Solve the equation tan θ + tan 2 θ + tan 3 θ = 0 .

Given, tan θ + tan 2 θ + tan 3 θ = 0 ⇒ tan θ + tan 2 θ + tan θ + 2 θ = 0 ⇒ tan θ + tan 2 θ + tan θ + tan 2 θ 1 - tan θ tan 2 θ = 0 ∵ tan A + B = tan A + tan B 1 - tan A tan B ⇒ tan θ + tan 2 θ 1 + 1 1 - tan θ tan 2 θ = 0 ⇒ tan θ + tan 2 θ 2 - tan θ tan 2 θ = 0 tan θ tan 2 θ ≠ 1 Hence, either tan θ + tan 2 θ = 0 or 2 - tan θ tan 2 θ = 0 If tan 2 θ + tan θ = 0 then tan 2 θ = - tan θ = tan ( π - θ ) ⇒ 2 θ = n π + π - θ ∵ tan α = tan β , α = m π + β , m ∈ I ⇒ θ = 1 3 π n + 1 , n ∈ I If 2 - tan θ 2 tan θ 1 - tan 2 θ = 0 ∵ tan 2 A = 2 tan A 1 - tan 2 A ⇒ 2 - 2 tan 2 θ - 2 tan 2 θ = 0 ⇒ 2 - 4 tan 2 θ = 0 ⇒ tan 2 θ = 1 2 ⇒ θ = n π ± tan - 1 1 2 , n ∈ I ∵ tan 2 α = tan 2 β , α = m π ± β , m ∈ I

Solve the equation tan θ + tan 2 θ + 3 tan θ tan 2 θ = 3 .

tan θ + tan 2 θ + 3 tan θ tan 2 θ = 3 ⇒ tan θ + tan 2 θ = 3 1 - tan θ tan 2 θ ⇒ tan θ + tan 2 θ 1 - tan θ tan 2 θ = 3 ⇒ tan θ + 2 θ = 3 ∵ tan A + B = tan A + tan B 1 - tan A tan B ⇒ tan 3 θ = 3 ⇒ 3 θ = n π + π 3 ∵ tan α = tan β ⇒ α = n π + β , n ∈ I ⇒ θ = 1 3 n π + π 3 , n ∈ I

If sin ( π cosθ ) = cos ( πsinθ ) , prove that cos θ ± π 4 = 1 2 2 .

sin ( πcos θ ) = cos ( π sin θ ) or cos π 2 - π cos θ = cos π sin θ or π 2 - π cos θ = 2 nπ ± π sin θ ∵ cosα = cosβ ⇒ α = 2 nπ ± β , n ∈ I or 1 2 - cos θ = 2 n ± sin θ or cos θ ± sin θ = 1 2 - 2 n Divided both the sides by 2 , we get 1 2 cos θ ± 1 2 sin θ = 1 - 4 n 2 2 or cos θ ± π 4 = 1 - 4 n 2 2 Now, - 1 ≤ 1 - 4 n 2 2 ≤ 1 ⇒ - 2 2 ≤ 1 - 4 n ≤ 2 2 ⇒ 2 2 + 1 4 ≥ n ≥ 1 - 2 2 4 Here, n is an integer So, n = 0 is the only possible value. ∴ cos θ ± π 4 = 1 2 2

If sin ( π cot θ ) = cos ( π tan θ ) , prove that either cosec 2 θ or cot 2 θ is equal to n + 1 4 , where n is a positive or negative integer.

sin ( πcot θ ) = cos ( πtan θ ) or cos π 2 - πcot θ = cos ( πtan θ ) ∵ cos θ = cos α ⇒ θ = 2 n π ± α , n ∈ I or πtan θ = 2 nπ ± π 2 - πcot θ , n ∈ I Taking the + sign and dividing by π , we get tan θ = 2 n + 1 2 - cot θ or sin θ cos θ + cos θ sin θ = 2 n + 1 2 or sin 2 θ + cos 2 θ sin θ cos θ = 2 n + 1 4 or 1 2 sin θ cos θ = n + 1 4 or 1 sin 2 θ = n + 1 4 or cosec 2 θ = n + 1 4 Taking the - v e sign into consideration, we get tan θ = 2 n - 1 2 + cot θ or sin θ cos θ - cos θ sin θ = 2 n - 1 2 or cos 2 θ - sin 2 θ 2 sin θ cos θ = 1 4 - n or cos 2 θ sin 2 θ = 1 4 - n or cot 2 θ = 1 4 - n

E M B I B E

Plane Trigonometry Part 1>Chapter 9 - Identities and Trigonometrical Equations>EXAMPLES XXI

S L Loney Solutions for Chapter: Identities and Trigonometrical Equations, Exercise 2: EXAMPLES XXI

Author:S L Loney

S L Loney Mathematics Solutions for Exercise - S L Loney Solutions for Chapter: Identities and Trigonometrical Equations, Exercise 2: EXAMPLES XXI

Attempt the practice questions on Chapter 9: Identities and Trigonometrical Equations, Exercise 2: EXAMPLES XXI with hints and solutions to strengthen your understanding. Plane Trigonometry Part 1 solutions are prepared by Experienced Embibe Experts.

Questions from S L Loney Solutions for Chapter: Identities and Trigonometrical Equations, Exercise 2: EXAMPLES XXI with Hints & Solutions

HARD

JEE Main

IMPORTANT

Plane Trigonometry Part 1>Chapter 9 - Identities and Trigonometrical Equations>EXAMPLES XXI>Q 35

Solve the equation $4 \cot 2 θ = \cot^{2} θ - \tan^{2} θ$ .

MEDIUM

JEE Main

IMPORTANT

Plane Trigonometry Part 1>Chapter 9 - Identities and Trigonometrical Equations>EXAMPLES XXI>Q 36

Solve the equation $3 \tan (θ - 15 °) = \tan (θ + 15 °)$ .

HARD

JEE Main

IMPORTANT

Plane Trigonometry Part 1>Chapter 9 - Identities and Trigonometrical Equations>EXAMPLES XXI>Q 37

Solve the equation $\Rightarrow (1 - \tan^{2} θ) {(1 - \tan θ)}^{2} = 0$ .

HARD

JEE Main

IMPORTANT

Plane Trigonometry Part 1>Chapter 9 - Identities and Trigonometrical Equations>EXAMPLES XXI>Q 38

Solve the equation $\tan θ + \tan 2 θ + \sqrt{3} \tan θ \tan 2 θ = \sqrt{3}$ .

HARD

JEE Main

IMPORTANT

Plane Trigonometry Part 1>Chapter 9 - Identities and Trigonometrical Equations>EXAMPLES XXI>Q 39

Solve the equation $\sin 3 α = 4 \sin α \sin (x + α) \sin (x - α)$ . $(\sin α \neq 0)$

HARD

JEE Main

IMPORTANT

Plane Trigonometry Part 1>Chapter 9 - Identities and Trigonometrical Equations>EXAMPLES XXI>Q 40

Prove that the equation $x^{3} - 2 x + 1 = 0$ is satisfied by putting for $x$ , either of the values $\sqrt{2} \sin 45^{°}$ , $2 \sin 18^{°}$ and $2 \sin 234^{°}$ .

HARD

JEE Main

IMPORTANT

Plane Trigonometry Part 1>Chapter 9 - Identities and Trigonometrical Equations>EXAMPLES XXI>Q 41

If $\sin (π cosθ) = \cos (πsinθ)$ , prove that $\cos (θ \pm \frac{π}{4}) = \frac{1}{2 \sqrt{2}}$ .

HARD

JEE Main

IMPORTANT

Plane Trigonometry Part 1>Chapter 9 - Identities and Trigonometrical Equations>EXAMPLES XXI>Q 42

If $\sin (π \cot θ) = \cos (π \tan θ)$ , prove that either $cosec 2 θ$ or $\cot 2 θ$ is equal to $n + \frac{1}{4}$ , where $n$ is a positive or negative integer.

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