If sin 4 α a + cos 4 α b = 1 a + b , then prove that sin 6 α a 2 + cos 6 α b 2 = 1 ( a + b ) 2 . .

Given: sin 4 α a + cos 4 α b = 1 a + b Using cos 2 A = 1 - sin 2 A ⇒ sin 4 α a + cos 2 α 2 b = 1 a + b ⇒ sin 4 α a + 1 - sin 2 α 2 b = 1 a + b ⇒ sin 4 α a + 1 + sin 4 α - 2 sin 2 α b = 1 a + b ⇒ b s in 4 α + a 1 + sin 4 α - 2 sin 2 α a b = 1 a + b ⇒ a + b s in 4 α + a - a sin 2 α - a sin 2 α a b = 1 a + b ⇒ a + b 2 sin 4 α + a a + b - 2 a a + b sin 2 α = a b ⇒ a + b 2 sin 4 α - 2 a a + b sin 2 α + a 2 = 0 ⇒ a + b sin 2 α - a 2 = 0 ⇒ a + b sin 2 α - a = 0 ⇒ a + b sin 2 α = a ⇒ sin 2 α = a a + b ⇒ cos 2 α = 1 - a a + b 2 = a 2 + b 2 - a 2 a + b 2 = b a + b Consider L.H.S: sin 6 α a 2 + cos 6 α b 2 ⇒ sin 2 α 3 a 2 + cos 2 α 3 b 2 ⇒ a a + b 3 a 2 + b a + b 3 b 2 ⇒ a 3 a 2 a + b 3 + b 3 b 2 a + b 3 ⇒ a a + b 3 + b a + b 3 ⇒ a + b a + b 3 ⇒ 1 a + b 2 Hence, proved.

If 4 tanθ = 3 , evaluate 4 sinθ - cosθ + 1 4 sinθ + cosθ - 1 .

Given that, tan θ = 3 4 ∴ tan 2 θ = 9 16 We know that, sec 2 θ = 1 + tan 2 θ . ∴ sec 2 θ = 1 + 9 16 = 25 16 ∴ cos 2 θ = 16 25 ∴ cos θ = 4 5 We know that, sin 2 θ = 1 - cos 2 θ ∴ sin 2 θ = 1 - 16 25 = 9 25 ⇒ sin θ = 3 5 Now, 4 sin θ - cos θ + 1 4 sin θ + cos θ - 1 = 4 × 3 5 - 4 5 + 1 4 × 3 5 + 4 5 - 1 = 12 - 4 + 5 12 + 4 - 5 = 13 11 .

Prove that: sin A - 2 sin 3 A 2 cos 3 A - cos A = tan A .

To prove: sin A - 2 sin 3 A 2 cos 3 A - cos A = tan A Taking L . H . S . sin A - 2 sin 3 A 2 cos 3 A - cos A = sin A 1 - 2 sin 2 A cos A 2 cos 2 A - 1 = sin A cos A sin 2 A + cos 2 A - 2 sin 2 A 2 cos 2 A - sin 2 A + cos 2 A [Since sin 2 A + cos 2 A = 1 ] = sin A cos A sin 2 A + cos 2 A - 2 sin 2 A 2 cos 2 A - sin 2 A - cos 2 A = tan A cos 2 A - sin 2 A cos 2 A - sin 2 A = tan A = R . H . S . . [proved]

E M B I B E

Mathematics Crash Course (Based on Revised Syllabus-2023)>Chapter 8 - Introduction to Trigonometry>Exercise

Embibe Experts Solutions for Chapter: Introduction to Trigonometry, Exercise 1: Exercise

Author:Embibe Experts

Embibe Experts Mathematics Solutions for Exercise - Embibe Experts Solutions for Chapter: Introduction to Trigonometry, Exercise 1: Exercise

Attempt the free practice questions on Chapter 8: Introduction to Trigonometry, Exercise 1: Exercise with hints and solutions to strengthen your understanding. Mathematics Crash Course (Based on Revised Syllabus-2023) solutions are prepared by Experienced Embibe Experts.

Questions from Embibe Experts Solutions for Chapter: Introduction to Trigonometry, Exercise 1: Exercise with Hints & Solutions

EASY

10th CBSE

IMPORTANT

Mathematics Crash Course (Based on Revised Syllabus-2023)>Chapter 8 - Introduction to Trigonometry>Exercise>Q 6

If $\sin θ = cosec θ$ and $0 \leq θ \leq π$ , the value of $θ$ is

MEDIUM

10th CBSE

IMPORTANT

Mathematics Crash Course (Based on Revised Syllabus-2023)>Chapter 8 - Introduction to Trigonometry>Exercise>Q 16

A kite string is $48 m$ long. As the wind blew, the angle between the kite and the ground went from $27$ degrees to $54$ degrees. Determine the increase in the vertical height of the kite above the ground.

Question Image

HARD

10th CBSE

IMPORTANT

Mathematics Crash Course (Based on Revised Syllabus-2023)>Chapter 8 - Introduction to Trigonometry>Exercise>Q 82

If $\frac{\sin^{4} α}{a} + \frac{\cos^{4} α}{b} = \frac{1}{a + b}$ , then prove that $\frac{\sin^{6} α}{a^{2}} + \frac{\cos^{6} α}{b^{2}} = \frac{1}{(a + b)^{2}} .$ .