Trigonometry class 10 exercise 8.4 solutions

How to use Basic Trigonometric Identities Class 10 for solving Trigonometry problems with NCERT Ex 8.4 solutions How to use Basic trigonometric identities class 10 for solving trigonometry problems explained with example problems and NCERT Class 10 Ex 8.4 solved. Now we'll cover in this 4th and last part of Introduction to trigonometry class 10, • • Example Problems on Basic trigonometric identities Class 10 with quick solution. • Solution to Exercise 8.4 NCERT solutions Trigonometric Identities Class 10. You may move directly to any of the above sections by clicking its link and return by clicking on browser back button. First let us recall the concept of a mathematical identity, If an equation is true for all values of the variables involved, the equation is called an identity. In a trigonometric identity, • Each side of the identity equation will be an expression in trigonometric ratios, and, • For all values of the variables—the angles and the trigonometric ratio values—the two expressions in LHS and RHS will be equal. In earlier parts of this chapter, we have mentioned about the first and most basic trigonometric identity—$\sin^2 A+\cos^2 A=1$. Now we'll prove the truth of the identity and derive more basic trigonometric identities. Proof of $\sin^2 A+\cos^2 A=1$ The proof of this equation is very simple and you can do it yourself. We'll use the right-angled $\triangle ABC$ in the following diagram for explanation. First we'll use the geometric resource of Pythagoras ...

• NCERT Solutions Menu Toggle • NCERT Solutions for Class 12 • NCERT Solutions for Class 11 • NCERT Solutions for Class 10 • NCERT Solutions for Class 9 • NCERT Solutions for Class 8 • NCERT Solutions for Class 7 • NCERT Solutions for Class 6 • NCERT Books • TS Grewal Solutions • MCQ Questions Menu Toggle • NCERT MCQ Question 1. Express the trigonometric ratios sin A, sec A and tan A in terms of cot A. Solution: We know that Question 2. Write all the other trigonometric ratios of ∠A in terms of sec A. Solution: We know that Question 3. Evaluate: (i) \(\frac\) [Hint: Simplify LHS and RHS separately] Solution: L.H.S. Post navigation

Shaalaa.com has the CBSE Mathematics Class 10 Maths CBSE solutions in a manner that help students grasp basic concepts better and faster. The detailed, step-by-step solutions will help you understand the concepts better and clarify any confusion. Further, we at Shaalaa.com provide such solutions so students can prepare for written exams. NCERT textbook solutions can be a core help for self-study and provide excellent self-help guidance for students. Concepts covered in Using NCERT Class 10 Maths solutions Introduction to Trigonometry exercise by students is an easy way to prepare for the exams, as they involve solutions arranged chapter-wise and also page-wise. The questions involved in NCERT Solutions are essential questions that can be asked in the final exam. Maximum CBSE Class 10 Maths students prefer NCERT Textbook Solutions to score more in exams. Get the free view of Chapter 8, Introduction to Trigonometry Class 10 Maths additional questions for Mathematics Class 10 Maths CBSE, and you can use Shaalaa.com to keep it handy for your exam preparation.

Download NCERT Solutions for Class 10 Maths Chapter 8 Exercise 8.4 Introduction to trigo in NCERT Solutions for Class 10 Maths Chapter 8 Exercise 8.4 If you need solutions in Hindi, Click for Class 10 Maths Exercise 8.4 Solutions in Hindi Medium To get the solutions in English, Click for Practice Questions for Exams Trigonometric identities based questions Question 1: Prove that (sec A – tan A)² (1 + sin A) = 1- sin A. Question 2: Find the value of [sin A – sin B] / [cos A + cos B] + [cos A – cos B] / [sin A + sin B]. [Answer: 0] Question 3: If 0⁰ < θ < 90⁰, prove that √[cosec θ – 1]/[cosec θ + 1] + √[cosec θ + 1]/[cosec θ – 1] = 2 sec θ. Question 4: If 4 sin θ = 3, find the value of x if √[cosec² θ – cot² θ]/[sec² θ – 1] + 2cot θ = √7 / x + cos θ. [Answer: x = 4/3] Question 5: If tan θ + sin θ = m and tan θ – sin θ = n, show that m² – n² = 4 √mn. Question 6: If sin θ + cos θ = p and sec θ + cosec θ = q, show that q(p² – 1) = 2p. Question 7: If a cos θ + b sin θ = m and a sin θ – b cos θ = n, prove that a² + b² = m² + n². Question 8: If x = r sin A cos C, y = r sin A sin C and z = r cos A, prove that r² = x² + y² + z². Question 9: Without using trigonometric tables, evaluate the following: (cos² 25⁰ + cos² 65⁰) + cosec θ .sec (90⁰ – θ) – cot θ .tan (90⁰ – θ) [Answer: 2]. Question 10: Prove that (a) (i) cos (73⁰ + θ) = sin (17⁰ – θ) (ii) tan (30⁰ – θ) = cot(60⁰ + θ) (iii)cosec (53⁰ – θ) = sec (37⁰ + θ) (b) (i) tan (55⁰ – θ) – cot (35⁰ + θ) = 0 (ii) cosec (65⁰ + θ) – sec (25...

Chapter Name NCERT Maths Exemplar Solutions for Chapter8 Introduction to Trigonometry and its Equation Exercise 8.4 Book Name NCERT Exemplar for Class 10 Maths Other Exercises • Exercise 8.1 • Exercise 8.2 • Exercise 8.3 Related Study NCERT Solutions for Class 10 Maths Exercise 8.4 Solutions Long Answer Questions 1.If cosecθ + cotθ = p, then prove that cosθ = (p 2– 1)/(p 2 + 1) . Solution As, cosecθ + cot θ = p 2. Prove that √(sec 2θ + cosec 2θ) = tan θ + cot θ. Solution Taking, L.H.S = √(sec 2θ + cosec 2θ) Since, 3. The angle of elevation of the top of a tower from certain point is 30°. If the observer moves 20 metres towards the tower, the angle of elevation of the top increases by 15°. Find the height of the tower. Solution Let us take PR = h meter, the height of the tower. The distance between the observer and tower is QR = (20 +x) m, where QR = QS + SR = 20 + x ∠PQR = 30° ∠ PSR = θ In ∆PQR, tan 30° = h/(20 +x ) ⇒ 1/√3 = h/(20 + x) On cross multiplication, We get, 20 + x = √3h ⇒ x = √3h - 20 ...eq. (i) In∆PSR, tanθ = h/x As, angle of elevation increases by 15° when the observer moves 20 m towards the tower. So, θ = 30° + 15° = 45° Therefore, tan 45° = h/x ⇒ 1 = h/x ⇒ h = x Putting x = h in eq. (i), h = √3h - 20 ⇒√3h - h = 20 ⇒ h(√3 - 1) = 20 On rationalizing we get, h = 10(√3 + 1) The required height of the tower is 10(√3 + 1) meter. 4. If 1 + sin 2θ = 3sin θ cosθ, then prove that tan θ = 1 or 1/2. Solution We have, 1 + sin 2θ = 3sin θ cosθ on dividing L.H.S and R.H.S ...

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1. Express the trigonometric ratios sin A, sec A and tan A in terms of cot A. Solution: To convert the given trigonometric ratios in terms of cot functions, use trigonometric formulas We know that, cosec 2A – cot 2A = 1 cosec 2A = 1+ cot 2A Since cosec function is the inverse of sin function, it is written as 1/sin 2A = 1+ cot 2A Now, rearrange the terms, it becomes sin 2A = 1/(1+cot 2A) Now, take square roots on both sides, we get sin A = ±1/(√(1+cot 2A) The above equation defines the sin function in terms of cot function Now, to express sec function in terms of cot function, use this formula sin 2A = 1/ (1+cot 2A) Now, represent the sin function as cos function 1 – cos 2A = 1/ (1+cot 2A) Rearrange the terms, cos 2A = 1 – 1/(1+cot 2A) ⇒cos 2A =(1-1+cot 2A)/(1+cot 2A) Since sec function is the inverse of cos function, ⇒ 1/sec 2A = cot 2A/(1+cot 2A) Take the reciprocal and square roots on both sides, we get ⇒ sec A = ±√ (1+cot 2A)/cotA Now, to express tan function in terms of cot function tan A = sin A/cos A and cot A = cos A/sin A Since cot function is the inverse of tan function, it is rewritten as tan A = 1/cot A 2. Write all the other trigonometric ratios of ∠A in terms of sec A. Solution: Cos A function in terms of sec A: sec A = 1/cos A ⇒ cos A = 1/sec A sec A function in terms of sec A: cos 2A+ sin 2A = 1 Rearrange the terms sin 2A = 1 – cos 2A sin 2A = 1 – (1/sec 2A) sin 2A = (sec 2A-1)/sec 2A sin A = ± √(sec 2A-1)/sec A cosec A function in terms of sec A: sin A = 1...

Shaalaa.com has the CBSE Mathematics Class 10 Maths CBSE solutions in a manner that help students grasp basic concepts better and faster. The detailed, step-by-step solutions will help you understand the concepts better and clarify any confusion. Further, we at Shaalaa.com provide such solutions so students can prepare for written exams. NCERT textbook solutions can be a core help for self-study and provide excellent self-help guidance for students. Concepts covered in Using NCERT Class 10 Maths solutions Introduction to Trigonometry exercise by students is an easy way to prepare for the exams, as they involve solutions arranged chapter-wise and also page-wise. The questions involved in NCERT Solutions are essential questions that can be asked in the final exam. Maximum CBSE Class 10 Maths students prefer NCERT Textbook Solutions to score more in exams. Get the free view of Chapter 8, Introduction to Trigonometry Class 10 Maths additional questions for Mathematics Class 10 Maths CBSE, and you can use Shaalaa.com to keep it handy for your exam preparation.

1. Express the trigonometric ratios sin A, sec A and tan A in terms of cot A. Solution: To convert the given trigonometric ratios in terms of cot functions, use trigonometric formulas We know that, cosec 2A – cot 2A = 1 cosec 2A = 1+ cot 2A Since cosec function is the inverse of sin function, it is written as 1/sin 2A = 1+ cot 2A Now, rearrange the terms, it becomes sin 2A = 1/(1+cot 2A) Now, take square roots on both sides, we get sin A = ±1/(√(1+cot 2A) The above equation defines the sin function in terms of cot function Now, to express sec function in terms of cot function, use this formula sin 2A = 1/ (1+cot 2A) Now, represent the sin function as cos function 1 – cos 2A = 1/ (1+cot 2A) Rearrange the terms, cos 2A = 1 – 1/(1+cot 2A) ⇒cos 2A =(1-1+cot 2A)/(1+cot 2A) Since sec function is the inverse of cos function, ⇒ 1/sec 2A = cot 2A/(1+cot 2A) Take the reciprocal and square roots on both sides, we get ⇒ sec A = ±√ (1+cot 2A)/cotA Now, to express tan function in terms of cot function tan A = sin A/cos A and cot A = cos A/sin A Since cot function is the inverse of tan function, it is rewritten as tan A = 1/cot A 2. Write all the other trigonometric ratios of ∠A in terms of sec A. Solution: Cos A function in terms of sec A: sec A = 1/cos A ⇒ cos A = 1/sec A sec A function in terms of sec A: cos 2A+ sin 2A = 1 Rearrange the terms sin 2A = 1 – cos 2A sin 2A = 1 – (1/sec 2A) sin 2A = (sec 2A-1)/sec 2A sin A = ± √(sec 2A-1)/sec A cosec A function in terms of sec A: sin A = 1...

Download NCERT Solutions for Class 10 Maths Chapter 8 Exercise 8.4 Introduction to trigo in NCERT Solutions for Class 10 Maths Chapter 8 Exercise 8.4 If you need solutions in Hindi, Click for Class 10 Maths Exercise 8.4 Solutions in Hindi Medium To get the solutions in English, Click for Practice Questions for Exams Trigonometric identities based questions Question 1: Prove that (sec A – tan A)² (1 + sin A) = 1- sin A. Question 2: Find the value of [sin A – sin B] / [cos A + cos B] + [cos A – cos B] / [sin A + sin B]. [Answer: 0] Question 3: If 0⁰ < θ < 90⁰, prove that √[cosec θ – 1]/[cosec θ + 1] + √[cosec θ + 1]/[cosec θ – 1] = 2 sec θ. Question 4: If 4 sin θ = 3, find the value of x if √[cosec² θ – cot² θ]/[sec² θ – 1] + 2cot θ = √7 / x + cos θ. [Answer: x = 4/3] Question 5: If tan θ + sin θ = m and tan θ – sin θ = n, show that m² – n² = 4 √mn. Question 6: If sin θ + cos θ = p and sec θ + cosec θ = q, show that q(p² – 1) = 2p. Question 7: If a cos θ + b sin θ = m and a sin θ – b cos θ = n, prove that a² + b² = m² + n². Question 8: If x = r sin A cos C, y = r sin A sin C and z = r cos A, prove that r² = x² + y² + z². Question 9: Without using trigonometric tables, evaluate the following: (cos² 25⁰ + cos² 65⁰) + cosec θ .sec (90⁰ – θ) – cot θ .tan (90⁰ – θ) [Answer: 2]. Question 10: Prove that (a) (i) cos (73⁰ + θ) = sin (17⁰ – θ) (ii) tan (30⁰ – θ) = cot(60⁰ + θ) (iii)cosec (53⁰ – θ) = sec (37⁰ + θ) (b) (i) tan (55⁰ – θ) – cot (35⁰ + θ) = 0 (ii) cosec (65⁰ + θ) – sec (25...