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Selina solutions concise maths class 9 icse Solutions for Chapter 23 chapter 23 trigonometrical ratios of standard angles

Exercises in Chapter 23 chapter 23 trigonometrical ratios of standard angles Grade 9
Questions in Exercise 23(A)

Find the value of:

(i) sin 30° cos 30°

(ii) tan 30° tan 60°

(iii) cos2 60° + sin2 30°

(iv) cosec2 60° -tan2 30°

(v) sin2 30° + cos2 30° + cot2 45°

(vi) cos2 60° + sec2 30° + tan2 45° .

Find the value of :

(i) tan2 30° + tan2 45° + tan2 60°

(ii) Question 2 Image - Selina Solutions CONCISE Maths - Class 9 ICSE chapter Chapter 23- Trigonometrical Ratios of Standard Angles

(iii) 3 sin2 30° + 2 tan2 60° -5 cos2 45° .

Prove that:

(i) sin 60° cos 30° + cos 60° . sin 30° = 1

(ii) cos 30° . cos 60° -sin 30° . sin 60° = 0

(iii) cosec2 45° -cot2 45° = 1

(iv) cos2 30° -sin2 30° = cos 60° .

(v) Question 3 Image - Selina Solutions CONCISE Maths - Class 9 ICSE chapter Chapter 23- Trigonometrical Ratios of Standard Angles

(vi) 3 cosec2 60° -2 cot2 30° + sec2 45° = 0.

Prove that:

(i) sin 60° = 2 sin 30° cos 30° .

(ii) 4 (sin4 30° + cos4 60° )-3 (cos2 45° -sin2 90° ) = 2

(i) If sin x = cos x and x is acute, state the value of x.

(ii) If sec A = cosec A and 0° A 90° , state the value of A.

(iii) If tanθ = cotθ and 0° <= θ <= 90° , state the value of θ.

(iv) If sin x = cos y; write the relation between x and y, if both the angles x and y are acute.

(i) If sin x = cos y, then x + y = 45° ; write true of false.

(ii) secθ . Cotθ = cosecθ ; write true or false.

(iii) For any angle, state the value of : Sin2θ + cos2θ .

Questions in Exercise 23(B)

Given A = 60° and B = 30° , prove that:

(i) sin (A + B) = sin A cos B + cos A sin B

(ii) cos (A + B) = cos A cos B - sin A sin B

(iii) cos (A - B) = cos A cos B + sin A sin B

(iv) tan (A - B) = Question 1 Image - Selina Solutions CONCISE Maths - Class 9 ICSE chapter Chapter 23- Trigonometrical Ratios of Standard Angles

If A =30° , then prove that:

(i) sin 2A = 2sin A cos A = 𝟐𝒕𝒂𝒏𝑨/ 𝟏+𝒕𝒂𝒏𝟐𝑨

(ii) cos 2A = cos2A - sin2A= 𝟏−𝒕𝒂𝒏𝟐𝑨/ 𝟏+𝒕𝒂𝒏𝟐𝑨

(iii) 2 cos2 A - 1 = 1 - 2 sin2A

(iv) sin 3A = 3 sin A - 4 sin3A

If A = B = 45° , show that:

(i) sin (A - B) = sin A cos B - cos A sin B

(ii) cos (A + B) = cos A cos B - sin A sin B

If A = 30° ; show that:

(i) sin 3 A = 4 sin A sin (60° -A) sin (60° + A)

(ii) (sin A - cos A)2 = 1 - sin 2A

(iii) cos 2A = cos4 A - sin4 A

(iv) 𝟏−𝒄𝒐𝒔 𝟐𝑨/ 𝒔𝒊𝒏 𝟐𝑨 = 𝒕𝒂𝒏𝑨

(v) 𝟏+𝒔𝒊𝒏𝟐𝑨+𝒄𝒐𝒔𝟐𝑨/ 𝒔𝒊𝒏𝑨+𝒄𝒐𝒔 𝑨 = 𝟐𝒄𝒐𝒔 𝑨

(vi) 4 cos A cos (60° -A). cos (60° + A) = cos 3A

(vii) 𝑐𝑜𝑠3𝐴−𝑐𝑜𝑠3𝐴/ 𝑐𝑜𝑠𝐴

  • 𝑠𝑖𝑛3𝐴−𝑠𝑖𝑛3𝐴/ 𝑠𝑖𝑛𝐴 = 3
Questions in Exercise 23(C)

If 2 sin x° -1 = 0 and xo is an acute angle; find :

(i) sin x °

(ii) x °

(iii) cos x ° and tan xo .

If 4 sin2 𝜽 - 1= 0 and angle 𝜽 is less than 90° , find the value of 𝜽 and hence the value of cos2 𝜽 + tan2 𝜽.

If sin 3A = 1 and 0 <= A <= 90° , find:

(i) sin A

(ii) cos 2A

(iii) 𝑡𝑎𝑛2𝐴 − 1/ 𝑐𝑜𝑠2A

(i) If sin x + cos y = 1 and x = 30° , find the value of y.

(ii) If 3 tan A - 5 cos B= √𝟑 and B = 90° , find the value of A.

From the given figure, find:

(i) cos x °

(ii) x °

(iii) 𝟏 𝒕𝒂𝒏𝟐𝒙° − 𝟏 𝒔𝒊𝒏𝟐𝒙°

(iv) Use tan xo , to find the value of y.

Question 9 Image - Selina Solutions CONCISE Maths - Class 9 ICSE chapter Chapter 23- Trigonometrical Ratios of Standard Angles

Solve for x:

(i) 2 cos 3x - 1 = 0

(ii) Cos 𝒙/ 𝟑 − 𝟏 = 0

(iii) sin (x + 10o ) = ½

(iv) cos (2x - 30° ) = 0

(v) 2 cos (3x - 15° ) = 1

(vi) tan2 (x - 5 ° ) = 3

(vii) 3 tan2 (2x - 20° ) = 1

(viii) Cos ( 𝒙/ 𝟐 +𝟏𝟎) = √𝟑/ 𝟐

(ix) sin2 x + sin2 30° = 1

(x) cos2 30° + cos2 x = 1

(xi) cos2 30° + sin2 2x = 1

(xii) sin2 60° + cos2 (3x- 9 ° ) = 1

In triangle ABC, angle B = 90° , AB = y units, BC = √𝟑 units, AC = 2 units and angle A = xo , find:

(i) sin x°

(ii) x °

(iii) tan x°

(iv) use cos x ° to find the value of y.

Question 14 Image - Selina Solutions CONCISE Maths - Class 9 ICSE chapter Chapter 23- Trigonometrical Ratios of Standard Angles

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