CBSE Solutions For Class 10 Mathematics Chapter 8 Introduction To Trigonometry

Class 10 Maths Introduction To Trigonometry

Question 1. In ΔABC, ∠B = 90° and, Sin A = \(\frac{3}{5}\) , then find all other trigonometric rotiss for ∠A.
Solution:

Sin A = \(\frac{3}{5}=\frac{perpendicular}{\text { hypo }}\)

ΔABC

∠B=90°, BC=3t, AC=5k

AB² + BC²= AC²

AB² = AC² – BC²

= (25k²) – (9k²)

AB² = 16k²

AB = 4K

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Cos A = \(=\frac{\text { base }}{\text { hypotenuse }}=\frac{A B}{A C}=\frac{4 k}{5 k}=\frac{4}{5}\)

tan A = \(\frac{\text {perpendicular }}{\text {base }}=\frac{B C}{A B}=\frac{3 k}{4 k}=\frac{3}{4}\)

Cosec A = \(\frac{\text { hypotenuse }}{\text { perpendicular }}=\frac{A C}{B C}=\frac{5 k}{3 k}=\frac{5}{3}\)

Sec A = \(\frac{\text { hypotenuse }}{\text { base }}=\frac{A C}{A B}=\frac{5 k}{4 k}=\frac{5}{4}\)

Cot A = \(\frac{\text { base }}{\text { perpendicular }}=\frac{A B}{B C}=\frac{4 k}{3 k}=\frac{4}{3}\)

Question 2. In ΔABC, ∠B = 90° and Cos A = \(\frac{9}{41}\), then find all other trigonometric ratios for ∠A.

Solution:

Cos A = \(\frac{9}{41}\)

ΔABC

∠B=90°, AB=9k, AC = 41k

AB² + BC² = AC²

BC² = AC² – AB²

BC² = (41k) = (9k)²

BC² = 1681k² – 81k²

BC² = 1600k²

BC = 40k

Sin A = \(\frac{\text { perpendicular }}{\text { hypotemuse }}=\frac{B C}{A C}=\frac{40}{41}\)

tan A = \(\frac{\text { perpendicular }}{\text { base }}=\frac{B C}{A B}=\frac{40}{9}\)

Cosec A = \(\frac{\text { hypotenuse }}{\text { perpendicular }}=\frac{A C}{B C}=\frac{41}{40}\)

SecA = \(\frac{\text { hypotenusc }}{\text { base }}=\frac{A C}{A B}=\frac{41}{9}\)

Cot A = \(\frac{\text { base }}{\text { perpendicular }}=\frac{A B}{B C}=\frac{9}{40}\)

Question 3. In ΔABC, ∠A= 90° and tan B= \(\frac{5}{6}\), then find all other trigonometric ratios for ∠B.

Solution:

ΔABC, ∠Ạ = 90°

tan B = \(\frac{\text { Perpendicular }}{\text { Base }}=\frac{5}{6}\)

AC = 6K, BC=5K

AB² = AC²+BC²

AB² = (6k)² + (5k)²

AB² = 36k² +25k²

AB² = 61k²

AB = \(\sqrt{61 k}\)

Sin B = \(\frac{\text { Perpendicular }}{\text { hyotenuse }}=\frac{B C}{A B}=\frac{5 K}{\sqrt{61 K}}=\frac{5}{\sqrt{61}}\)

Cos B = \(\frac{\text { base }}{\text { hypotenuse }}=\frac{A C}{A B}=\frac{6 k}{\sqrt{61 k}}=\frac{6}{\sqrt{61}}\)

Cosec B = \(\frac{\text { hypotenuse }}{\text {Perpendicular}}=\frac{A B}{B C}=\frac{\sqrt{61}}{5}\)

Sec B = \(\frac{\text { hypotenuse }}{\text { base }}=\frac{A B}{A C}=\frac{\sqrt{61}}{6}\)

Cot B = \(\frac{\text { base }}{\text { perpendicular }}=\frac{A C}{B C}=\frac{6}{5}\)

Question 4. In ΔPQR, ∠R = 90° and Cosec P =\(\frac{13}{5}\), then find all trignometric ratio for ∠Q.

Solution:

ΔPQR, ∠R=90°

Cosec P = \(=\frac{\text { hypotenuse }}{\text { perpendicular }}=\frac{13}{5}=\frac{P Q}{Q R}\)

PR² + QR² = PQ²

PQ = 13K, QR = 5k

PR² = PQ² – QR²

= (13k)²– (5k)²

PR² = 169k²-25k²

PR² = 144k²

PR = 12k

Sin Q = \(\frac{\text { perpendicular }}{\text { hypotenuesc }}=\frac{Q R}{P Q}=\frac{5 K}{13 K}=\frac{5}{13}\)

Cos Q = \(\frac{\text { base }}{\text { hypdenuse }}=\frac{P R}{P Q}=\frac{12 k}{13 k}=\frac{12}{13}\)

tan Q = \(\frac{\text { perpendicular }}{\text { base }}=\frac{Q R}{P R}=\frac{5 k}{12 k}=\frac{5}{12}\)

Sec Q = \(\frac{\text { hypotenuse }}{\text { base }}=\frac{P Q}{P R}=\frac{13 k}{12 k}=\frac{13}{12}\)

Cot Q = \(=\frac{\text { base }}{\text { perpendicular }}=\frac{P R}{Q R}=\frac{12 k}{5 k}=\frac{12}{5}\)

Question 5. In ΔABC, ∠C = 90° and Sec B = \(\frac{5}{4}\), then find all other trigonometric ratios for ∠B.

Solution:

SecB = \(\frac{\text { hypo }}{\text { base }}=\frac{5}{4}=\frac{A B}{A C}\)

AB = 5K, AC = 4K

BC² = AB² – AC²

BC² = (5k)² = (4k)²

BC² = 25k² – 16k²

BC² = 9k²

BC = 3K

SinB = \(\frac{\text { perpendiular }}{\text { hypo }}=\frac{B C}{A C}=\frac{3 k}{4 k}=\frac{3}{4}\)

COS B = \(\frac{\text { base }}{\text { hypo }}=\frac{A B}{A C}=\frac{5 k}{4 k}=\frac{5}{4}\)

tan B = \(=\frac{\text { perpendicular }}{\text { base }}=\frac{B C}{A B}=\frac{3 k}{5 k}=\frac{3}{5}\)

Cosec B = \(\frac{\text { hypo }}{\text { base }}=\frac{A C}{B C}=\frac{4 k}{3 k}=\frac{4}{3}\)

Cot B = \(=\frac{\text { base }}{\text { perpendicular }}=\frac{A B}{B C}=\frac{5 k}{3 k}=\frac{5}{3}\)

Question 6. If tanθ = 2, then they find the value of \(\frac{2 \sin \theta \cos \theta}{\cos ^2 \theta-\sin ^2 \theta}\)

Solution:

tanθ = 2 = \(\frac{\sin \theta}{\cos \theta}=\frac{2}{1}\)

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

⇒ \(\frac{2(2)(1)}{(1)^2-(2)^2}\)

⇒ \(\frac{4}{1-4}\)

⇒ \(\frac{-4}{3}\)

Question 7. If tanθ = \(\tan \theta=\frac{a}{b},\), then find the value of \(\frac{2 \sin \theta-3 \cos \theta}{2 \sin \theta+3 \cos \theta}\)

Solution:

⇒ \(\tan \theta=\frac{a}{b}\)

⇒ \(\frac{\sin \theta}{\cos \theta}=\frac{a}{b}\)

⇒ \(\frac{2 \sin \theta-3 \cos \theta}{2 \sin \theta+3 \cos \theta}\)

⇒ \(\frac{2 a-3 b}{2 a+3 b}\)

Question 8. If tanθ = \(\frac{3}{5}\), then find the Value of \(\frac{2 \sin \theta-3 \cos \theta}{2 \sin \theta+3 \cos \theta}\)

Solution:

⇒ \(\tan \theta=\frac{3}{5}\)

⇒ \(\frac{\sin \theta}{\cos \theta}=\frac{3}{5}\)

⇒ \(\frac{3 \sin \theta+4 \cos \theta}{3 \sin \theta-4 \cos \theta}\)

⇒ \(\frac{3(3)+4(5)}{3(3)-4(5)}\)

⇒ \(\frac{9+20}{9-20}\)

⇒ \(\frac{-29}{11}\)

Question 9. Find the value of Sin 60° Cos 30° – Cos 60° sin 30°.

Solution:

Sin 60° Cos 30°- Cos 60° Sin 30°

⇒ \(\left(\frac{\sqrt{3}}{2}\right)\left(\frac{\sqrt{3}}{2}\right)-\left(\frac{1}{2}\right)\left(\frac{1}{2}\right)\)

⇒ \(\left(\frac{\sqrt{3}}{2}\right)^2-\left(\frac{1}{2}\right)^2\)

⇒ \(\frac{3}{4}-\frac{1}{4}\)

⇒ \(\frac{12-4}{16}\)

⇒ \(\frac{8}{16}\)

⇒ \(\frac{1}{2}\)

Question 10. Find the Value of \(\frac{2 \tan 30^{\circ}}{1-\tan ^2 30^{\circ}}\)

Solution:

⇒ \(\frac{2 \tan 30^{\circ}}{1-\tan ^2 30^{\circ}}\)

⇒ \(\frac{2\left(\frac{1}{\sqrt{3}}\right)}{1-\left(\frac{1}{\sqrt{3}}\right)^2}\)

⇒ \(\frac{\frac{2}{\sqrt{3}}}{1-\frac{1}{3}}\)

⇒ \(\frac{\frac{2}{\sqrt{3}}}{\frac{3-1}{3}}\)

⇒ \(\frac{\frac{2}{\sqrt{3}}}{\frac{2}{3}}\)

⇒ \(\sqrt{3}\)

Question 11. Find the value of 2sin 30° Cos30°

Solution:

2 sin 30° Cos 30°

⇒ \(2\left(\frac{1}{2}\right)\left(\frac{\sqrt{3}}{2}\right)\)

⇒ \(2\left(\frac{\sqrt{3}}{4}\right)\)

⇒ \(\frac{\sqrt{3}}{2}\)

Question 12. Show that: 5 Cos² 60° + 4 sec² 30° tan² 4s° + Cos² 90° = \(\frac{67}{12}\)

Solution:

5 Cos² 60° + 4 Sec² 30° – tan² 45° + Cos² 90°

⇒ \(5\left(\frac{1}{2}\right)^2+4\left(\frac{2}{\sqrt{3}}\right)^2-1+(0)^2\)

⇒ \(5\left(\frac{1}{4}\right)+4\left(\frac{4}{3}\right)-1\)

⇒ \(\frac{5}{4}+\frac{16}{3}-1\)

⇒ \(\frac{3 \times 5+4 \times 16-12}{12}\)

⇒ \(\frac{15+64-12}{12}\)

⇒ \(\frac{67}{12}\)

Question 13. Find the value of 4sin² 30° + tan² 60° + Sec² 45°

Solution:

4sin² 30°+ tan² 60°+ Sec² 45°.

⇒ \(4\left(\frac{1}{2}\right)^2+(\sqrt{3})^2+(\sqrt{2})^2\)

⇒ \(4\left(\frac{1}{4}\right)+3+2\)

= 1+3+2

= 6

Question 14. Find the Value of \(\frac{\sin 30^{\circ}}{\cos ^2 45^{\circ}}-\tan ^2 60^{\circ}+3 \cos 90^{\circ}+\sin 0^{\circ}\)

Solution:

⇒ \(\frac{\sin 30^{\circ}}{\cos ^2 45^{\circ}}-\tan ^2 60^{\circ}+3 \cos 90^{\circ}+\sin 0^{\circ}\)

⇒ \(\frac{\frac{1}{2}}{\left(\frac{1}{\sqrt{2}}\right)^2}-(\sqrt{3})^2+3(0)+0\)

⇒ \(\frac{\frac{1}{2}}{\frac{1}{2}}-3\)

⇒ 1 – 3 = -2

Question 15. Evaluate: \(\frac{\sin ^2 30^{\circ}+\sin ^2 45^{\circ}-4 \cot ^2 60^{\circ}}{2 \sin 30^{\circ} \cos 30^{\circ}+\frac{1}{2} \tan 60^{\circ}}\)

Solution:

⇒ \(\frac{\sin ^2 30^{\circ}+\sin ^2 45^{\circ}-4 \cot ^2 60^{\circ}}{2 \sin 30^{\circ} \cos 30^{\circ}+\frac{1}{2} \tan 60^{\circ}}\)

⇒ \(\frac{\left(\frac{1}{2}\right)^2+\left(\frac{1}{\sqrt{2}}\right)^2-4\left(\frac{1}{\sqrt{3}}\right)^2}{2\left(\frac{1}{2}\right)\left(\frac{\sqrt{3}}{2}\right)+\frac{1}{2}(\sqrt{3})}\)

⇒ \(\frac{\frac{1}{4}+\frac{1}{2}-4\left(\frac{1}{3}\right)}{\frac{\sqrt{3}}{2}+\frac{\sqrt{3}}{2}}\)

⇒ \(\frac{\frac{1}{4}+\frac{1}{2}-\frac{4}{3}}{\frac{2 \sqrt{3}}{4}}\)

⇒ \(\frac{\frac{1}{6}-\frac{4}{3}}{\frac{2 \sqrt{3}}{4}}\)

⇒ \(\frac{\frac{1-8}{6}}{\frac{2 \sqrt{3}}{4}}\)

⇒ \(\frac{\frac{-7}{6}}{\frac{2 \sqrt{3}}{4}}\)

⇒ \(\frac{\frac{-7}{6}}{\frac{\sqrt{3}}{2}}\)

⇒ \(\frac{-7}{12 \sqrt{3}}\)

Question 16. Show that: \(\sin 60^{\circ}=\frac{2 \tan 30^{\circ}}{1+\tan ^2 30^{\circ}}\)

Solution:

⇒ \(\sin 60^{\circ}=\frac{2 \tan 30^{\circ}}{1+\tan ^2 30^{\circ}}\)

⇒ \(\frac{\sqrt{3}}{2}=\frac{2\left(\frac{1}{\sqrt{3}}\right)}{1+\left(\frac{1}{\sqrt{3}}\right)^2}\)

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

⇒ \(\frac{\sqrt{3}}{2}=\frac{\frac{2}{\sqrt{3}}}{\frac{4}{3}}\)

⇒ \(\frac{\sqrt{3}}{2}=\frac{1}{\frac{2}{\sqrt{3}}}\)

⇒ \(\frac{\sqrt{3}}{2}=\frac{\sqrt{3}}{2}\)

Question 17. Show that: Cos² 60°- Sin² 60° = -Sin30°

Solution:

Cos² 60°-sin²60°= -Sin30°

⇒ \(\frac{1}{4}-\frac{3}{4}= -\frac{1}{2}\)

⇒ \(\frac{-2}{4}=\frac{-1}{2}\)

⇒ \(\frac{-1}{2}=\frac{-1}{2}\)

Question 18. Show that: tan 60° = \(\tan 60^{\circ}=\frac{2 \tan 30^{\circ}}{1-\tan ^2 30^{\circ}}\)

Solution:

⇒ \(\tan 60^{\circ}=\frac{2 \tan 30^{\circ}}{1-\tan ^2 30^{\circ}}\)

⇒ \(\sqrt{3}=\frac{2\left(\frac{1}{\sqrt{3}}\right)}{1-\left(\frac{1}{\sqrt{3}}\right)^2}\)

⇒ \(\sqrt{3}=\frac{2 / \sqrt{3}}{1-\frac{1}{3}}\)

⇒ \(\sqrt{3}=\frac{2/ \sqrt{3}}{\frac{3-1}{3}}\)

⇒ \(\sqrt{3}=\frac{2 / \sqrt{3}}{2 / 3}\)

⇒ \(\sqrt{3}=\frac{1}{1 / \sqrt{3}}\)

⇒ \(\sqrt{3}=\sqrt{3}\)

Question 19. Show that: \({Cos} 30^{\circ}=\sqrt{\frac{1+\cos 60^{\circ}}{2}}\)

Solution:

⇒ \({Cos} 30^{\circ}=\sqrt{\frac{1+\cos 60^{\circ}}{2}}\)

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

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

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

⇒ \(\frac{\sqrt{3}}{2}=\sqrt{\frac{\frac{3}{2}}{2}}\)

Question 20. Evaluate: \(\frac{\tan 45^{\circ}}{2 \sin 30^{\circ}-\cos 60^{\circ}}\)

Solution:

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

⇒ \(\frac{1}{2\left(\frac{1}{2}\right)-\frac{1}{2}}\)

⇒ \(\frac{1}{1-\frac{1}{2}}\)

⇒ \(\frac{1}{\frac{2-1}{2}}\)

⇒ \(\frac{1}{\frac{1}{2}}\)

= 2

Question 21. If A = 45°, then show that: Cos 2A = Cos²A – Sin²A

Solution:

Given A = 45°

Cos2(45°) = Cos²(45°)-Sin²(45°)

Cos 90° = \(\left(\frac{1}{\sqrt{2}}\right)^2-\left(\frac{1}{\sqrt{2}}\right)^2\)

0 = \(\frac{1}{2}-\frac{1}{2}\)

0 = 0

Question 22. If A=30° and B=60°, then show that: Cos (A+B) = CosA CosB – Sin A Sin B

Solution:

Given A= 30°, B=60°

Cos (30°+60°) = Cos 30° Cos 60° – Sin 30° Sin60°

⇒ \(\cos 90^{\circ}=\frac{\sqrt{3}}{2}\left(\frac{1}{2}\right)-\left(\frac{1}{2}\right) \frac{\sqrt{3}}{2}\)

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

0=0

Question 23. If A = 30°, then show that: \(\tan 2 A=\frac{2 \tan A}{1-\tan ^2 A}\)

Solution:

Given A = 30°

tan 2(30°) = \(\frac{2 \tan 30^{\circ}}{1-\tan ^2 30^{\circ}}\)

tan 60° = \(\frac{2\left(\frac{1}{\sqrt{3}}\right)}{1-\left(\frac{1}{\sqrt{3}}\right)^2}\)

⇒ \(\sqrt{3}=\frac{2 / \sqrt{3}}{1-\frac{1}{3}}\)

⇒ \(\sqrt{3}=\frac{2 / \sqrt{3}}{\frac{3-1}{3}}\)

⇒ \(\sqrt{3}=\frac{2/\sqrt{3}}{2 / 3}\)

⇒ \(\sqrt{3}=\sqrt{3}\)

Question 24. If \(\cos \theta=\frac{\sqrt{3}}{2}\), then find the value of Sin 3θ.

Solution:

⇒ \(\cos \theta=\frac{\sqrt{3}}{2}\)

Cos θ = Cos30°

θ = 30°

= Sin 3θ

= Sin 3(30°)

= Sin 90°

= 1

Question 25. If tan (A+B) = \(\sqrt{3}\) and Sin (A-B) = \(\frac{1}{2}\), then find the value of tan (2A-3B)

Solution:

tan (A+B)= \(\sqrt{3}\)

Sin (A-B)= \(\frac{1}{2}\)

tan (2A-3B)

tan (A+B)= \(\sqrt{3}\)

A+B = 60°

Sin (A-B) = Sin 30° =) A-B = 30°

Acting equations

⇒ \(A=\frac{90^{\circ}}{2}=45^{\circ}\)

Subtracting equations

⇒ \(B=\frac{30^{\circ}}{2}\)

B = 15°

tan (2A-3B)

= tan (2(45°)-3((5°))

= tan (90°-45°)

= tan 45° = 1

Question 26. If A+B = 90° and tan A = \(\sqrt{3}\), then find the Value of B.

Solution:

A+B=90°

tan A = \(\sqrt{3}\)

tan A = tan 60°

A = 60°

60°+B=90°

B = 30°

Question 27. If A-B = 30° and Sin A = \(\frac{\sqrt{3}}{2}\), then find the value of B.

Solution:

A-B = 30°

Sin A = \(\frac{\sqrt{3}}{2}\)

Sin A = Sin 60°

A = 60°

60°-B = 30°

B = 30°

Question 28. \(\frac{1}{1+\tan ^2 \theta}+\frac{1}{1+\cos ^2 \theta}=1\)

Solution:

⇒ \(\frac{1}{1+\tan ^2 \theta}+\frac{1}{1+\cot ^2 \theta}\)

⇒ \(\frac{1}{1+\frac{\sin ^2 \theta}{\cos ^2 \theta}}+\frac{1}{1+\frac{\cos ^2 \theta}{\sin ^2 \theta}}\)

⇒ \(\frac{1}{\frac{\cos ^2 \theta+\sin ^2 \theta}{\cos ^2 \theta}}+\frac{1}{\frac{\sin ^2 \theta+\cos ^2 \theta}{\sin ^2 \theta}}\)

⇒ \(\frac{\cos ^2 \theta+\sin ^2 \theta}{1}\)

= 1 = R.H.S

Question 29. \(\sin ^2 \theta+\frac{1}{1+\tan ^2 \theta}=1\)

Solution:

⇒ \(\sin ^2 \theta+\frac{1}{1+\tan ^2 \theta}\)

⇒ \(\sin ^2 \theta+\frac{1}{1+\frac{\sin ^2 \theta}{\cos ^2 \theta}}\)

⇒ \(\sin ^2 \theta+\frac{\cos ^2 \theta}{1}\)

⇒ \(\sin ^2 \theta+\cos ^2 \theta\)

= 1 = R.H.S

Question 30. \(1+\frac{\cos ^2 \theta}{\sin ^2 \theta}-{cosec}^2 \theta=0\)

Solution:

⇒ \(1+\frac{\cos ^2 \theta}{\sin ^2 \theta}-{cosec}^2 \theta \Rightarrow \text { L.H.S }\)

⇒ \(1+\frac{\cos ^2 \theta}{\sin ^2 \theta}-\frac{1}{\sin ^2 \theta}\)

⇒ \(\frac{\sin ^2 \theta+\cos ^2 \theta-1}{\sin ^2 \theta}\)

⇒ \(\frac{1-1}{\sin ^2 \theta}=0=\text { R.H.S }\)

Question 31. \(\frac{1+\tan ^2 \theta}{{cosec}^2 \theta}=\tan ^2 \theta\)

Solution:

⇒ \(\text { L.H.S }=\frac{1+\tan ^2 \theta}{{cosec}^2 \theta}\)

⇒ \(\frac{1+\frac{\sin ^2 \theta}{\cos ^2 \theta}}{\frac{1}{\sin ^2 \theta}}\)

⇒ \(\frac{\frac{\cos ^2 \theta+\sin ^2 \theta}{\cos ^2 \theta}}{\frac{1}{\sin ^2 \theta}}\)

⇒ \(=\frac{1}{\cos ^2 \theta} \times \frac{\sin ^2 \theta}{1}\)

⇒ \(\tan ^2 \theta=\text { R.HS }\)

Question 32. Prove that: Cosec A-Cot A = \(\frac{1}{{cosec} A+\cot A}\)

Solution:

L.H.S = CosecA – Cota = \(({cosec} A-\cot A) \cdot \frac{({cosec} A+\cot A)}{({cosec} A+\cot A)}\)

⇒ \(\frac{{cosec}^2 A-\cot ^2 A}{{cosec} A+\cot A}\)

⇒ \(\frac{1}{{cosec} A+\cot A}\)

= R.H.S

Question 33. Prove that \(\frac{\sec A+1}{\tan A}=\frac{\tan A}{{Sec} A-1}\)

Solution:

L.H.S = \(\frac{\sec A+1}{\tan A}=\frac{\sec A+1}{\tan A} \times \frac{\sec A-1}{\sec A-1}\)

⇒ \(\frac{\sec ^2 A-1}{\tan A(\sec A-1)}=\frac{\tan ^2 A}{\tan A(\sec A-1)}\)

⇒ \(\frac{\tan A}{(\sec A-1)}\)

= R.H.S

Question 34. Check whether the equation \(\frac{\tan \phi+\sin \phi}{\tan \phi-\sin \phi}=\frac{\sec \phi+1}{\sec \phi-1}\) is an identity or not?

Solution:

L.H.S = \(\frac{\tan \phi+\sin \phi}{\tan \phi-\sin \phi}\)

⇒ \(\frac{\frac{\sin \phi}{\cos \phi}+\sin \phi}{\frac{\sin \phi}{\cos \phi}-\sin \phi}\)

⇒ \(\frac{\sin \phi \sec \phi+\sin \phi}{\sin \phi \sec \phi-\sin \phi}\)

⇒ \(\frac{\sin \phi(\sec \phi+1)}{\sin \phi(\sec \phi-1)}\)

⇒ \(\frac{\sec \phi+1}{\sec \phi-1}\)

= R.H.S

Question 35. If \(\cos ^2 30^{\circ}+\cos ^2 45^{\circ}+\cos ^2 60^{\circ}=x\), then find the value of “x”.

Solution:

⇒ \(\cos ^2 30^{\circ}+\cos ^2 45^{\circ}+\cos ^2 60^{\circ}=x\)

⇒ \(x=\left(\frac{\sqrt{3}}{2}\right)^2+\left(\frac{1}{\sqrt{2}}\right)^2+\left(\frac{1}{2}\right)^2\)

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

⇒ \(x=\frac{6+4+2}{8}\)

⇒ \(x=\frac{12}{8}\)

⇒ \(x=\frac{3}{2}\)

Question 36. Without using trigonometric tables, evaluate:

1. \(\frac{\sin 11^{\circ}}{\cos 79^{\circ}}\)

Solution:

⇒ \(\frac{\sin 11^{\circ}}{\cos 79^{\circ}}=\frac{\sin 11^{\circ}}{\left.\cos \left(90^{\circ}-79\right)^{\circ}\right)}=\frac{\sin 11^{\circ}}{\sin 11^{\circ}}=1\)

2. \(\frac{\sec 15^{\circ}}{{cosec} 75^{\circ}}\)

Solution:

⇒ \(\frac{\sec 15^{\circ}}{{cosec} 75^{\circ}}=\frac{\sec 15^{\circ}}{{cosec}\left(90^{\circ}-75^{\circ}\right)}=\frac{\sec 15^{\circ}}{{cosec} 15^{\circ}}=1\)

3. \(\frac{\tan 54^{\circ}}{\cot 36^{\circ}}\)

Solution:

⇒ \(\frac{\tan 54^{\circ}}{\cot 36^{\circ}}=\frac{\tan 54^{\circ}}{\cot \left(90^{\circ}-36^{\circ}\right)}=\frac{\tan 54^{\circ}}{\cot \left(90^{\circ}-36^{\circ}\right)}=\frac{\tan 54^{\circ}}{\tan 54{ }^{\circ}}=1\)

Question 37. Evaluate: tan42° – Cot 48°

Solution:

tan 42°-Cot 48° = tan 42°- Cot (90°- 42°)

tan42°-tan 42°

= 0

Question 38. Evaluate: Sex 36° – Cosec 54°

Solution:

Sec 36°- Cosec 54° = Sec 36° – Cosec (90°-36°)

= Sec 36° – Sec 36°

= 0

Question 39. Prove that: Sin 42°Cos 48°+ Sin 48° Cos42° = 1

Solution:

L.H.S= Sin 42° Cos48° + Sin48° Cos 42°

= Sin 42°cos (90°-42°)+ Sin(90° – 42°) Cos 42°

= Sin 42° Sin 42° + Cos42° Cos42°

Sin² 42° + Cos² 42°

= 1

Question 40. If Sin 3A = Cos(A-26°) where 3A is an acute angle, then find the value of A.

Solution:

Given that,

Sin 3A = Cos (A-26°)

Cos (90°-3A) = Cos (A-26°)

90°-3A = A-26°

-4A = -116°

A = 29°