WBBSE Solutions For Class 10 Maths Trigonometry Chapter 3 Trigonometric Ratios Of Complementary

Trigonometry Chapter 3 Trigonometric Ratios Of Complementary

Trigonometric ratios of complementary angle:

⇔ sin (90° – θ) = cos θ

⇔ cos (90° – θ) = sin θ

⇔ tan (90° – θ) — cot θ

⇔ cot (90° – θ) = tan θ

⇔ sec (90° – θ) = cosec θ

⇔ cosec (90° – θ) = sec θ

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Trigonometry Chapter 3 Trigonometric Ratios Of Complementary True Or False

Example 1. The value of cos 54° and sin 36° are equal.

Solution: cos 54° = cos (90° – 36°)

= sin 36°

∴ The statement is true.

Example 2. The simplified value of (sin 12° – cos 78°) is 1.

Solution: sin 12° – cos 78°

= sin 12° – cos (90° – 12°)

= sin 12° – sin 12° = 0

∴ The statement is false.

Example 3. If tan 3θ = cot 2θ and 3θ is a positive acute angle, then the value of θ is 18°.

Solution: tan 3θ= cot 2θ

⇒ tan 3θ = tan (90° – 2θ)

⇒ 3θ = 90° – 20

⇒ 3θ + 2θ = 90°

⇒ 5θ = 90°

⇒ θ = 18°

∴ The statement is True.

Example 4. If tan 5θ.tan 4θ = 1 and 5θ is positive acute angle ; then the value of sin 3θ is \(\frac{1}{\sqrt{2}}\)

Solution: tan 5θ.tan 4θ= 1

⇒ tan 5θ = \(\frac{1}{\tan 4 \theta}\)

⇒ tan 5θ = cot 4θ

⇒ tan 5θ= tan (90° – 4θ)

⇒ 5θ = 90° – 4θ

⇒ 9θ = 90°

⇒ θ = 10°

sin 3θ = sin 3 x 10° = sin 30° = \(\frac{1}{2}\)

∴ The statement is False.

Class 10 Maths Trigonometry Chapter 3 Solutions

Example 5. If cosθ = \(\frac{5}{13}\), then the value of cos (90° – θ) is \(\frac{12}{13}\).

Solution: cos (90° – θ)

= sinθ = \(\sqrt{1-\cos ^2} \theta\)

= \(\sqrt{1-\left(\frac{5}{13}\right)^2}=\sqrt{1-\frac{25}{169}}\)

= \(\sqrt{\frac{144}{169}}=\frac{12}{13}\)

∴ The statement is True.

Trigonometry Chapter 3 Trigonometric Ratios Of Complementary Fill In The Blanks

Example 1. The value of (tan 15° x tan 45° x tan 60° x tan 75°) is ______

Solution: tan 15° x tan 45° x tan 60° x tan 75°

= tan 15° x tan 45d x tan 60° x tan (90° – 15°)

= tan 15° x 1 x √3 x cot 15°

= tan 15° x √3 x \(\frac{1}{\tan 15^{\circ}}\) = √3

∴ Answer is √3

Example 2. The value of (sin 12° x cos 18° x sec 78° x cosec 72°) is ______

Solution: sin 12° x cos 18° x sec 78° x cosec 72°

= sin 12° x cos 18° x sec (90° – 12°) x cosec (90° – 18°)

= sin 12° x cos 18° x cosec 12° x sec 18°

= sin 12° x cos 12° x \(\frac{1}{\sin 12^{\circ}}\) x \(\frac{1}{\cos 18^{\circ}}\)

= 1

Example 3. If A and B are complementary to each other, sin A = ______

Solution: A + B = 90°

sin A = sin (90° – B) = cos B

Wbbse Class 10 Maths Trigonometry Solutions

Example 4. The value of sec 52° sin 38° is ______

Solution: sec 52° sin 38°

= sec 52° sin (90° – 52°)

= sec 52° cos 52°

= sec 52° x \(\frac{1}{\sec 52^{\circ}}\) = 1

∴ Answer is 1.

Example 5. If α + β = 90°, then the value of (1 – tanα.tanβ) is _______

Solution: 1 – tanα.tanβ

= 1 – tanα.tan (90° – α)

= 1 – tanα.cotα

= 1-1 = 0

∴ Answer is zero.

Trigonometric Identities Class 10 Solutions

Trigonometry Chapter 3 Trigonometric Ratios Of Complementary Short Answer Type Question

Example 1. If sin 10θ = cos 8θ and 10θ is a positive acute angle, find the value of tan 9θ.

Solution: sin 10θ = cos 8θ

⇒ sin 10θ = sin (90° – 8θ)

⇒ 10θ = 90° – 8θ

⇒ 18θ = 90°

⇒ θ = 5°

∴ tan 9θ = tan 9 x 5° = tan 45° = 1

Example 2. If tan 4θ x tan 6θ = 1 and 6θ is a positive acute angle, then find the value of θ.

Solution: tan 4θ x tan 6θ= 1

⇒ tan 4θ = \(\frac{1}{\tan 6 \theta}\)

⇒ tan 4θ = cot 6θ

⇒ tan 4θ = tan (90° – 6θ)

⇒ 4θ = 90° – 6θ

⇒ 10θ = 90°

⇒ θ = 9°

Example 3. Find the value of \(\frac{2 \sin ^2 63^{\circ}+1+2 \sin ^2 27^{\circ}}{3 \cos ^2 17^{\circ}-2+3 \cos ^2 73^{\circ}}\)

Solution: \(\frac{2 \sin ^2 63^{\circ}+1+2 \sin ^2 27^{\circ}}{3 \cos ^2 17^{\circ}-2+3 \cos ^2 73^{\circ}}\)

= \(\frac{2 \sin ^2 63^{\circ}+1+2 \sin ^2\left(90^{\circ}-63^{\circ}\right)}{3 \cos ^2 17^{\circ}-2+3 \cos ^2\left(90^{\circ}-17^{\circ}\right)}\)

= \(\frac{2 \sin ^2 63^{\circ}+1+2 \cos ^2 63^{\circ}}{3 \cos ^2 17^{\circ}-2+3 \sin ^2 17^{\circ}}\)

= \(\frac{2\left(\sin ^2 63^{\circ}+\cos ^2 63^{\circ}\right)+1}{3\left(\cos ^2 17^{\circ}+\sin ^2 17^{\circ}\right)-2}\)

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

Example 4. Find the value of tan 1° x tan 2° x tan 3° x…….. x tan 89°

Solution: tan 1° x tan 2° x tan 3° x …… x tan (90° – 2°) x tan (90° – 1°)

= tan 1° x tan 2° x tan 3° x …. x cot 2° x cot 1°

= (tan 1° x cot 1°) x (tan 2° x cot 2°) x ……..x (tan 44° x cot 44°) x tan 45°

= 1 x 1 = 1…… x 1 x 1 = 1

Example 5. If sec 5A = cosec (A + 36°) and 5A is a positive acute angle, then find the value of A.

Solution: sec 5A = cosec (A + 36°)

⇒ sec 5A = sec {90° – (A + 36°)}

⇒ 5A = 90° – A – 36°

⇒ 6A = 54°

⇒ A = 9°

Example 6. If sin (2θ + 45°) = cos (30° – θ) where (2θ + 45°) and (30° – θ) are positive acute angles then find the value of tan 40.

Solution: sin (2θ+ 45°) = cos (30° – θ)

⇒ sin (2θ + 45°) = sin {90° – (30° – θ)}

⇒ 2θ+ 45° = 90° – 30° + θ

⇒ 2θ – θ = 60° – 45°

⇒ θ = 15°

tan 4θ = tan 4 x 15° = tan 60° = √3

Class 10 Trigonometry Chapter 3 Solved Examples

Example 7. If tan θ = cot (n – 1) θ, then find the value of θ.

Solution: tan θ = cot (n – 1) θ

⇒ cot (90° – θ) = cot (nθ – θ)

⇒ 90° – θ = nθ – θ

⇒ nθ = 90°

⇒ θ = \(\frac{90^{\circ}}{n}\)

Example 8. Find the value of \(\sin ^2 \frac{\pi}{16}+\sin ^2 \frac{3 \pi}{16}+\sin ^2 \frac{5 \pi}{16}+\sin ^2 \frac{7 \pi}{16}\)

Solution: \(\sin ^2 \frac{\pi}{16}+\sin ^2 \frac{3 \pi}{16}+\sin ^2 \frac{5 \pi}{16}+\sin ^2 \frac{7 \pi}{16}\)

= \(\sin ^2 \frac{\pi}{16}+\sin ^2 \frac{3 \pi}{16}+\sin ^2\left(\frac{\pi}{2}-\frac{3 \pi}{16}\right)+\sin ^2\left(\frac{\pi}{2}-\frac{\pi}{16}\right)\)

= \(\sin ^2 \frac{\pi}{16}+\sin ^2 \frac{3 \pi}{16}+\cos ^2 \frac{3 \pi}{16}+\cos ^2 \frac{\pi}{16}\)

= \(\left(\sin ^2 \frac{\pi}{16}+\cos ^2 \frac{\pi}{16}\right)+\left(\sin ^2 \frac{3 \pi}{16}+\cos ^2 \frac{3 \pi}{16}\right)\)

= 1 + 1 = 2

Example 9. Find the value of tan 20° tan 35° tan 45° tan 55° tan 70°

Solution: tan 20° tan 35° tan 45° tan 55° tan 70°

= tan 20° tan 35° tan 45° tan (90° – 35°) tan (90° – 20°)

= tan 20° tan 35° x 1 x cot 35° x cot 20°

= tan 20° tan 35° x tan 35° x \(\frac{1}{\tan 35^{\circ}} \times \frac{1}{\tan 20^{\circ}}\) = 1

Wbbse Class 10 Trigonometry Notes

Example 10. If x sinθ  – y cosθ = 3 and x cosθ+ y sinθ = 4 then find the value of \(\sqrt{x^2+y^2}\).

Solution: (x sinθ – y cosθ)² + (x cosθ + y sinθ)² = 3² + 4²

⇒ x² sin² θ – 2xy sinθ cosθ + y²cos² θ+ x²cos² θ + 2xy sinθ cosθ + y² sin² θ = 25

⇒ x² (sin² θ + cos² θ) + y2 (sin² θ + cos² θ) = 25

⇒ x² x 1 + y² x 1 = 25

⇒ \(\sqrt{x^2+y^2}= \pm \sqrt{25}= \pm 5\)