WBBSE Solutions For Class 10 Maths Chapter 23 Trigonometric Ratios And Trigonometric Identities Exercise 23.2

Maths WBBSE Class 10 Solutions Chapter 23 Trigonometric Ratios And Trigonometric Identities Exercise 23.2

WBBSE Solutions For Class 10 Maths Chapter 23 Trigonometric Ratios And Trigonometric Identities Exercise 23.2

Question 1. In the window of our house, there is a ladder at an angle of 60° with the ground. If the ladder is 2√3 m long, then let us write by calculating the height of our window above the ground.

Solution:

Given

In the window of our house, there is a ladder at an angle of 60° with the ground. If the ladder is 2√3 m long

Let AB be the height of the window & AC be the ladder = 2√3 m.

∠ACB = 60°

∴ sin60° = AB/AC

 √3/2 = AB/2√3

or, 2 AB=2√3.√3 = 6

∴ AB 6/2 = 3 m.

WBBSE Solutions For Class 10 Maths Chapter 23 Trigonometric Ratios And Trigonometric Identities Exercise 23.2..

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WBBSE Solutions For Class 10 Maths Chapter 23 Trigonometric Ratios And Trigonometric Identities 1

 

Question 2. ABC is a right-angled triangle with its B being 1 right angle. If AB = 8√3 cm and BC = 8 cm, then let us write by calculating, the values of ∠ACB and <BAC.

Solution:

Given

ABC is a right-angled triangle with its B being 1 right angle. If AB = 8√3 cm and BC = 8 cm

tan ∠ACB = AB/BC

= 83 cm / 8cm

= √3

= tan 60°

∴ ∠ACB = 60°

∴ ∠BAC 180°= (90° +60°) 

= 180° – 150° 

= 30°

 

WBBSE Solutions For Class 10 Maths Chapter 23 Trigonometric Ratios And Trigonometric Identities 2

Question 3. In a right-angled triangle ABC, ∠B = 90°,∠A = 30°, and AC = 20 cm. Let us determine the lengths of the two sides BC and AB.

Solution:

In a right-angled triangle ABC, ∠B = 90°,∠A = 30°, and AC = 20 cm.

A 30°, B = 90°

∴ ∠C = 60°

sin60° = AB/AC

Or, √3/2 = AB/20

∴ AB = 20√3

= 10√3 cm.

Again, cos60° = BC/AC

1/2= BC/20

∴ BC = 20/2 

= 10cm.

WBBSE Solutions For Class 10 Maths Chapter 23 Trigonometric Ratios And Trigonometric Identities 3

 

Question 4. In a right-angled triangle PQR, ZQ = 90°, ZR = 45°; if PR = 3√2, then let us find out the lengths of the two sides PQ and QR.

Solution:

Given

In a right-angled triangle PQR, ZQ = 90°, ZR = 45°; if PR = 3√2,

In ΔPQR, Q= 90°, R = 45°

∴ ∠P 90° – 45° = 45°

sin45° = PQ/PR

1/√2 = PQ/3√2

∴ PQ = 3√2/√2 

= 3 units.

cos45° = QR/PR

1/√2 = QR/ 3√2

√2QR = 3√2

∴ QR = 3√2/√2

= 3 units.

 

WBBSE Solutions For Class 10 Maths Chapter 23 Trigonometric Ratios And Trigonometric Identities 4

Question 5. Let us determine the values of 

1. sin? 45° cosec² 60° + sec² 30° 

Solution: sin² 45°- cosec² 60° + sec² 30°

= (1/√2)² – (2/√3)² + (2/√3)²

= 1/2 – 4/3 + 4/3 

= 1/2

sin² 45°- cosec² 60° + sec² 30° = 1/2

 

2. sec² 45°- cot² 45°- sin² 30° – sin² 60° 

Solution: sec² 45°-cot² 45°- sin² 30° sin² 60°

= (√2)² -(1)² – (1/2)² – (√3/2)²

=2-1-1/4 -3/4

= 8-4-1-3 / 4

= 8-8/4

= 0/4

= 0

sec² 45°-cot² 45°- sin² 30° sin² 60° = 0

3. 3tan² 45°- sin² 60° – 1/3 cot² 30° – 1/8 sec² 45°

Solution: 3tan² 45°- sin² 60° – 1/3 cot² 30° – 1/8 sec² 45°

= 3(1)²- (√3 /2)² – 1/8(√2)²

= 3×1 3/4 – 1/3 x 3 – 1/8 x 2

= 3 – 3/4 – 1 – 1/4

= 12-3-4-1 / 4

= 12-8 / 4

=4/4

= 1.

3tan² 45°- sin² 60° – 1/3 cot² 30° – 1/8 sec² 45° = 1.


4. 4/
3 cot² 30° + 3 sin² 60°-2cosec² 60°-

Solution: 4/3 cot² 30° + 3 sin² 60° – 2cosec² 60°- tan² 30°

= 4/3(√3)²  + 3 (√3/2)² – 3/4(1/√3)

= 4/3 x 3 + 3/4 -2 4/3 x 3/4 x 1/3

= 4 + 9/4 – 8/3 – 1/4

= 48+27-32-3 / 12

= 75 – 35 / 12

= 40/12 

= 10/3

 = 3 1/3

4/3 cot² 30° + 3 sin² 60° – 2cosec² 60°- tan² 30°  = 3 1/3

WBBSE Solutions Guide Class 10

5. ⅓ cos 30°/ ½ sin45° + tan 60°/cos30°

Solution :  ⅓ cos 30°/ ½ sin45° + tan 60°/cos30°

= \(\frac{\frac{1}{3} \times \frac{\sqrt{3}}{2}}{\frac{1}{2} \times \frac{1}{\sqrt{2}}}+\frac{\sqrt{3}}{\frac{\sqrt{3}}{2}}=\frac{\frac{\sqrt{3}}{6}}{\frac{1}{2 \sqrt{3}}}+\frac{\frac{\sqrt{3}}{6}}{\frac{\sqrt{3}}{2}}=\frac{\sqrt{3}}{6} \times 2 \sqrt{2}+\sqrt{3} \times \frac{2}{\sqrt{3}}\)

= \(\frac{\sqrt{6}}{3}+2=\frac{\sqrt{6}+6}{3}=\frac{\sqrt{6}+6}{3}\)

6. cot²30°-2cos² 60°- 3/4 sec² 45°-4sin² 30°

Solution: cot²30°-2cos² 60°- 3/4 sec² 45°-4sin² 30°

= (√3)² -2(1/2)² – 3/4 (√2)²- 4(1/2)²

=3-2 x 1/4 – 3/4 x 2-4 x 1/4

= 3- 1/2 – 3/2 – 1

= 6-1-3-2 / 2

= 6-6 / 2

= 0.

cot²30°-2cos² 60°- 3/4 sec² 45°-4sin² 30° = 0.

7. sec² 60°-cot² 30°- 2tan30°cosec60° / 1+tan²30°

Solution: sec² 60° cot² 30°- 2tan30°cosec60° / 1+tan²30°

= \((2)^2-(\sqrt{3})^2-\frac{2 \times \frac{1}{\sqrt{9}} \times \frac{2}{\sqrt{3}}}{1+\left(\frac{1}{\sqrt{3}}\right)^2}\)

= \(4-3-\frac{\frac{4}{\sqrt{3}}}{1+\frac{1}{3}}=4-3-\frac{\frac{4}{3}}{\frac{4}{3}}=4-3-1=4-4=0\)

8. tan60° – tan30° / 1+tan 60° tan 30° + cos60° cos30° + sin60°sin30°

Solution: tan60° – tan30° / 1+tan 60° tan 30° + cos60° cos30° + sin60°sin30°

= \(\frac{\sqrt{3}-\frac{1}{\sqrt{3}}}{1+\sqrt{3} \cdot \frac{1}{\sqrt{3}}}+\frac{1}{2} \cdot \frac{\sqrt{3}}{2}+\frac{\sqrt{3}}{2} \cdot \frac{1}{2}\)

= \(\frac{\frac{3-1}{\sqrt{3}}}{1+1}+\frac{\sqrt{3}}{4}+\frac{\sqrt{3}}{4}=\frac{2}{\sqrt{3}} \times \frac{1}{2}+\frac{2 \sqrt{3}}{4}=\frac{1}{\sqrt{3}}+\frac{\sqrt{3}}{2}=\frac{2+3}{2 \sqrt{3}}=\frac{5}{2 \sqrt{3}}\)

 

9. 1-sin² 30°/1+sin² 30° x cos²60+ cos² 30°/cos ec²90°-cot² 90° ÷ (sin 60° tan30°)

Solution : 1-sin² 30°/1+sin² 30° x cos²60+ cos² 30°/cosec²90°-cot² 90° ÷ (sin 60° tan30°)

= \(\frac{1-\left(\frac{1}{2}\right)^2}{1+\left(\frac{1}{\sqrt{2}}\right)^2} \times \frac{\left(\frac{1}{2}\right)^2+\left(\frac{\sqrt{3}}{2}\right)^2}{1-0} \div\left(\frac{\sqrt{3}}{2} \cdot \frac{1}{\sqrt{3}}\right)\)

= \(\frac{1-\frac{1}{4}}{1+\frac{1}{2}} \times \frac{\frac{1}{4}+\frac{3}{4}}{1} \div\left(\frac{1}{2}\right)=\frac{3 / 4}{3 / 2} \times \frac{\frac{4}{4}}{1} \times 2\)

 

= 3/4 x 2/3 x 1 x 2

= 1/2 x 2

= 1

Question 6.

1. sin² 45° + cos² 45° = 1

Solution: sin² 45° + cos² 45° = 1

L.H.S. sin²45+ cos²45 = (1/√2)² + (1/√2)²

= 1/2 + 1/2 

= 1


2. cos60° =
cos²30° – sin²30°

Solution: cos60° = cos²30° – sin²30°

L.H.S. = cos60° = 1/2 

R.H.S = cos²30° – sin²30° 

= (√3/2)² – (1/2)²

= 3/4 – 1/4 

= 2/4 

= 1/2 

∴ L.H.S = R.H.S Proved.


3. 2 tan30°/1-tan²30° = √3

Solution:  2 tan30°/1-tan²30° = √3

L.H.S = \(\frac{2 \tan 30^{\circ}}{1-\tan ^2 30^{\circ}}=\frac{21 / \sqrt{3}}{1-\left(\frac{1}{\sqrt{3}}\right)^2}=\frac{2 / \sqrt{3}}{1-\frac{1}{3}}=\frac{2 / \sqrt{3}}{2 / 3}=\frac{2}{\sqrt{3}} \times \frac{3}{2}\)

 

= √3

∴ L.H.S = R.H.S Proved.


4.√(1+ cos30° / 1-cos30°)  = sec 60° + tan 60°

Solution:  √(1+ cos30° / 1-cos30°)  = sec 60° + tan 60°

L.H.S = \(\sqrt{\frac{1+\cos 30^{\circ}}{1-\cos 30^{\circ}}}=\sqrt{\frac{1+\frac{\sqrt{3}}{2}}{1-\frac{\sqrt{3}}{2}}}=\sqrt{\frac{2+\sqrt{3}}{2-\sqrt{3}}}=\sqrt{\frac{(2+\sqrt{3})^2}{(2-\sqrt{3})(2+\sqrt{3})}}\)

= \(\frac{2+\sqrt{3}}{4-3}=2+\sqrt{3}\)

 

= sec60° + tan60°

= 2+ √3

∴ L.H.S = R.H.S Proved.


5. 2 tan² 30°/1-tan²30° + sec² 45°- cot² 45° = sec 60°

Solution : 2 tan² 30°/1-tan²30° + sec² 45°- cot² 45° = sec 60°

L.H.S. = 2 tan² 30°/1-tan²30° + sec² 45°- cot² 45°

= \(\frac{2 \cdot\left(\frac{1}{\sqrt{3}}\right)^2}{2-\left(\frac{1}{\sqrt{3}}\right)^2}+(\sqrt{2})^2-(1 .)^2\)

= \(\frac{2 \cdot \frac{1}{3}}{1-\frac{1}{3}}-1-2-1\)

= \(\frac{2 / 3}{2 / 3}+1=\frac{2}{3} \quad \frac{3}{2}+1=1+1=2\)

 

R.H.S = sec 60° = 2

∴ L.H.S = R.H.S Proved.

6. tan² Π/4 sin Π/3  tanΠ/6 tan² Π/3 = 1 1/2 

Solution: tan²Π/4 sin Π/3  tanΠ/6 tan² Π/3 = 1 1/2

L.H.S. = tan2 Π/4 sin Π/3  tanΠ/6 tan2 Π/

= tan²45°sin60°tan30°tan260°

= (1)2. (√3/2) . 1/√3 . (√3)2

= 1/2 .3

= 3/2

= 1 1/2 = R.H.S.

.. L.H.S = R.H.S Proved. 

7.sinΠ/3 tanΠ/6 + + sinΠ/2 cosΠ/3 = 2sin²Π/4

Solution: sinΠ/3 tanΠ/6 + + sinΠ/2 cosΠ/3 = 2sin²Π/4

L.H.S. sin = Π/3 tanΠ/6 + sinΠ/2– cosΠ/3 

= sin60°tan30° + sin90°cos60°

= √3/2 . 1/√3 + 1 . 1/2

= 1/2 + 1/2  

R.H.S= 2sin² = 2. 

sin²45° = 2x (1/√2)²

= 2 x 1/2

= 1

∴ L.H.S = R.H.S Proved.

Question 7.

1. If x sin45° cos45° tan60° = tan² 45°- cos60°, then let us determine the value of x.

Solution: x sin45° cos45° tan60° = tan² 45° – cos60°

or, x . 1/√2 . 1/√2 .√3

= (1)² – 1/2 

= x√3/2

= 1/2

or, √3x = 1

2. If x sin 60° cos² 30° =  tan² 45° sec60° / cosec60° then let us determine the value of x.

Solution: x sin 60° cos² 30° = tan² 45° sec60° / cosec60°

or, \(x \frac{\sqrt{3}}{2}\left(\frac{\sqrt{3}}{2}\right)^2=\frac{(1)^2 \cdot 2}{2 / \sqrt{3}}\)

or, \(x \cdot \frac{\sqrt{3}}{2} \cdot \frac{3}{4}=\frac{2}{2 / \sqrt{3}}\)

or, \(x \frac{3 \sqrt{3}}{8}=\frac{2 \sqrt{3}}{2}\)

or, \(x=\frac{2 \sqrt{3}}{2} \quad \frac{8}{3 \sqrt{3}}=\frac{8}{3}\)

3. If x²= sin²30° + 4cot² 45°- sec² 60°, then let us determine the value of x. 

Solution: x² = sin²30° + 4cot² 45°- sec² 60°

or, x² = (1/2)² +4. (1)²- (1)²

or, x² = 1/4 + 4 – 4

x² = 1/4 

x = 1/2 

Question 8. If x tan 30° + y cot 60° = 0 and 2x – y tan 45° = 1, then let us write by calculating, the values of x and y.

Solution: x tan 30°+ y cot 60° = 0

or, x . 1/√3 + Y . 1/ √3 = 0

or, x/√3 + y/√3 = 0

or, or, x + y = 0——–(1)

Again 2x – y. 1 = 1

or, 2x – y = 1 ——-(2)

∴ x + y = 0

2x – y = 1

Adding 3x = 1

∴ X = 1/3

∴ Y = -X            ∴ Y = – 1/3


Question 9. If A B 45°, then let us justify:

1. sin (A + B) = sin A cos B + cos A sin B

Solution: L.H.S= sin (A + B)

=sin(45° +45°) 

= sin90° 

= 0

R.H.S sinA cosB + cosA sinB

= sin45° cos45° + cos45° sin45°

= 1/√2 x 1/√2 x 1/√2  x 1/√2

= 1/2 + 1/2

= 1

∴ L.H.S = R.H.S


2. cos (A + B) = cos A cos B-sin A sin B

Solution: L.H.S= cos (A + B)

= cos(45° + 45°) = cos90° = 0

R.H.S. cosA cosB – sinA sinB

= cos45 cos45 – sinA sinB

= 1/√2 . 1/√2 – 1/√2 . 1/√2

= 1/2 – 1/2

= 0

∴ L.H.S = R.H.S

Question 10.

1. In an equilateral triangle ABC, BD is a median. Let us prove that, tan ∠ABD = cot <BAD.

Solution: L.H.S. tan ∠ABD = P/B

= AD/BD

R.H.S. cot <BAD = B/P

= AD/BD

∴ L.H.S. = R.H.S.

WBBSE Solutions For Class 10 Maths Chapter 23 Trigonometric Ratios And Trigonometric Identities 13

2. In an isosceles triangle ABC, AB = AC and <BAC 90°; the bisector of ∠BAC intersects the side BC at point D.

Solution: L.H.S = sec ∠ACD/sin <CAD

= AC/CD/CD/AC

= AC/CD x  AC/CD

= AC²/CD²

R.H.S. cosec² ∠CAD

=(AC/CD)² = AC²/CD²

∴ L.H.S = R.H.S Proved.

WBBSE Solutions For Class 10 Maths Chapter 23 Trigonometric Ratios And Trigonometric Identities 14

11. Let us determine the value / values of 0(0°≤ 0≤ 90°), for which 2cos²θ-3cosθ +10 wil be true.

Solution: 2cos²θ-3cosθ + 1 = 0

or, 2cos²θ – 2cos θ – 1cosθ + 1 = 0

or, 2cosθ (cosθ-1) – 1 (cosθ-1)=0

Either cosθ1=0            ∴ cosθ = 1 = cos0°

∴ 0 = 0°

Or, 2cosθ – 1 = 0

∴ 2cosθ = 1

∴ cosθ = 1/2 = cos60°

∴ 0 = 60°

∴ Values of = 0° or 60°. 

Application 23. For 0° ≤ 0 ≤ 90°, let us write with reason whether sin = √3/2 and cos = 1/3 are possible or not.

Solution: sin θ = √3/2,  cos θ = 1/3

sin²θ + cos²θ = (√3/2)² + (1/3)²

= 3/4 + 1/9

= 27+4 / 36

= 31/36

∴ It is impossible as

we know sin² θ + cos² θ = 1

Application 26. I express cote and cosec e in terms of cos e. 

Solution: We know, sin² θ+ cos² θ= 1

∴ sin²θ= 1 – cos²θ

or, sin θ = √1-cos²θ

∴ cos θ = cos θ/sin θ = cos θ/√1-cos²θ

Again, sin θ= √1-cos²θ    ∴cosec θ = 1/sin θ

= 1/√1-cos²θ

Application 28. If tan θ = 3/4, then let us write by calculating, the value of (sinθ + cos θ) 

Solution: sec²θ = 1 + tan²θ = 1 +(4/3)²

= 1 + 16/9

= 9+16 / 9

= 25/9

= (5/3)²

Secθ= 5/3            ∴ cos θ = 3/5

sinθ = √1-cos²θ

= √1- 9 /25

= √25-9 / 25

= √16/25

= 4/5

∴ sinθ + cos θ = 3/5 + 4/5 

= 3+4 / 5

= 7/5

Application 34. If  5cote+cosec 0/5cot 0-cosec 0 = 7/3, then let us determine the value of cose.

Solution: 5cotθ+cosec θ/5cot θ-cosec θ = 7/3

= 35cotθ – 7cosecθ= 15cotθ + 3cosecθ

35cotθ – 15cotθ = 3cosecθ+7cosecθ

20cotθ = 10cosecθ

cot θ/cosecθ 10/20

cos θ/sinθ x sinθ = 1/2

∴ cosv =1/2


Application 35. From the two relations 2x= 3sine and 5y = 3cose, by eliminating, let us write the relation between x and y. 

Solution: 3sinθ = 2x   And 3cosθ = 5y

∴ sinθ=2x/3       ∴cosθ =5y/3

We know, sin² + cos² = 1

or, (2x/3)² + (5y/3)² = 1

or, 4x²/9 + 25y²/9 = 1

or, 4x²+ 25y² = 9.

Application 36. Let us eliminate from the two relations x = a sec e, y = b tan e. 

Solution: a secθ = x     And      b tan θ = y

∴ secθ= x/a                  ∴ tanθ = y/b

We know, sec²θ-tan²θ = 1

or, (x/a)² – (y/b)² = 1               ∴ x²/a² – y²/b² = 1

Application 37. If cosθ + secθ = 2, then let us determine the value of (cos¹¹θ + sec¹¹θ) [Let me do it myself]

Solution: cosθ + secθ = 2       or, cos²+1=2cos

or, cose + 1/cos Ꮎ = 2       

or, cos²0 – 2cose + 1 = 0            ∴(cose1)²= 0

        ∴ cose – 1 = 0

         sece= 1/cos 0 

          = 1/1

         = 1

cose + sece

= (cose)¹¹+ (sece)¹¹

=(1)¹¹+ (1)¹¹

=1+1

=2

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