Trigonometry proving questions for class 10

Solution : L.H.S : = (tan 2 θ - 1)/(tan 2 θ + 1) = [(sin 2 θ/cos 2 θ) - 1] / [(sin 2 θ/cos 2 θ) + 1] = [(sin 2θ - cos 2 θ) /cos 2 θ] / [(sin 2 θ + cos 2 θ) /cos 2 θ] = [(sin 2θ -cos 2θ)/cos 2θ] ⋅ [ cos 2 θ/1] = sin 2θ -cos 2θ = 1 - cos 2θ -cos 2θ = 1 - 2 cos 2 θ R.H.S Hence proved. Question 2 : Prove that Solution : [(1 + sin θ - cos θ)/(1 + sin θ + cos θ)] 2 =[(1 + sin θ) - cos θ] 2/[(1 + sin θ) + cos θ] 2 Expanding the numerator and denominator using the formula (a - b) 2 and (a + b) 2 =[(1+sinθ) 2+cos 2θ - 2(1+sin θ)cos θ] / [(1+sinθ) 2+cos 2θ + 2(1+sin θ)cos θ] Numerator : (1 + sinθ) 2+ cos 2θ - 2(1 + sin θ)cos θ = 1 + sin 2θ + 2sinθ + cos 2θ - 2cos θ - 2cos θ sin θ = 1 + (sin 2θ + cos 2θ) + 2sinθ - 2cos θ - 2cos θ sin θ = 2 + 2sinθ - 2cos θ - 2cos θ sin θ = 2(1 + sinθ - cos θ - cos θ sin θ) = 2 [(1 + sinθ) - cos θ(1 + sinθ)] = 2 [(1 + sinθ) (1 - cos θ)] -----(1) Denominator : [(1+sinθ) 2+cos 2θ + 2(1+sin θ)cos θ] = 1 + sin 2θ + 2sinθ + cos 2θ + 2cos θ + 2cos θ sin θ = 1 + (sin 2θ + cos 2θ) + 2sinθ + 2cos θ + 2cos θ sin θ = 2 + 2sinθ + 2cos θ + 2cos θ sin θ = 2(1 + sinθ + cos θ + cos θ sin θ) = 2 [(1 + sinθ) + cos θ(1 + sinθ)] = 2 [(1 + sinθ) (1 + cos θ)] -----(2) (1)/(2) = = (1 - cos θ)/ (1 + cos θ) Hence proved. Question 3 : If x sin 3 θ+ y cos 3 θ = sin θcos θand x sin θ= y cos θ, then prove that x 2 +y 2 = 1 . Solution : x sinθ (sin 2 θ) + (y cosθ)cos 2 θ = sinθcosθ x sinθ (sin 2 θ) + (x sinθ) cos 2 θ = sinθcosθ Since x sinθ = y cosθ x sinθ (sin 2 θ + cos 2 θ) = s...

Extra Questions for Class 10 Maths Introduction to Trigonometry with Answers Extra Questions for Class 10 Maths Chapter 8 Introduction to Trigonometry. According to new CBSE Exam Pattern, You can also download Class 10 Maths to help you to revise complete syllabus and score more marks in your examinations. Introduction to Trigonometry Class 10 Extra Questions Very Short Answer Type Question 1. From the given figure, find the value of x: Answer: In the given fig., only one side is known which is hypotenuse and side to be evaluated is BC which is perpendicular with reference to given angle ZA = 30°. ∴ sin 30° = \(\frac\) ___________ . Answer: 0 Filed Under:

Important Questions Class 10 Maths Chapter 9 Applications of Trigonometry Important questions for Class 10 Maths Chapter 9 Some Applications of Trigonometry are provided here for the board exams preparation. The questions are based on the new pattern of CBSE and are as per the revised syllabus. Students who are preparing CBSE 2022-2023 Maths exam are advised to practice these important questions of Some Applications of Trigonometry For Class 10 . Solving these questions will help students to score high marks in the questions asked from this chapter. Trigonometry has more applications in our daily existence, and hence, the chapter is crucial for the board exam and valuable in many other fields. Most of the questions from this chapter are also asked in the competitive exams such as JEE etc. Read more: • • • Below, we have provided the questions of Chapter 9 Applications of Trigonometry with the solutions. Students can als o find additional qu estions without solutions for their practice. Important Questions & Answers For Class 10 Maths Chapter 9 – Some Applications of Trigonometry Q.1: The shadow of a tower standing on level ground is found to be 40 m longer when the Sun’s altitude is 30° than when it is 60°. Find the height of the tower. Solution: Let AB be the tower and BC be the length of its shadow when the sun’s altitude (angle of elevation from the top of the tower to the tip of the shadow) is 60° and DB be the length of the shadow when the angle of elevation is...

C = tan ⁡ ( 40 ° ) csc ⁡ ( 40 ° ) csc ⁡ ( 50 ° ) sec ⁡ ( 60 ° ) D = sin ⁡ 2 ( 15 ° ) + sin ⁡ 2 ( 75 ° ) sin ⁡ ( 30 ° ) sin ⁡ ( 60 ° ) sin ⁡ ( 90 ° ) \begin C D ​ = csc ( 5 0 ° ) sec ( 6 0 ° ) tan ( 4 0 ° ) csc ( 4 0 ° ) ​ = sin ( 3 0 ° ) sin ( 6 0 ° ) sin ( 9 0 ° ) sin 2 ( 1 5 ° ) + sin 2 ( 7 5 ° ) ​ ​

Solution : L.H.S : = (tan 2 θ - 1)/(tan 2 θ + 1) = [(sin 2 θ/cos 2 θ) - 1] / [(sin 2 θ/cos 2 θ) + 1] = [(sin 2θ - cos 2 θ) /cos 2 θ] / [(sin 2 θ + cos 2 θ) /cos 2 θ] = [(sin 2θ -cos 2θ)/cos 2θ] ⋅ [ cos 2 θ/1] = sin 2θ -cos 2θ = 1 - cos 2θ -cos 2θ = 1 - 2 cos 2 θ R.H.S Hence proved. Question 2 : Prove that Solution : [(1 + sin θ - cos θ)/(1 + sin θ + cos θ)] 2 =[(1 + sin θ) - cos θ] 2/[(1 + sin θ) + cos θ] 2 Expanding the numerator and denominator using the formula (a - b) 2 and (a + b) 2 =[(1+sinθ) 2+cos 2θ - 2(1+sin θ)cos θ] / [(1+sinθ) 2+cos 2θ + 2(1+sin θ)cos θ] Numerator : (1 + sinθ) 2+ cos 2θ - 2(1 + sin θ)cos θ = 1 + sin 2θ + 2sinθ + cos 2θ - 2cos θ - 2cos θ sin θ = 1 + (sin 2θ + cos 2θ) + 2sinθ - 2cos θ - 2cos θ sin θ = 2 + 2sinθ - 2cos θ - 2cos θ sin θ = 2(1 + sinθ - cos θ - cos θ sin θ) = 2 [(1 + sinθ) - cos θ(1 + sinθ)] = 2 [(1 + sinθ) (1 - cos θ)] -----(1) Denominator : [(1+sinθ) 2+cos 2θ + 2(1+sin θ)cos θ] = 1 + sin 2θ + 2sinθ + cos 2θ + 2cos θ + 2cos θ sin θ = 1 + (sin 2θ + cos 2θ) + 2sinθ + 2cos θ + 2cos θ sin θ = 2 + 2sinθ + 2cos θ + 2cos θ sin θ = 2(1 + sinθ + cos θ + cos θ sin θ) = 2 [(1 + sinθ) + cos θ(1 + sinθ)] = 2 [(1 + sinθ) (1 + cos θ)] -----(2) (1)/(2) = = (1 - cos θ)/ (1 + cos θ) Hence proved. Question 3 : If x sin 3 θ+ y cos 3 θ = sin θcos θand x sin θ= y cos θ, then prove that x 2 +y 2 = 1 . Solution : x sinθ (sin 2 θ) + (y cosθ)cos 2 θ = sinθcosθ x sinθ (sin 2 θ) + (x sinθ) cos 2 θ = sinθcosθ Since x sinθ = y cosθ x sinθ (sin 2 θ + cos 2 θ) = s...

Extra Questions for Class 10 Maths Introduction to Trigonometry with Answers Extra Questions for Class 10 Maths Chapter 8 Introduction to Trigonometry. According to new CBSE Exam Pattern, You can also download Class 10 Maths to help you to revise complete syllabus and score more marks in your examinations. Introduction to Trigonometry Class 10 Extra Questions Very Short Answer Type Question 1. From the given figure, find the value of x: Answer: In the given fig., only one side is known which is hypotenuse and side to be evaluated is BC which is perpendicular with reference to given angle ZA = 30°. ∴ sin 30° = \(\frac\) ___________ . Answer: 0 Filed Under:

C = tan ⁡ ( 40 ° ) csc ⁡ ( 40 ° ) csc ⁡ ( 50 ° ) sec ⁡ ( 60 ° ) D = sin ⁡ 2 ( 15 ° ) + sin ⁡ 2 ( 75 ° ) sin ⁡ ( 30 ° ) sin ⁡ ( 60 ° ) sin ⁡ ( 90 ° ) \begin C D ​ = csc ( 5 0 ° ) sec ( 6 0 ° ) tan ( 4 0 ° ) csc ( 4 0 ° ) ​ = sin ( 3 0 ° ) sin ( 6 0 ° ) sin ( 9 0 ° ) sin 2 ( 1 5 ° ) + sin 2 ( 7 5 ° ) ​ ​

Important Questions Class 10 Maths Chapter 9 Applications of Trigonometry Important questions for Class 10 Maths Chapter 9 Some Applications of Trigonometry are provided here for the board exams preparation. The questions are based on the new pattern of CBSE and are as per the revised syllabus. Students who are preparing CBSE 2022-2023 Maths exam are advised to practice these important questions of Some Applications of Trigonometry For Class 10 . Solving these questions will help students to score high marks in the questions asked from this chapter. Trigonometry has more applications in our daily existence, and hence, the chapter is crucial for the board exam and valuable in many other fields. Most of the questions from this chapter are also asked in the competitive exams such as JEE etc. Read more: • • • Below, we have provided the questions of Chapter 9 Applications of Trigonometry with the solutions. Students can als o find additional qu estions without solutions for their practice. Important Questions & Answers For Class 10 Maths Chapter 9 – Some Applications of Trigonometry Q.1: The shadow of a tower standing on level ground is found to be 40 m longer when the Sun’s altitude is 30° than when it is 60°. Find the height of the tower. Solution: Let AB be the tower and BC be the length of its shadow when the sun’s altitude (angle of elevation from the top of the tower to the tip of the shadow) is 60° and DB be the length of the shadow when the angle of elevation is...