Question

Question 3 plss...i need it fast

Answer 1

Answer:

they both have the same answer that is 1

hope it helps

Answer 2

Answer- The above question is from the chapter 'Introduction to Trigonometry'.

Trigonometry- The branch of Mathematics which helps in dealing with measure of three sides of a right-angled triangle is called Trigonometry.

Trigonometric Ratios:

sin θ  = Perpendicular/Hypotenuse

cos θ = Base/Hypotenuse

tan θ = Perpendicular/Base

cosec θ = Hypotenuse/Perpendicular

sec θ = Hypotenuse/Base

cot θ = Base/Perpendicular

Also, tan θ = sin θ/cos θ and cot θ = cos θ/sinθ.

Trigonometric Identities:

1. sin²θ + cos²θ = 1

2. sec²θ - tan²θ = 1

3. cosec²θ - cot²θ = 1

Complementary Angles in Trigonometry:

1. sin θ = (90° - cos θ)

2. cos θ = (90° - sin θ)

3. cosec θ = (90° - sec θ)

4. sec θ = (90° - cosec θ)

5. tan θ = (90° - cot θ)

6. cot θ = (90° - tan θ)

Given question: Evaluate-

(i) $\dfrac{sin^{2} \: 63^{\circ} \: + sin^{2} \: 27^{\circ}}{cos^{2} \: 17^{\circ} \: + cos^{2} \: 73^{\circ}}$

(ii) sin 25° cos 65° + cos 25° sin 65°

Solution: i) We know that 63° + 27° = 90°

63° = 90° - 27°

Also, 17° + 73° = 90°

17° = 90° - 73°

$\dfrac{sin^{2} \: 63^{\circ} \: + sin^{2} \: 27^{\circ}}{cos^{2} \: 17^{\circ} \: + cos^{2} \: 73^{\circ}}$

= $\dfrac{sin^{2} \: (90^{\circ} \: - \: 27^{\circ}) \: + sin^{2} \: 27^{\circ}}{cos^{2} \: (90^{\circ} \: - \: 73^{\circ}) \: + cos^{2} \: 73^{\circ}}$

= $\dfrac{cos^{2} \: 27^{\circ} \: + sin^{2} \: 27^{\circ}}{sin^{2} \: 73^{\circ} \: + cos^{2} \: 73^{\circ}}$

= $\dfrac{1}{1}$      [∵ sin²θ + cos²θ = 1]

= 1

∴ $\dfrac{sin^{2} \: 63^{\circ} \: + sin^{2} \: 27^{\circ}}{cos^{2} \: 17^{\circ} \: + cos^{2} \: 73^{\circ}}$ = 1

(ii) We know that 65° + 25° = 90°

65° = 90°- 25°

25° = 90° - 65°

sin 25° cos 65° + cos 25° sin 65°

= sin (90° - 65°) cos 65° + cos (90° - 65°) sin 65°

= cos 65° × cos 65° + sin 65° × sin 65°

= cos² 65° + sin² 65°        [∵ sin²θ + cos²θ = 1]

= 1

∴ sin 25° cos 65° + cos 25° sin 65° = 1