Selina Solutions Concise Maths Class 10 Chapter 21 Trigonometrical Identities Exercise 21(C)

Selina Solutions Class 10 Maths Chapter 21 Trigonometrical Identities

The science related to the measurement of triangles is called trigonometry. Trigonometric ratios and its relations are used to prove trigonometric identities. In addition, trigonometric ratios of complementary angles and the use of trigonometric tables are other topics covered in this chapter. Since this chapter lays the foundation for high grade mathematics, students should gain a strong grip on this chapter. For this purpose, www.mathspdfsolution.co  has created Selina Solutions for Class 10 Mathematics prepared by expert faculty with vast academic experience. It also improves students' problem-solving skills, which are important from the exam point of view. Selina Solutions for Class 10 Mathematics Chapter 21 Trigonometric Identification PDFs are available practice-wise in the link below.

Exercise 21(A) Solutions

Exercise 21(B) Solutions

Exercise 21(C) Solutions

Exercise 21(D) Solutions

Exercise 21(E) Solutions

Selina Solutions Concise Maths Class 10 Chapter 21 Trigonometrical Identities Exercise 21(C)

Question 1. Show that:

(i) tan 10o tan 15o tan 75o tan 80o = 1

Solution:

Taking, tan 10o tan 15o tan 75o tan 80o

= tan (90o – 80o) tan (90o – 75o) tan 75o tan 80o

= cot 80o cot 75o tan 75o tan 80o

= 1 [Since, tan θ x cot θ = 1]

(ii) sin 42o sec 48o + cos 42o cosec 48o = 2

Solution:

Taking, sin 42o sec 48o + cos 42o cosec 48o

= sin 42o sec (90o – 42o) + cos 42o cosec (90o – 42o)

= sin 42o cosec 42o + cos 42o sec 42o

= 1 + 1 [Since, sin θ x cosec θ = 1 and cos θ x sec θ = 1]

= 2

(iii) sin 26o/ sec 64o + cos 26o/ cosec 64o = 1

Solution:

Taking,

Question 2. Express each of the following in terms of angles between 0°and 45°:

(i) sin 59°+ tan 63°

(ii) cosec 68°+ cot 72°

(iii) cos 74°+ sec 67°

Solution:

(i) sin 59°+ tan 63°

= sin (90 – 31)°+ tan (90 – 27)°

= cos 31°+ cot 27°

(ii) cosec 68°+ cot 72°

= cosec (90 – 22)°+ cot (90 – 18)°

= sec 22°+ tan 18°

(iii) cos 74°+ sec 67°

= cos (90 – 16)°+ sec (90 – 23)°

= sin 16°+ cosec 23°

Question 3. Show that:

Solution:

= sin A cos A – sin3 A cos A – cos3 A sin A

= sin A cos A – sin A cos A (sin2 A + cos2 A)

= sin A cos A – sin A cos A (1) [Since, sin2 A + cos2 A = 1]

= 0

Question 4. For triangle ABC, show that:

(i) sin (A + B)/ 2 = cos C/2

(ii) tan (B + C)/ 2 = cot A/2

Solution:

We know that, in triangle ABC

∠A + ∠B + ∠C = 180o [Angle sum property of a triangle]

(i) Now,

(∠A + ∠B)/ 2 = 90o – ∠C/ 2

So,

sin ((A + B)/ 2) = sin (90o – C/ 2)

= cos C/ 2

(ii) And,

(∠C + ∠B)/ 2 = 90o – ∠A/ 2

So,

tan ((B + C)/ 2) = tan (90o – A/ 2)

= cot A/ 2

Question 5. Evaluate:

Solution:

(i)

(ii) 3 cos 80o cosec 10o + 2 cos 59o cosec 31o

= 3 cos (90 – 10)o cosec 10o + 2 cos (90 – 31)o cosec 31o

= 3 sin 10o cosec 10o + 2 sin 31o cosec 31o

= 3 + 2 = 5

(iii) sin 80o/ cos 10o + sin 59o sec 31o

= sin (90 – 10)o/ cos 10o + sin (90 – 31)o sec 31o

= cos 10o/ cos 10o + cos 31o sec 31o

= 1 + 1 = 2

(iv) tan (55o – A) – cot (35o + A)

= tan [90o – (35o + A)] – cot (35o + A)

= cot (35o + A)] – cot (35o + A)

= 0

(v) cosec (65o + A) – sec (25o – A)

= cosec [90o – (25o – A)] – sec (25o – A)

= sec (25o – A) – sec (25o – A)

= 0

(vi)

(vii)

= 1 – 2 = -1

(viii)

(ix) 14 sin 30o + 6 cos 60o – 5 tan 45o

= 14 (1/2) + 6 (1/2) – 5(1)

= 7 + 3 – 5

= 5

Question 6. A triangle ABC is right angled at B; find the value of (sec A. cosec C – tan A. cot C)/ sin B

Solution:

As, ABC is a right angled triangle right angled at B

So, A + C = 90o

(sec A. cosec C – tan A. cot C)/ sin B

= (sec (90o – C). cosec C – tan (90o – C). cot C)/ sin 90o

= (cosec C. cosec C – cot C. cot C)/ 1 = cosec2 C – cot2 C

= 1 [Since, cosec2 C – cot2 C = 1]

Exercise 21(A) Solutions

Exercise 21(B) Solutions

Exercise 21(C) Solutions

Exercise 21(D) Solutions

Exercise 21(E) Solutions

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