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

Trigonometry Chapter 2 Trigonometric Ratios And Trigonometric Identities

Trigonometric ratios:

⇔ In ΔABC, ∠ABC = 90° and ∠ACB = θ.

 

⇔ sin θ = \(\frac{\text { perpendicular }}{\text { hypotenuse }}=\frac{\mathrm{AB}}{\mathrm{AC}}\)

⇔ cos θ = \(\frac{\text { base }}{\text { hypotenuse }}=\frac{\mathrm{BC}}{\mathrm{AC}}\)

⇔ tan θ = \(\frac{\text { perpendicular }}{\text { base }}=\frac{\mathrm{AB}}{\mathrm{BC}}\)

⇔ cosec θ = \(\frac{\text { hypotenuse }}{\text { perpendicular }}=\frac{\mathrm{AC}}{\mathrm{AB}}\)

⇔ sec θ = \(\frac{\text { hypotenuse }}{\text { base }}=\frac{\mathrm{AC}}{\mathrm{BC}}\)

⇔ cot θ = \(\frac{\text { base }}{\text { perpendicular }}=\frac{\mathrm{BC}}{\mathrm{AB}}\)

∴ sin θ = \(\frac{1}{cosec \theta}\)

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⇒ sin θ.cosec θ = 1

⇔ cos θ = \(\frac{1}{sec \theta}\)

⇒ cos θ.sec θ = 1

⇔ tan θ = \(\frac{1}{cot \theta}\)

⇒ tan θ.cot θ = 1

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⇔ \(\tan \theta=\frac{\mathrm{AB}}{\mathrm{BC}}=\frac{\frac{\mathrm{AB}}{\mathrm{AC}}}{\frac{\mathrm{BC}}{\mathrm{CA}}}=\frac{\sin \theta}{\cos \theta}\)

⇔ \(\cot \theta=\frac{\cos \theta}{\sin \theta}\)

Class 10 Maths Trigonometry Chapter 2 Solutions

Some Trigonometric identities of an acute angle:

1. sin² θ+ cos² θ = 1
2. sec² θ= 1 + tan² θ
3. cosec² θ = 1 + cot² θ

Following tables gives the values of trigonometric ratios :

Trigonometry Chapter 2 Trigonometric Ratios And Trigonometric Identities True Or False

Example 1. The value of tan A is always greater than 1.

Solution:

 

tan A = \(\frac{\mathrm{BC}}{\mathrm{AC}}\)

tan A = \(\frac{\mathrm{BC}}{\mathrm{BC}}\) [If BC = AC] = 1

tan A < 1 [If BC < AC]

tan A > 1 [If BC > AC]

So the statement is false.

Example 2. The value of cot A is always less tan 1.

Solution:

 

cot A = \(\frac{\mathrm{AC}}{\mathrm{BC}}\)

cotA = \(\frac{\mathrm{AC}}{\mathrm{AC}}\) [If AC = BC]

cot A = 1

cot A < 1 [If AC < BC]

cot A > 1 [If AC > BC]

∴ So the statement is false.

Wbbse Class 10 Maths Trigonometry Solutions

Example 3. For an angle θ, it may be possible that, sin θ = \(\frac{4}{3}\)

Solution:

 

sin θ = \(\frac{\mathrm{AB}}{\mathrm{AC}}\)

In a right-angled triangle hypotenuse is always greater than each of other two sides

∴ AB < AC  i.e. \(\frac{\mathrm{AB}}{\mathrm{AC}}\) < 1

∴ sin θ < 1

∴ The statement is false.

Example 4. For an angle α, it may be possible that, sec α = \(\frac{12}{5}\)

Solution:

 

sec α = \(\frac{\mathrm{AC}}{\mathrm{BC}}\)

as AC > BC [hypotenuse > side]

i.e. \(\frac{\mathrm{AC}}{\mathrm{BC}}\) > 1  ∴ sec α > 1

∴ The statement is true.

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Example 5. For an angle β, it may be possible that, cosec β= \(\frac{5}{13}\)

Solution:

 

cosec β = \(\frac{\mathrm{AC}}{\mathrm{AB}}\)

and AC > AB i.e. \(\frac{\mathrm{AC}}{\mathrm{AB}}\) > 1 ∴ cosec β > 1

∴ The statement is false.

Example 6. For an angle θ, is may be possible that, cos θ = \(\frac{3}{5}\)

Solution:

 

cos θ = \(\frac{\mathrm{BC}}{\mathrm{AC}}\)

BC < AC

i.e, \(\frac{\mathrm{BC}}{\mathrm{AC}}\) < 1 ∴ cos θ < 1

∴ The statement is true.

Example 7. If 0° ≤ α ≤ 90°, then the least value of (sec² α + cos² α) is 2.

Solution: sec² α + cos² α

= \(\frac{1}{\cos ^2 \alpha}+\cos ^2 \alpha\)

= \(\left(\frac{1}{\cos \alpha}-\cos \alpha\right)^2+2 \cdot \frac{1}{\cos \alpha} \cdot \cos \alpha\)

= \(\left(\frac{1}{\cos \alpha}-\cos \alpha\right)^2+2\)

\(\text { as } 0^{\circ} \leq \alpha \leq 90^{\circ}\)

 

∴ \(\left(\frac{1}{\cos \alpha}-\cos \alpha\right)^2 \geq 0\)

∴ The lease value of (sec² α + cos² α)

∴ The statements is true.

Example 8. The value of (cos 0° x cos 1° x cos 2° x cos 3° x…..x cos 90°) is 1.

Solution: (cos 0° x cos 1° x cos 2° x cos 3° x …… x cos 90°)

= cos 0° x cos 1° x cos 2° x ……. x 0 = 0

∴ The statement is false.

Class 10 Trigonometry Chapter 2 Solved Examples

Trigonometry Chapter 2 Trigonometric Ratios And Trigonometric Identities Fill In The Blanks

Example 1. The value of \(\left(\frac{4}{\sec ^2 \theta}+\frac{1}{1+\cot ^2 \theta}+3 \sin ^2 \theta\right)\) is ________

Solution: \(\frac{4}{\sec ^2 \theta}+\frac{1}{1+\cot ^2 \theta}+3 \sin ^2 \theta\)

= 4 cos² θ + \(\frac{1}{cosec^2 \theta}\) + 3sin² θ

= 4 cos² θ + sin² θ + 3 sin² θ = 4 cos² θ + 4 sin² θ

= 4 (cos² θ + sin² θ) = 4 x 1 =4

∴ Answer is 4.

Example 2. If sin (θ – 30°) = \(\frac{1}{2}\), then the value of cos θ is ______

Solution: sin (θ – 30°) = \(\frac{1}{2}\)

⇒ sin (θ – 30°) = sin 30°

⇒ θ – 30° = 30° ⇒ θ = 60°

cos θ = cos 60° = \(\frac{1}{2}\)

∴ Answer is \(\frac{1}{2}\)

Example 3. If cos² θ – sin² θ = \(\frac{1}{2}\), then the value of cos⁴ θ – sin⁴ θ is _______

Solution: cos⁴ θ- sin⁴ θ

= (cos² θ)² – (sin² θ)²

= (cos² θ + sin² θ) (cos² θ – sin² θ)

= 1 x \(\frac{1}{2}\) = \(\frac{1}{2}\)

∴ Answer is \(\frac{1}{2}\)

Wbbse Class 10 Trigonometry Notes

Example 4. The value of \(\left(\sin \frac{\pi}{3} \cos \frac{\pi}{6}+\cos \frac{\pi}{3} \sin \frac{\pi}{6}\right)\) is _____

Solution: sin\(\frac{\pi}{3}\)cos\(\frac{\pi}{6}\)+ cos\(\frac{\pi}{3}\)sin\(\frac{\pi}{6}\)

= sin\(\frac{180^{\circ}}{3}\)cos\(\frac{180^{\circ}}{6}\) + cos\(\frac{180^{\circ}}{3}\)sin\(\frac{180^{\circ}}{6}\)

= sin 60° cos 30° + cos 60° sin 30°

= \(\frac{\sqrt{3}}{2}\) x \(\frac{\sqrt{3}}{2}\) + \(\frac{1}{2}\) x \(\frac{1}{2}\) = \(\frac{3}{4}\) + \(\frac{1}{4}\) = 1

∴ Answer is 1.

Example 5. If a cos θ = 4 and a sec θ = 9, then the value of a is _______

Solution: a cos θ.a sec θ = 4 x 9

⇒ a² x 1 = 36

⇒ a = ± 6

∴ Answer is ± 6.

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Trigonometry Chapter 2 Trigonometric Ratios And Trigonometric Identities Short Answer Type Questions

Example 1. If r cos θ = 2√3, r sin θ = 2, and 0° < θ < 90°, then determine the value of both r and θ.

Solution: r cos θ = 2√3

r sin θ = 2

r² cos² θ + r² sin² θ = (2√3)² + (2)²

⇒ r² (cos² θ + sin² θ) = 12 + 4

⇒ r² x 1 = 16  ⇒ r = √16 = 4

r sin θ = 2

4 sin θ = 2

⇒ sin θ = \(\frac{2}{4}\) = \(\frac{1}{2}\)

⇒ sin θ = sin 30° ⇒ θ = 30°

∴ r = 2, θ = 30°

Example 2. If sin A + sin B = 2 where 0° ≤ A ≤ 90° and 0° ≤ B ≤ 90°, then find out the value of (cos A + cos B)

Solution: sin A + sin B = 2 = 1 + 1

⇔ sin A + sin B = sin 90° + sin 90°

⇔ sin A = sin 90°

⇔ A = 90°

⇔ Sin B = sin 90°

⇔ B = 90°

⇔ cos A + cos B = cos 90° + cos 90° = 0 + 0 = 0

Example 3. If 0° < θ < 90°, then calculate the least value of (9 tan² θ + 4 cot² θ).

Solution: 9 tan² θ +4 cot² θ

= (3 tan θ)² + (2 cot θ)²

= (3 tan θ – 2 cot θ)² + 2.3 tan θ.2 cot θ

= (3 tan θ- 2 cot θ)² + 12.1 [tan θ = 1/cot θ]

= (3 tan θ – 2 cot θ)² + 12

⇔ The least value of any perfect square of the expression is 0 (zero).

⇔ So, the least value of (3 tan θ – 2 cot θ) is zero.

∴ The least value of (9 tan² θ + 4 cot² θ) is 12.

Example 4. Calculate the value of (sin⁶ α + cos⁶ α + 3 sin²α + cos²α ).

Solution: sin⁶α + cos⁶α + 3 sin²α cos²α

= (sin²α)³ + (cos²α)³ + 3 sin²α cos²α

= (sin²α + cos²α)³ – 3.sin²α cos²α (sin²α + cos²α) + 3 sin²α cos²α

= (1)³ – 3 sin²α cos²α (1) + 3 sin²α cot²α

= 1-3 sin²α cos²α + 3 sin²α cos²α = 1

Example 5. If cosec² θ = 2 cot ² θ and 0° < θ < 90°, then determine the value of θ.

Solution: cosec² θ = 2 cot θ

⇒ 1 + cot² θ = 2 cot θ

⇒ cot² θ – 2 cot θ + 1 = 0

⇒ (cot θ – 1)² = 0  ⇒ cot θ – 1 = 0 ⇒ cot θ = 1

⇒ cot θ = cot 45° ⇒ θ = 45°

Class 10 Maths Trigonometry Important Questions

Example 6. If \(\frac{\cos \theta+\sin \theta}{\cos \theta-\sin \theta}=\frac{\sqrt{3}+1}{\sqrt{3}-1}\) then find the value of

Solution: \(\frac{\cos \theta+\sin \theta}{\cos \theta-\sin \theta}=\frac{\sqrt{3}+1}{\sqrt{3}-1}\)

⇒ \(\frac{\cos \theta+\sin \theta+\cos \theta-\sin \theta}{\cos \theta+\sin \theta-\cos \theta+\sin \theta}=\frac{\sqrt{3}+1+\sqrt{3}-1}{\sqrt{3}+1-\sqrt{3}+1}\) [By componendo-dividend process]

⇒ \(\frac{2 \cos \theta}{2 \sin \theta}=\frac{2 \sqrt{3}}{2}\)

⇒ \(\cot \theta=\sqrt{3}\)

⇒ cot θ = cot 30°

⇒ θ = 30°

Example 7. If \(\tan ^2 \frac{\pi}{4}-\sin ^2 \frac{\pi}{6}=x \sec ^2 \frac{\pi}{4} \cos ^2 \frac{\pi}{3}\) then find the value of x.

Solution: \(x \sec ^2 \frac{\pi}{4} \cos ^2 \frac{\pi}{3}=\tan ^2 \frac{\pi}{4}-\sin ^2 \frac{\pi}{6}\)

⇒ \(x \sec ^2 45^{\circ} \cos ^2 60^{\circ}=\tan ^2 45^{\circ}-\sin ^2 30^{\circ}\)

⇒ \(x \times(\sqrt{2})^2 \times\left(\frac{1}{2}\right)^2=(1)^2-\left(\frac{1}{2}\right)^2\)

⇒ x x 2 x \(\frac{1}{4}\) = 1 – \(\frac{1}{4}\)

⇒ \(\frac{x}{2}\) = \(\frac{3}{4}\)

⇒ x = \(\frac{3}{2}\)

Example 8. If tan θ + sin θ = a and tan θ – sin θ = b, then find the value of \(\frac{a^2-b^2}{\sqrt{a b}}\)

Solution: a² – b² = (tan θ + sin θ)² – (tan θ – sin θ)²

= 4 tan θ.sin θ

\(\sqrt{a b}=\sqrt{(\tan \theta+\sin \theta)(\tan \theta-\sin \theta)}=\sqrt{\tan ^2 \theta-\sin ^2 \theta}\)

= \(\sqrt{\frac{\sin ^2 \theta}{\cos ^2 \theta}-\sin ^2 \theta}\)

= \(\sqrt{\sin ^2 \theta\left(\frac{1}{\cos ^2 \theta}-1\right)}=\sin \theta \sqrt{\frac{1-\cos ^2 \theta}{\cos ^2 \theta}}\)

= \(\sin \theta \cdot \sqrt{\frac{\sin ^2 \theta}{\cos ^2 \theta}}=\sin \theta \cdot \sqrt{\tan ^2 \theta}=\sin \theta \cdot \tan \theta\)

\(\frac{a^2-b^2}{\sqrt{a b}}=\frac{4 \sin \theta \tan \theta}{\sin \theta \cdot \tan \theta}=4\)

Class 10 Maths Trigonometry Important Questions

Example 9. If a sinθ + b cosθ = c, then find the value of (a cosθ – b sinθ).

Solution: a sinθ+ b cosθ = c

⇒ (a sinθ + bcosθ)² = c²

⇒ a² sin² θ + b² cos² θ + 2ab sinθcosθ= c²

⇒  a² (1 – cos² θ) + b² (1 – sin² θ) + 2ab si θ cosθ = c²

⇒ a² – a² cos² θ + b² – b² sin² θ + 2ab sinθ cosθ = c²

⇒ a² + b² – c² = a² cos² θ+ b² sin² θ- 2ab cosθ sinθ

⇒ a² + b² – c² = (acosθ – bsinθ)²

⇒ acosθ – bsinθ = ± \(\sqrt{a^2+b^2-c^2}\)

Example 10. If x = a sinθ and y = b tanθ, then find the value of \(\left(\frac{a^2}{x^2}-\frac{b^2}{y^2}\right)\).

Solution: a sinθ = x

⇒ \(\frac{a}{x}=\frac{1}{\sin \theta}={cosec} \theta\)

b tan θ = y

⇒ \(\frac{b}{y}=\frac{1}{\tan \theta}=\cot \theta\)

⇒ \(\frac{a^2}{x^2}-\frac{b^2}{y^2}={cosec}^2 \theta-\cot ^2 \theta=1\)

Example 11. If 0° < θ< 90° and 2 cos² θ + 3 sinθ = 3 then find the value of θ.

Solution: 2 cos² θ+ 3 sinθ = 3

⇒ 2 (1 – sin² θ) + 3 sinθ= 3

⇒ 2-2 sin² θ + 3 sinθ = 3

⇒ 2 sin² θ – 3 sinθ + 1 = 0

⇒ 2 sin² θ – 2 sinθ – sinθ + 1 = 0

⇒ 2 sinθ (sinθ – 1) – 1 (sinθ – 1) = 0

⇒ (sinθ – 1) (2 sinθ – 1) = 0

either sinθ – 1 =0

⇒ sinθ = 1

2 sinθ – 1 = 0

⇒ 2 sinθ = 1

⇒ sinθ = sin90°

⇒ sinθ = \(\frac{1}{2}\)

⇒ θ = 90°

⇒ sinθ = sin 30°

⇒ θ = 30°

As 0° < θ < 90°

∴ The value of θ is 30°

Example 12. If θ is positive acute angle then find the least value of (25 sec² θ + 49 cos² θ)

Solution: (25 sec² θ + 49 cos² θ)

= (5 secθ)² + (7 cosθ)²

= (5 secθ – 7 cosθ)² + 2.5 secθ 7 cosθ

=(5secθ -7cosθ)² +70secθ.\(\frac{1}{\sec \theta}\)

= (5 secθ – 7 cosθ)² + 70

(5 secθ – 7 cosθ)² ≥ 0

∴ Least value of (25 sec² θ + 49 cos² θ) is 70.

Example 13. If sinθ + sin² θ + sin³ θ = 1 then find the value of (cos⁶ θ- 4 cos⁴ θ + 8 cos² θ).

Solution: sinθ+ sin² θ+ sin³ θ = 1

⇒ sinθ + sin³ θ = 1 – sin² θ

⇒ sinθ (1 + sin² θ) = cos² θ

⇒ sin² θ (1 + sin² θ)² = cos⁴ θ [squaring both side]

⇒ (1 – cos² θ) (1 + 1 – cos² θ)² = cos⁴ θ

⇒ (1 -cos² θ) (2 – cos² θ)² = cos⁴ θ

⇒ (1- cos² θ) (4 – 4 cos² θ + cos⁴ θ) = cos⁴ θ

⇒ 4 – 4 cos² θ + cos⁴ θ – 4 cos² θ + 4 cos⁴ θ- cos⁶ θ = cos⁴ θ

⇒ cos⁶ θ – 4 cos⁴ θ + 8 cos² θ = 4

Class 10 Maths Trigonometry Chapter 2 Solutions

Example 14. In ΔABC each angle is a positive acute angle. If cos (B + C – A) = 0 and sin (C + A – B) = \(\frac{\sqrt{3}}{2}\), then find the value of A, B, and C.

Solution: cos (B + C – A) = 0

⇒ cos (B + C – A) = cos 90°

⇒ B + C – A = 90° ……(1)

sin (C + A – B) = \(\frac{\sqrt{3}}{2}\)

⇒ sin (C + A – B) = sin 60°

⇒ C + A- B = 60° ……..(2)

(1) + (2), we get

B + C- A + C + A- B= 90° + 60°

⇒ 2C = 150°  ⇒ C = 75°

From (2), 75° + A -B = 60°

⇒ A – B = – 15°……(3)

Again, In ΔABC,

A + B + C = 180°

A + B + 75° = 180°

⇒ A + B = 105°….(4)

A + B + A- B = 105°- 15°

⇒ 2A = 90°

⇒  A = 45°

∴ B = 105° – 45° = 60°.