CBSE Class 11 Mathematics Trigonometric Functions Worksheet Set 06

CBSE Class 11 Mathematics Worksheet - Trigonometric Functions (6). Students can download these worksheets and practice them. This will help them to get better marks in examinations. Also refer to other worksheets for the same chapter and other subjects too. Use them for better understanding of the subjects.

Question. If \( A, B, C, D \) be the angles of a cyclic quadrant lateral, then show that \( \cos(180^\circ - A) + \cos(180^\circ - B) + \cos(180^\circ + C) - \sin(90^\circ + D) = 0 \)?
Answer: We have \( A + C = 180^\circ \) and \( B + D = 180^\circ \) (property of cyclic quadrilateral)
L.H.S. \( \cos(180^\circ - A) + \cos(180^\circ - B) + \cos(180^\circ + C) - \sin(90^\circ + D) \)
\( = \cos C - \cos B - \cos C - \cos D \) ........... \( (A + C = 180^\circ) \)
\( = -\cos(180^\circ - D) - \cos D \)
\( = \cos D - \cos D \) ........... \( (B + D = 180^\circ) \)
\( = 0 \) R.H.S (proved)

Question. Find \( x \) from the equation: \( x \cot(90^\circ + \theta) + x \tan(90^\circ + \theta) \sin \theta + \csc(90^\circ + \theta) = 0 \)
Answer: \( x \cot(90^\circ + \theta) + x \tan(90^\circ + \theta) \sin \theta + \csc(90^\circ + \theta) = 0 \)
\( \implies -x \tan \theta - x \cot \theta \cdot \sin \theta + \sec \theta = 0 \)
\( \implies -x \frac{\sin \theta}{\cos \theta} - \frac{\cos \theta}{\sin \theta} \cdot \sin \theta + \sec \theta = 0 \)
\( \implies -x \frac{\sin \theta}{\cos \theta} - \cos \theta + \frac{1}{\cos \theta} = 0 \)
\( \implies -x \sin \theta - \cos^2 \theta + 1 = 0 \)
\( \implies -x \sin \theta + \sin^2 \theta = 0 \)
\( \implies -x \sin \theta = -\sin^2 \theta \)
\( \implies x = \sin \theta \) ans.

Question. Show that (3 angle & sum of two equal to third) \( \tan(3x) - \tan(2x) - \tan x = \tan(3x) \tan(2x) \tan x \)?
Answer: We have, \( 3x = 2x + x \)
\( \implies \tan(3x) = \tan(2x + x) \)
\( \implies \tan(3x) = \frac{\tan(2x) + \tan x}{1 - \tan(2x) \tan x} \)
\( \implies \tan(3x) [1 - \tan(2x) \tan x] = \tan(2x) + \tan x \)
\( \implies \tan(3x) - \tan(3x) \tan(2x) \tan x = \tan(2x) + \tan x \)
\( \implies \tan(3x) - \tan(2x) - \tan x = \tan(3x) \tan(2x) \tan x \) (proved)

Question. Show that, \( \cot(2x) \cot x - \cot(3x) \cot x - \cot(3x) \dots \cot(2x) = 1 \)
Answer: We have, \( 3x = 2x + x \)
\( \implies \cot(3x) = \cot(2x + x) \)
\( \implies \cot(3x) = \frac{\cot(2x) \cdot \cot x - 1}{\cot(2x) + \cot x} \)
\( \implies \cot(3x) [\cot(2x) + \cot x] = \dots \cot(2x) \cdot \cot x - 1 \)
\( \implies \cot(3x) \cdot \dots \cot(2x) + \cot(3x) \cdot \dots \cot x = \cot(2x) \cdot \cot x - 1 \)
\( \implies 1 = \dots \cot(2x) \cdot \cot x - \cot(3x) \cdot \dots \cot(2x) - \dots \cot(3x) \cdot \dots \cot x \) (proved)

Question. Show that, \( \tan(70^\circ) = 2\tan(50^\circ) + \tan(20^\circ) \)?
Answer: We have, \( 70^\circ = 50^\circ + 20^\circ \)
\( \implies \tan(70^\circ) = \tan(50^\circ + 20^\circ) \)
\( \implies \tan(70^\circ) = \frac{\tan(50^\circ) + \tan(20^\circ)}{1 - \tan(50^\circ) \tan(20^\circ)} \)
\( \implies \tan(70^\circ) [1 - \tan(50^\circ) \tan(20^\circ)] = \tan(50^\circ) + \tan(20^\circ) \)
\( \implies \tan(70^\circ) - \tan(70^\circ) \tan(50^\circ) \tan(20^\circ) = \tan(50^\circ) + \tan(20^\circ) \)
\( \implies \tan(70^\circ) - \tan(90^\circ - 20^\circ) \cdot \tan(50^\circ) \tan(20^\circ) = \tan(50^\circ) + \tan(20^\circ) \)
\( \implies \tan(70^\circ) - \cot(20^\circ) \cdot \tan(50^\circ) \tan(20^\circ) = \tan(50^\circ) + \tan(20^\circ) \)
\( \implies \tan(70^\circ) - \tan(50^\circ) = \tan(50^\circ) + \tan(20^\circ) \)
\( \implies \tan(70^\circ) = 2\tan(50^\circ) + \tan(20^\circ) \) (proved)

Question.
1. Find the value of \( \tan\left(\frac{13\pi}{12}\right) \)?
2. Show that \( \tan(189^\circ) = \frac{\cos(36^\circ) - \sin(36^\circ)}{\cos(36^\circ) + \sin(36^\circ)} \)?
Answer:
1. \( \tan\left(\frac{13\pi}{12}\right) = \tan(195^\circ) \)
\( = \tan(180^\circ + 15^\circ) = \tan(15^\circ) \)
\( = \tan(45^\circ - 30^\circ) \)
\( = \frac{\tan(45^\circ) - \tan(30^\circ)}{1 + \tan(45^\circ) \tan(30^\circ)} \)
\( = \frac{1 - \frac{1}{\sqrt{3}}}{1 + \frac{1}{\sqrt{3}}} \)
\( = \frac{\sqrt{3} - 1}{\sqrt{3} + 1} \)
\( = \frac{\sqrt{3} - 1}{\sqrt{3} + 1} \times \frac{\sqrt{3} - 1}{\sqrt{3} - 1} \)
\( = \frac{3 + 1 - 2\sqrt{3}}{3 - 1} \)
\( = \frac{4 - 2\sqrt{3}}{2} \)
\( = 2 - \sqrt{3} \) ans.

2. L.H.S. \( \tan(189^\circ) = \tan(180^\circ + 9^\circ) \)
\( = \tan(9^\circ) \)
\( = \tan(45^\circ - 36^\circ) \)
\( = \frac{\tan(45^\circ) - \tan(36^\circ)}{1 + \tan(45^\circ) \tan(36^\circ)} \)
\( = \frac{1 - \tan(36^\circ)}{1 + \tan(36^\circ)} \)
\( = \frac{1 - \frac{\sin(36^\circ)}{\cos(36^\circ)}}{1 + \frac{\sin(36^\circ)}{\cos(36^\circ)}} \)
\( = \frac{\cos(36^\circ) - \sin(36^\circ)}{\cos(36^\circ) + \sin(36^\circ)} \) ans.

Question. If \( \sin a = \frac{3}{5} \); \( 0 < a < \frac{\pi}{2} \) and \( \cos b = -\frac{12}{13} \); \( \pi < b < \frac{3\pi}{2} \). Find the value of \( \tan(a - b) \)?
Answer: We have, \( \sin a = \frac{3}{5} \)
\( \cos^2 a = 1 - \sin^2 a \)
\( = 1 - \frac{9}{25} \)
\( \implies \cos^2 a = \frac{16}{25} \)
\( \implies \cos a = \frac{4}{5} \) (\( \because a \to \) 1st quadrant)
Given: \( \cos b = -\frac{12}{13} \)
\( \implies \sin^2 b = 1 - \cos^2 b = 1 - \frac{144}{169} \)
\( = \frac{25}{169} \)
\( \implies \sin b = -\frac{5}{13} \) (\( \because b \to \) 3rd quadrant)
\( \therefore \tan a = \frac{\sin a}{\cos a} = \frac{3/5}{4/5} = \frac{3}{4} \)
\( \tan b = \frac{\sin b}{\cos b} = \frac{-5/13}{-12/13} = \frac{5}{12} \)
Now, \( \tan(a - b) = \frac{\tan a - \tan b}{1 + \tan a \tan b} \)
\( = \frac{\frac{3}{4} - \frac{5}{12}}{1 + \frac{3}{4} \times \frac{5}{12}} \)
\( = \frac{\frac{36 - 20}{48}}{\frac{48 + 15}{48}} \)
\( = \frac{16}{63} \) ans.

Question.
1. Show that, \( \cos\left(\frac{\pi}{4} - a\right) \cdot \cos\left(\frac{\pi}{4} - b\right) - \sin\left(\frac{\pi}{4} - a\right) \cdot \sin\left(\frac{\pi}{4} - b\right) = \sin(a + b) \)
2. Show that, \( \sin(n + 1)a \cdot \sin(n + 2)a + \cos(n + 1)a \cdot \cos(n + 2)a = \cos a \)
Answer:
1. L.H.S. \( \cos\left(\frac{\pi}{4} - a\right) \cdot \cos\left(\frac{\pi}{4} - b\right) - \sin\left(\frac{\pi}{4} - a\right) \cdot \sin\left(\frac{\pi}{4} - b\right) \)
\( = \cos\left(\frac{\pi}{4} - a + \frac{\pi}{4} - b\right) \) ............. {using formula \( \cos(x + y) \)}
\( = \cos\left(\frac{\pi}{2} - (a + b)\right) = \sin(a + b) \)

2. L.H.S. \( \sin(n + 1)a \cdot \sin(n + 2)a + \cos(n + 1)a \cdot \cos(n + 2)a \)
\( = \cos((n + 1)a - (n + 2)a) \) ....... {using formula \( \cos(x - y) \)}
\( = \cos(na + a - na - 2a) = \cos(-a) = \cos a \) ans.

Question. If \( a + b = \frac{\pi}{4} \), show that, \( (1 + \tan a)(1 + \tan b) = 2 \)?
Answer: We have, \( a + b = \frac{\pi}{4} \)
\( \implies \tan(a + b) = \tan\left(\frac{\pi}{4}\right) \)
\( \implies \frac{\tan a + \tan b}{1 - \tan a \tan b} = 1 \)
\( \implies \tan a + \tan b = 1 - \tan a \tan b \)
\( \implies \tan a \tan b + \tan a + \tan b = 1 \)
Adding 1 on both sides
\( \implies \tan a \tan b + \tan a + \tan b + 1 = 1 + 1 \)
\( \implies \tan a (\tan b + 1) + 1(\tan b + 1) = 2 \)
\( \implies (1 + \tan a)(1 + \tan b) = 2 \) (proved)

Question. If \( \cos(\alpha + \beta) = \frac{4}{5} \) and \( \sin(\alpha - \beta) = \frac{5}{13} \) and \( \alpha, \beta \) lie between \( 0^\circ \) and \( 45^\circ \). Show that \( \tan(2\alpha) = \frac{56}{33} \)
Answer: Since, \( \alpha \) and \( \beta \) lie between \( 0^\circ \) and \( 45^\circ \)
\( \therefore -45^\circ < (\alpha - \beta) < 45^\circ \) and \( 0^\circ < (\alpha + \beta) < 90^\circ \)
\( \therefore \cos(\alpha - \beta) \) and \( \sin(\alpha + \beta) \) are both positive
Now, \( \sin(\alpha + \beta) = \sqrt{1 - \cos^2(\alpha + \beta)} = \sqrt{1 - \frac{16}{25}} = \frac{3}{5} \) and
\( \cos(\alpha - \beta) = \sqrt{1 - \sin^2(\alpha - \beta)} = \sqrt{1 - \frac{25}{169}} = \frac{12}{13} \)
\( \therefore \tan(\alpha + \beta) = \frac{\sin(\alpha+\beta)}{\cos(\alpha+\beta)} = \frac{3/5}{4/5} = \frac{3}{4} \) and \( \tan(\alpha - \beta) = \frac{\sin(\alpha-\beta)}{\cos(\alpha-\beta)} = \frac{5/13}{12/13} = \frac{5}{12} \)
Now, \( \tan(2\alpha) = \tan\{(\alpha + \beta) + (\alpha - \beta)\} \)
\( = \frac{\tan(\alpha + \beta) + \tan(\alpha - \beta)}{1 - \tan(\alpha + \beta) \tan(\alpha - \beta)} \)
\( = \frac{\frac{3}{4} + \frac{5}{12}}{1 - \frac{3}{4} \times \frac{5}{12}} \)
\( = \frac{\frac{36 + 20}{48}}{\frac{48 - 15}{48}} = \frac{56}{33} \)
\( \implies \tan(2\alpha) = \frac{56}{33} \) ans.