin \beta}{\sin \beta} \right) \end{array}\]Using \( \sin \beta = \sin(90^\circ - \alpha) = \cos \alpha \):
\[\begin{array}{l} = \frac{\cos \beta}{\cos \alpha} (1 + \sin \beta) \\ = \text{R.H.S.} \quad \textbf{[Proved]} \end{array}\](iii)
\[\begin{array}{l} \text{L.H.S.} = \frac{\sec \alpha}{\cos \alpha} - \cot^2 \beta \\ = \sec \alpha \cdot \frac{1}{\cos \alpha} - \cot^2(90^\circ - \alpha) \\ = \sec \alpha \cdot \sec \alpha - \tan^2 \alpha \\ = \sec^2 \alpha - \tan^2 \alpha \\ = 1 = \text{R.H.S.} \quad \textbf{[Proved]} \end{array}\]Proof:
Given \( \sin 17^\circ = \frac{x}{y} \).
First, we find \( \cos 17^\circ \):
\[\begin{array}{l} \cos 17^\circ = \sqrt{1 - \sin^2 17^\circ} = \sqrt{1 - \left(\frac{x}{y}\right)^2} \\ = \sqrt{\frac{y^2 - x^2}{y^2}} = \frac{\sqrt{y^2 - x^2}}{y} \end{array}\]Therefore, \( \sec 17^\circ = \frac{1}{\cos 17^\circ} = \frac{y}{\sqrt{y^2 - x^2}} \).
Also, \( \sin 73^\circ = \sin(90^\circ - 17^\circ) = \cos 17^\circ = \frac{\sqrt{y^2 - x^2}}{y} \).
Now, substitute these into the L.H.S.:
\[\begin{array}{l} \text{L.H.S.} = \sec 17^\circ - \sin 73^\circ \\ = \frac{y}{\sqrt{y^2 - x^2}} - \frac{\sqrt{y^2 - x^2}}{y} \\ = \frac{y \cdot y - \sqrt{y^2 - x^2} \cdot \sqrt{y^2 - x^2}}{y\sqrt{y^2 - x^2}} \\ = \frac{y^2 - (y^2 - x^2)}{y\sqrt{y^2 - x^2}} \\ = \frac{y^2 - y^2 + x^2}{y\sqrt{y^2 - x^2}} \\ = \frac{x^2}{y\sqrt{y^2 - x^2}} \\ = \text{R.H.S.} \quad \textbf{[Proved]} \end{array}\]Proof:
\[\begin{array}{l} \text{L.H.S.} = \sec^2 12^\circ - \frac{1}{\tan^2 78^\circ} \\ = \sec^2 12^\circ - \cot^2 78^\circ \quad [\because \frac{1}{\tan\theta} = \cot\theta] \\ = \sec^2 12^\circ - \cot^2(90^\circ - 12^\circ) \\ = \sec^2 12^\circ - \tan^2 12^\circ \quad [\because \cot(90^\circ-\theta) = \tan\theta] \\ = 1 = \text{R.H.S.} \quad \textbf{[Proved]} \end{array}\]Proof:
Given \( \angle A + \angle B = 90^\circ \), which means \( \angle B = 90^\circ - \angle A \).
\[\begin{array}{l} \text{L.H.S.} = 1 + \frac{\tan A}{\tan B} \\ = 1 + \frac{\tan A}{\tan(90^\circ - A)} \\ = 1 + \frac{\tan A}{\cot A} \quad [\because \tan(90^\circ-\theta) = \cot\theta] \\ = 1 + \tan A \cdot \tan A \quad [\because \frac{1}{\cot A} = \tan A] \\ = 1 + \tan^2 A \\ = \sec^2 A = \text{R.H.S.} \quad \textbf{[Proved]} \end{array}\]Proof:
Solving the Left Hand Side (L.H.S.):
\[\begin{array}{l} \text{L.H.S.} = \operatorname{cosec}^2 22^\circ \cot^2 68^\circ \\ = \operatorname{cosec}^2 22^\circ \cdot \cot^2(90^\circ - 22^\circ) \\ = \operatorname{cosec}^2 22^\circ \cdot \tan^2 22^\circ \\ = \frac{1}{\sin^2 22^\circ} \times \frac{\sin^2 22^\circ}{\cos^2 22^\circ} \\ = \frac{1}{\cos^2 22^\circ} \\ = \sec^2 22^\circ \end{array}\]Solving the Right Hand Side (R.H.S.):
\[\begin{array}{l} \text{R.H.S.} = \sin^2 22^\circ + \sin^2 68^\circ + \cot^2 68^\circ \\ = \sin^2 22^\circ + \sin^2(90^\circ - 22^\circ) + \cot^2 68^\circ \\ = (\sin^2 22^\circ + \cos^2 22^\circ) + \cot^2 68^\circ \\ = 1 + \cot^2 68^\circ \quad [\because \sin^2\theta + \cos^2\theta = 1] \\ = \operatorname{cosec}^2 68^\circ \\ = \operatorname{cosec}^2(90^\circ - 22^\circ) \\ = \sec^2 22^\circ \end{array}\]Since L.H.S. = R.H.S., the identity is [Proved].
Proof:
Given \( \angle P + \angle Q = 90^\circ \), which means \( \angle Q = 90^\circ - \angle P \).
Therefore, \( \cos Q = \cos(90^\circ - P) = \sin P \).
Substitute this into the L.H.S.:
\[\begin{array}{l} \text{L.H.S.} = \sqrt{\frac{\sin P}{\cos Q} - \sin P \cos Q} \\ = \sqrt{\frac{\sin P}{\sin P} - \sin P \cdot \sin P} \\ = \sqrt{1 - \sin^2 P} \end{array}\]Using the identity \( \sin^2 P + \cos^2 P = 1 \):
\[\begin{array}{l} = \sqrt{\cos^2 P} \\ = \cos P \\ = \text{R.H.S.} \quad \textbf{[Proved]} \end{array}\]Proof:
Taking the Left Hand Side (L.H.S.) of the equation:
\[\begin{array}{l} \text{L.H.S.} = \cot 12^\circ \cot 38^\circ \cot 52^\circ \cot 78^\circ \cot 60^\circ \end{array}\]We group the terms whose angles are complementary (sum to \( 90^\circ \)):
\( 12^\circ \) and \( 78^\circ \) are complementary because \( 12^\circ + 78^\circ = 90^\circ \).
\( 38^\circ \) and \( 52^\circ \) are complementary because \( 38^\circ + 52^\circ = 90^\circ \).
Using the complementary angle identity \( \cot(90^\circ - \theta) = \tan\theta \):
\[\begin{array}{l} \cot 78^\circ = \cot(90^\circ - 12^\circ) = \tan 12^\circ \\ \cot 52^\circ = \cot(90^\circ - 38^\circ) = \tan 38^\circ \end{array}\]Substituting these back into the expression along with the standard value \( \cot 60^\circ = \frac{1}{\sqrt{3}} \):
\[\begin{array}{l} \text{L.H.S.} = (\cot 12^\circ \cdot \tan 12^\circ) \cdot (\cot 38^\circ \cdot \tan 38^\circ) \cdot \frac{1}{\sqrt{3}} \end{array}\]Since \( \cot\theta \cdot \tan\theta = 1 \):
\[\begin{array}{l} = (1) \cdot (1) \cdot \frac{1}{\sqrt{3}} \\ = \frac{1}{\sqrt{3}} = \text{R.H.S.} \quad \textbf{[Proved]} \end{array}\](i) \( \tan \angle ABC = \cot \angle ACO \)
(ii) \( \sin^2 \angle BCO + \sin^2 \angle ACO = 1 \)
(iii) \( \operatorname{cosec}^2 \angle CAB - 1 = \tan^2 \angle ABC \)
Initial Setup & Explanation:
Since AOB is the diameter of the circle, the angle formed in the semicircle is a right angle. Therefore, \( \angle ACB = 90^\circ \).
In the right-angled triangle ABC, the sum of the other two acute angles is \( 90^\circ \). So, \( \angle CAB + \angle ABC = 90^\circ \).
In triangle AOC, sides OA and OC are radii of the same circle. Thus, \( OA = OC \), which implies that angles opposite to these sides are equal. So, \( \angle OAC = \angle OCA \), which can be written as \( \angle CAB = \angle ACO \).
Similarly, in triangle BOC, sides OB and OC are radii. Thus, \( OB = OC \), which implies \( \angle OBC = \angle OCB \), which can be written as \( \angle ABC = \angle BCO \).
Since \( \angle ACB = 90^\circ \), we know that \( \angle ACO + \angle BCO = 90^\circ \).
Proof for (i):
We need to show \( \tan \angle ABC = \cot \angle ACO \).
From our setup, we know \( \angle ABC = \angle BCO \).
Since \( \angle BCO + \angle ACO = 90^\circ \), we can write \( \angle BCO = 90^\circ - \angle ACO \).
Therefore, \( \angle ABC = 90^\circ - \angle ACO \).
\[\begin{array}{l} \text{L.H.S.} = \tan \angle ABC \\ = \tan(90^\circ - \angle ACO) \\ = \cot \angle ACO = \text{R.H.S.} \quad \textbf{[Proved]} \end{array}\]Proof for (ii):
We need to show \( \sin^2 \angle BCO + \sin^2 \angle ACO = 1 \).
We established that \( \angle BCO + \angle ACO = 90^\circ \), which means \( \angle BCO = 90^\circ - \angle ACO \).
\[\begin{array}{l} \text{L.H.S.} = \sin^2 \angle BCO + \sin^2 \angle ACO \\ = \sin^2(90^\circ - \angle ACO) + \sin^2 \angle ACO \\ = \cos^2 \angle ACO + \sin^2 \angle ACO \\ = 1 = \text{R.H.S.} \quad \textbf{[Proved]} \end{array}\]Proof for (iii):
We need to show \( \operatorname{cosec}^2 \angle CAB - 1 = \tan^2 \angle ABC \).
We established that \( \angle CAB + \angle ABC = 90^\circ \), which means \( \angle CAB = 90^\circ - \angle ABC \).
\[\begin{array}{l} \text{L.H.S.} = \operatorname{cosec}^2 \angle CAB - 1 \\ = \operatorname{cosec}^2(90^\circ - \angle ABC) - 1 \\ = \sec^2 \angle ABC - 1 \end{array}\]Using the standard trigonometric identity \( \sec^2\theta - 1 = \tan^2\theta \):
\[\begin{array}{l} = \tan^2 \angle ABC = \text{R.H.S.} \quad \textbf{[Proved]} \end{array}\](i) \( \tan \angle ACD = \cot \angle ACB \)
(ii) \( \tan^2 \angle CAD + 1 = \frac{1}{\sin^2 \angle BAC} \)
Initial Setup & Explanation:
In the rectangular figure ABCD, all interior angles are right angles. Specifically, the corner angles are \( 90^\circ \).
Therefore, at vertex C, the angle is \( \angle BCD = 90^\circ \). The diagonal AC divides this angle into two parts, so \( \angle ACD + \angle ACB = 90^\circ \).
Similarly, at vertex A, the angle is \( \angle BAD = 90^\circ \). The diagonal AC divides this angle into two parts, so \( \angle CAD + \angle BAC = 90^\circ \).
Proof for (i):
We need to show \( \tan \angle ACD = \cot \angle ACB \).
Since \( \angle ACD + \angle ACB = 90^\circ \), we can express one angle in terms of the other: \( \angle ACD = 90^\circ - \angle ACB \).
\[\begin{array}{l} \text{L.H.S.} = \tan \angle ACD \\ = \tan(90^\circ - \angle ACB) \end{array}\]Using the complementary angle formula \( \tan(90^\circ - \theta) = \cot\theta \):
\[\begin{array}{l} = \cot \angle ACB = \text{R.H.S.} \quad \textbf{[Proved]} \end{array}\]Proof for (ii):
We need to show \( \tan^2 \angle CAD + 1 = \frac{1}{\sin^2 \angle BAC} \).
Since \( \angle CAD + \angle BAC = 90^\circ \), we can write \( \angle CAD = 90^\circ - \angle BAC \).
\[\begin{array}{l} \text{L.H.S.} = \tan^2 \angle CAD + 1 \\ = \tan^2(90^\circ - \angle BAC) + 1 \\ = \cot^2 \angle BAC + 1 \end{array}\]Using the standard trigonometric identity \( 1 + \cot^2\theta = \operatorname{cosec}^2\theta \):
\[\begin{array}{l} = \operatorname{cosec}^2 \angle BAC \\ = \frac{1}{\sin^2 \angle BAC} = \text{R.H.S.} \quad \textbf{[Proved]} \end{array}\]📘 12. Very short answer type questions (V.S.A)
(A) M.C.Q.Explanation:
Using the complementary angle formulas \( \cos\theta = \sin(90^\circ - \theta) \) and \( \sin\theta = \cos(90^\circ - \theta) \):
\[\begin{array}{l} \sin 43^\circ \cos 47^\circ + \cos 43^\circ \sin 47^\circ \\ = \sin 43^\circ \cos(90^\circ - 43^\circ) + \cos 43^\circ \sin(90^\circ - 43^\circ) \\ = \sin 43^\circ \sin 43^\circ + \cos 43^\circ \cos 43^\circ \\ = \sin^2 43^\circ + \cos^2 43^\circ \end{array}\]We know the identity \( \sin^2\theta + \cos^2\theta = 1 \):
\[\begin{array}{l} = 1 \end{array}\]Correct Option: (b) 1
Explanation:
Using the complementary angle formulas \( \cot(90^\circ - \theta) = \tan\theta \) and \( \tan(90^\circ - \theta) = \cot\theta \):
\[\begin{array}{l} \frac{\tan 35^\circ}{\cot 55^\circ} + \frac{\cot 78^\circ}{\tan 12^\circ} \\ = \frac{\tan 35^\circ}{\cot(90^\circ - 35^\circ)} + \frac{\cot 78^\circ}{\tan(90^\circ - 78^\circ)} \\ = \frac{\tan 35^\circ}{\tan 35^\circ} + \frac{\cot 78^\circ}{\cot 78^\circ} \\ = 1 + 1 \\ = 2 \end{array}\]Correct Option: (c) 2
Explanation:
Using the complementary angle formula \( \cos x = \sin(90^\circ - x) \):
\[\begin{array}{l} \cos(40^\circ+\theta) \\ = \sin(90^\circ - (40^\circ+\theta)) \\ = \sin(90^\circ - 40^\circ - \theta) \\ = \sin(50^\circ - \theta) \end{array}\]Now substitute this back into the expression:
\[\begin{array}{l} \{\cos(40^\circ+\theta) - \sin(50^\circ-\theta)\} \\ = \sin(50^\circ - \theta) - \sin(50^\circ-\theta) \\ = 0 \end{array}\]Correct Option: (c) 0
Explanation:
In any triangle ABC, the sum of the three angles is \( 180^\circ \):
\[\begin{array}{l} A + B + C = 180^\circ \\ B + C = 180^\circ - A \end{array}\]Divide both sides by 2:
\[\begin{array}{l} \frac{B+C}{2} = \frac{180^\circ - A}{2} \\ \frac{B+C}{2} = 90^\circ - \frac{A}{2} \end{array}\]Now, take the sine of both sides:
\[\begin{array}{l} \sin\left(\frac{B+C}{2}\right) = \sin\left(90^\circ - \frac{A}{2}\right) \end{array}\]Using the complementary angle formula \( \sin(90^\circ - \theta) = \cos\theta \):
\[\begin{array}{l} = \cos\frac{A}{2} \end{array}\]Correct Option: (b) \( \cos\frac{A}{2} \)
Explanation:
Given \( A + B = 90^\circ \), which means \( B = 90^\circ - A \).
We need to find the value of \( \cot B \):
\[\begin{array}{l} \cot B = \cot(90^\circ - A) \end{array}\]Using the complementary angle formula \( \cot(90^\circ - \theta) = \tan\theta \):
\[\begin{array}{l} \cot B = \tan A \end{array}\]We are given \( \tan A = \frac{3}{4} \). Therefore:
\[\begin{array}{l} \cot B = \frac{3}{4} \end{array}\]Correct Option: (a) \( \frac{3}{4} \)
📘 12. Very short answer type questions (V.S.A)
(B) True or FalseThe value of \( \cos 54^\circ \) and \( \sin 36^\circ \) are equal.
Explanation:
Using the complementary angle formula \( \cos\theta = \sin(90^\circ - \theta) \):
\[\begin{array}{l} \cos 54^\circ = \sin(90^\circ - 54^\circ) \\ \cos 54^\circ = \sin 36^\circ \end{array}\]Thus, their values are exactly equal.
Answer: True
The simplified value of \( (\sin 12^\circ - \cos 78^\circ) \) is 1.
Explanation:
Using the complementary angle formula \( \cos\theta = \sin(90^\circ - \theta) \):
\[\begin{array}{l} \sin 12^\circ - \cos 78^\circ \\ = \sin 12^\circ - \sin(90^\circ - 78^\circ) \\ = \sin 12^\circ - \sin 12^\circ \\ = 0 \end{array}\]The simplified value is 0, not 1.
Answer: False
📘 12. Very short answer type questions (V.S.A)
(C) Fill in the blanksExplanation:
We group the complementary angles and use standard values:
\[\begin{array}{l} (\tan 15^\circ \times \tan 75^\circ) \times \tan 45^\circ \times \tan 60^\circ \end{array}\]Using the complementary angle formula \( \tan(90^\circ - \theta) = \cot\theta \):
\[\begin{array}{l} = (\tan 15^\circ \times \cot 15^\circ) \times \tan 45^\circ \times \tan 60^\circ \end{array}\]Since \( \tan\theta \times \cot\theta = 1 \), \( \tan 45^\circ = 1 \), and \( \tan 60^\circ = \sqrt{3} \):
\[\begin{array}{l} = (1) \times 1 \times \sqrt{3} \\ = \sqrt{3} \end{array}\]Answer: \( \sqrt{3} \)
Explanation:
We use the complementary angle formulas \( \sec(90^\circ - \theta) = \operatorname{cosec}\theta \) and \( \operatorname{cosec}(90^\circ - \theta) = \sec\theta \):
\[\begin{array}{l} \sec 78^\circ = \operatorname{cosec}(90^\circ - 78^\circ) = \operatorname{cosec} 12^\circ \\ \operatorname{cosec} 72^\circ = \sec(90^\circ - 72^\circ) = \sec 18^\circ \end{array}\]Substitute these back into the expression and group terms:
\[\begin{array}{l} \sin 12^\circ \times \cos 18^\circ \times \operatorname{cosec} 12^\circ \times \sec 18^\circ \\ = (\sin 12^\circ \times \operatorname{cosec} 12^\circ) \times (\cos 18^\circ \times \sec 18^\circ) \end{array}\]Since \( \sin\theta \times \operatorname{cosec}\theta = 1 \) and \( \cos\theta \times \sec\theta = 1 \):
\[\begin{array}{l} = (1) \times (1) \\ = 1 \end{array}\]Answer: 1
Explanation:
If A and B are complementary angles, their sum is \( 90^\circ \).
\[\begin{array}{l} A + B = 90^\circ \\ A = 90^\circ - B \end{array}\]Taking the sine of both sides:
\[\begin{array}{l} \sin A = \sin(90^\circ - B) \end{array}\]Using the complementary angle formula \( \sin(90^\circ - \theta) = \cos\theta \):
\[\begin{array}{l} \sin A = \cos B \end{array}\]Answer: \( \cos B \)
📘 13. Short answer type questions :
Exercise 24Explanation:
Given equation:
\[\begin{array}{l} \sin 10\theta = \cos 8\theta \end{array}\]We know the complementary angle relationship: \( \cos \alpha = \sin(90^\circ - \alpha) \). Substituting this into the right side:
\[\begin{array}{l} \sin 10\theta = \sin(90^\circ - 8\theta) \end{array}\]Comparing the angles (since \( 10\theta \) is acute):
\[\begin{array}{l} 10\theta = 90^\circ - 8\theta \\ 10\theta + 8\theta = 90^\circ \\ 18\theta = 90^\circ \\ \theta = \frac{90^\circ}{18} = 5^\circ \end{array}\]Now, we need to find the value of \( \tan 9\theta \):
\[\begin{array}{l} \tan 9\theta = \tan(9 \times 5^\circ) \\ = \tan 45^\circ \\ = \mathbf{1} \end{array}\]Explanation:
Given equation:
\[\begin{array}{l} \tan 4\theta \times \tan 6\theta = 1 \end{array}\]We can rewrite this by dividing both sides by \( \tan 4\theta \):
\[\begin{array}{l} \tan 6\theta = \frac{1}{\tan 4\theta} \\ \tan 6\theta = \cot 4\theta \end{array}\]We know the complementary angle relationship: \( \cot \alpha = \tan(90^\circ - \alpha) \). Applying this:
\[\begin{array}{l} \tan 6\theta = \tan(90^\circ - 4\theta) \end{array}\]Comparing the angles:
\[\begin{array}{l} 6\theta = 90^\circ - 4\theta \\ 6\theta + 4\theta = 90^\circ \\ 10\theta = 90^\circ \\ \theta = \frac{90^\circ}{10} \\ \theta = \mathbf{9^\circ} \end{array}\]Explanation:
We use the complementary angle relationships: \( \sin 27^\circ = \cos(90^\circ - 27^\circ) = \cos 63^\circ \) and \( \cos 73^\circ = \sin(90^\circ - 73^\circ) = \sin 17^\circ \).
Substitute these into the expression:
\[\begin{array}{l} = \frac{2\sin^2 63^\circ + 1 + 2\cos^2 63^\circ}{3\cos^2 17^\circ - 2 + 3\sin^2 17^\circ} \end{array}\]Group the terms to apply the identity \( \sin^2\alpha + \cos^2\alpha = 1 \):
\[\begin{array}{l} = \frac{2(\sin^2 63^\circ + \cos^2 63^\circ) + 1}{3(\cos^2 17^\circ + \sin^2 17^\circ) - 2} \\ = \frac{2(1) + 1}{3(1) - 2} \\ = \frac{2 + 1}{3 - 2} \\ = \frac{3}{1} = \mathbf{3} \end{array}\]Explanation:
The given sequence is a product of tangents from \( 1^\circ \) to \( 89^\circ \).
We can pair the terms from the beginning and the end. Using the complementary property \( \tan(90^\circ - \theta) = \cot\theta \):
\[\begin{array}{l} \tan 89^\circ = \tan(90^\circ - 1^\circ) = \cot 1^\circ \\ \tan 88^\circ = \tan(90^\circ - 2^\circ) = \cot 2^\circ \end{array}\]And so on.
Now, substitute these back into the sequence and group them:
\[\begin{array}{l} = (\tan 1^\circ \times \cot 1^\circ) \times (\tan 2^\circ \times \cot 2^\circ) \times \dots \times \tan 45^\circ \end{array}\]We know that \( \tan\theta \times \cot\theta = 1 \) and \( \tan 45^\circ = 1 \).
\[\begin{array}{l} = (1) \times (1) \times \dots \times 1 \\ = \mathbf{1} \end{array}\]Explanation:
Given equation:
\[\begin{array}{l} \sec 5A = \operatorname{cosec}(A + 36^\circ) \end{array}\]Using the complementary angle identity \( \sec\theta = \operatorname{cosec}(90^\circ - \theta) \), we can write \( \sec 5A \) as:
\[\begin{array}{l} \operatorname{cosec}(90^\circ - 5A) = \operatorname{cosec}(A + 36^\circ) \end{array}\]Since both sides have the cosecant function and the angles are positive acute angles, we can equate the angles:
\[\begin{array}{l} 90^\circ - 5A = A + 36^\circ \end{array}\]Solving for \( A \):
\[\begin{array}{l} 90^\circ - 36^\circ = A + 5A \\ 54^\circ = 6A \\ A = \frac{54^\circ}{6} \\ A = \mathbf{9^\circ} \end{array}\]
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