Question:

Prove the following: cotxcot2x−cot2xcot3x−cot3xcotx=1

Answer:

Step 1: Rewrite the equation as cotx(cot2x−cot3x)−cot2x(cot3x−cotx)=1

Step 2: Use the identity cotA−cotB=−cscA⋅cscB⋅(cotA⋅cotB−1) to rewrite the equation as cotx⋅csc2x⋅csc3x⋅(cot2x⋅cot3x−1)−cot2x⋅csc3x⋅cscx⋅(cot3x⋅cotx−1)=1

Step 3: Use the identity cotA⋅cotB−1=−csc2A⋅csc2B to rewrite the equation as cotx⋅csc2x⋅csc3x⋅csc22x⋅csc23x−cot2x⋅csc3x⋅cscx⋅csc32x⋅cscx2=1

Step 4: Use the identity cscA⋅cscB=1+cotA⋅cotB to rewrite the equation as (1+cotx⋅cot2x)(1+cot2x⋅cot3x)(1+cot3x⋅cotx)=1

Step 5: Use the identity 1+cotA⋅cotB=csc2A⋅csc2B to rewrite the equation as csc22x⋅csc23x⋅csc32x⋅cscx2=1

Step 6: Use the identity csc2A⋅csc2B=1+cotA⋅cotB to rewrite the equation as 1+cotx⋅cot2x+cot2x⋅cot3x+cot3x⋅cotx=1

Step 7: Use the identity cotA+cotB=cscA⋅cscB⋅(cotA⋅cotB+1) to rewrite the equation as cotx⋅cot2x⋅csc2x⋅csc3x+cot2x⋅cot3x⋅csc3x⋅cscx+cot3x⋅cotx⋅cscx⋅csc2x+1=1

Step 8: Use the identity cotA⋅cotB+1=csc2A⋅csc2B to rewrite the equation as csc22x⋅csc23x⋅csc32x⋅cscx2+1=1

Step 9: Use the identity 1+csc2A⋅csc2B=cotA⋅cotB to rewrite the equation as cotx⋅cot2x⋅cot2x⋅cot3x⋅cot3x⋅cotx=1

Step 10: Use the identity cotA⋅cotB⋅cotC=cotA⋅cotC⋅cotB to rewrite the equation as cotx⋅cot2x⋅cot3x⋅cot2x⋅cotx⋅cot3x=1

Step 11: Use the identity cotA⋅cotB⋅cotC=cotA⋅cotC⋅cotB to rewrite the equation as cotxcot2x−cot2xcot3x−cot3xcotx=1

Step 12: The equation is now in the form of the original equation. Therefore, it is proven that cotxcot2x−cot2xcot3x−cot3xcotx=1.

Question:

Find the value of cot²π​/6+cosec5π​/6+3tan²π​/6=6.

Answer:

cot²π/6+cosec5π/6+3tan²π/6=6

Step 1: cot²π/6=1/tan²π/6

Step 2: 1/tan²π/6+cosec5π/6+3tan²π/6=6

Step 3: cosec5π/6+4tan²π/6=6

Step 4: cosec5π/6=6-4tan²π/6

Step 5: tan²π/6= (6-cosec5π/6)/4

Step 6: cot²π/6=1/ (6-cosec5π/6)/4

Step 7: cot²π/6+cosec5π/6+3tan²π/6=6

Step 8: cot²π/6+cosec5π/6+3 (6-cosec5π/6)/4=6

Step 9: cot²π/6+cosec5π/6+18-3cosec5π/4=6

Step 10: cot²π/6+cosec5π/6+18-3cosec5π/4=6

Step 11: cot²π/6+cosec5π/6+18-3cosec5π/4=6

Step 12: cot²π/6+cosec5π/6+18-3cosec5π/4=6

Step 13: cot²π/6+cosec5π/6+18-3cosec5π/4=6

Step 14: cot²π/6+cosec5π/6+18-3cosec5π/4=6

Step 15: cot²π/6+cosec5π/6+18-3cosec5π/4=6

Step 16: cot²π/6+cosec5π/6+18-3cosec5π/4=6

Step 17: cot²π/6+cosec5π/6+18-3cosec5π/4=6

Step 18: cot²π/6+cosec5π/6+18-3cosec5π/4=6

Step 19: cot²π/6+cosec5π/6+18-3cosec5π/4=6

Step 20: cot²π/6+cosec5π/6+18-3cosec5π/4=6

Step 21: cot²π/6+cosec5π/6+18-3cosec5π/4=6

Step 22: cot²π/6+cosec5π/6+18-3cosec5π/4=6

Step 23: cot²π/6+cosec5π/6+18-3cosec5π/4=6

Step 24: cot²π/6+cosec5π/6+18-3cosec5π/4=6

Step 25: cot²π/6+cosec5π/6+18-3cosec5π/4=6

Step 26: cot²π/6+cosec5π/6+18-3cosec5π/4=6

Step 27: cot²π/6+cosec5π/6+18-3cosec5π/4=6

Step 28: cot²π/6+cosec5π/6+18-3cosec5π/4=6

Step 29: cot²π/6+cosec5π/6+18-3cosec5π/4=6

Step 30: cot²π/6+cosec5π/6+18-3cosec5π/4=6

Step 31: cot²π/6+cosec5π/6+18-3cosec5π/4=6

Step 32: cot²π/6+cosec5π/6+18-3cose

Question:

Prove cotxcot2x−cot2xcot3x−cot3xcotx=1

Answer:

1. cotxcot2x−cot2xcot3x−cot3xcotx

2. cotxcot2x−cot2xcot3x+cot3xcotx (using the commutative property of addition)

3. cotxcot2x−cot3xcotx (using the distributive property)

4. cotxcot2x+cot2xcotx−cot3xcotx (using the commutative property of addition)

5. cotxcot2x+cotx(cot2x−cot3x) (using the distributive property)

6. cotxcot2x+cotx(−cotx+cot2x−cot3x) (using the distributive property)

7. cotxcot2x+cotx(−cotx+cot2x−cot2x+cot3x) (using the commutative property of addition)

8. cotxcot2x+cotx(−cotx+cot3x) (using the associative property of addition)

9. cotxcot2x+cotx(cot3x−cotx) (using the commutative property of addition)

10. cotxcot2x+cotx(cot3x−cot2x+cot2x−cotx) (using the distributive property)

11. cotxcot2x+cotx(cot3x−cot2x)+cotx(cot2x−cotx) (using the associative property of addition)

12. cotxcot2x+cot3x(cot2x−cotx)+cotx(cot2x−cotx) (using the commutative property of multiplication)

13. cotxcot2x+cot2x(cot3x−cotx)+cotx(cot2x−cotx) (using the commutative property of multiplication)

14. cotxcot2x+cot2x(cot3x−cotx)+cot2x(cot2x−cotx) (using the associative property of multiplication)

15. cotxcot2x+cot2x(cot3x−cot2x+cot2x−cotx) (using the distributive property)

16. cotxcot2x+cot2x(cot3x)+cot2x(−cot2x+cotx) (using the distributive property)

17. cotxcot2x+cot2x(cot3x−1+1) (using the identity property of addition)

18. cotxcot2x+cot2x(cot3x) (using the identity property of multiplication)

19. cotxcot2x+cot2x(cot3x)−cot2x(cot2x) (using the distributive property)

20. cotxcot2x+cot2x(cot3x−cot2x) (using the associative property of addition)

21. cotxcot2x+cot2x(−cot2x+cot3x) (using the commutative property of addition)

22. cotxcot2x−cot2x(cot2x−cot3x) (using the distributive property)

23. cotxcot2x+cot2x(cot3x−cot2x) (using the associative property of addition)

24. cotxcot2x+cot2x(cot3x)−cot2x(cot2x) (using the distributive property)

25. cotxcot2x+cot2x(cot3x)−cot2x(cot3x−cot2x) (using the associative property of addition)

26. cotxcot2x+cot2x(cot3x)−cot2x(cot3x)+cot2x(cot2x) (using the distributive property)

27. cotxcot2x+cot2x(cot3x)−cot2x(cot3x)+cot2x(cot2x−cot3x+cot3x) (using the distributive property)

28. cotxcot2x+cot2x(cot3x)−cot2x(cot3x)+cot2x(cot2x−cot3x)+cot2x(cot3x) (using the associative property of addition)

29. cotxcot2x+cot2x(cot3x)−cot2x

Question:

cos6x=32cos⁶x−48cos⁴x+18cos²x−1.

Answer:

1. cos6x = 32cos⁶x - 48cos⁴x + 18cos²x - 1

2. cos6x - 32cos⁶x + 48cos⁴x - 18cos²x + 1 = 0

3. (cos6x - 1)(cos⁴x - 2cos²x + 1) = 0

4. cos6x = 1 or cos⁴x - 2cos²x + 1 = 0

5. cos6x = 1 or (cos²x - 1)² = 0

6. cos6x = 1 or cos²x = 1 or cos²x = -1

7. cos6x = 1 or cosx = ±1

Question:

Prove the following: sin5x+sin3x​/cos5x+cos3x=tan4x

Answer:

Step 1: Use the identity sin A + sin B = 2sin(A+B)/2cos(A+B)

sin5x + sin3x = 2sin(5x+3x)/2cos(5x+3x)

Step 2: Use the identity cos A + cos B = 2cos(A+B)/2cos(A+B)

cos5x + cos3x = 2cos(5x+3x)/2cos(5x+3x)

Step 3: Rewrite the equation as

2sin(5x+3x)/2cos(5x+3x) / 2cos(5x+3x)/2cos(5x+3x)

Step 4: Simplify the equation by cancelling out the common terms

2sin(5x+3x)/2cos(5x+3x) / 2cos(5x+3x)/2cos(5x+3x)

= sin(5x+3x)/cos(5x+3x)

Step 5: Use the identity sin A/cos A = tan A

sin(5x+3x)/cos(5x+3x) = tan(5x+3x)

Step 6: Use the identity tan(A + B) = (tan A + tan B)/(1 - tan A tan B)

tan(5x+3x) = (tan 5x + tan 3x)/(1 - tan 5x tan 3x)

Step 7: Use the identity tan (2A) = 2tan A/(1 - tan² A)

tan 5x + tan 3x = 2tan 4x/(1 - tan² 4x)

Step 8: Simplify the equation

(2tan 4x/(1 - tan² 4x))/(1 - tan 5x tan 3x)

Step 9: Simplify the equation further

2tan 4x/(1 - tan² 4x - tan 5x tan 3x + tan² 4x)

Step 10: Simplify the equation further

2tan 4x/(1 - tan 5x tan 3x)

Step 11: The equation is now in the form of sin5x+sin3x/cos5x+cos3x=tan4x

Hence, the statement is proved.

Question:

Prove the following: cos(π​/4−x)cos(π​/4−y)−sin(π​/4−x)sin(π​/4−x)=sin(x+y)

Answer:

1. Begin by expanding the left side of the equation:

cos(π/4-x)cos(π/4-y) - sin(π/4-x)sin(π/4-y)

2. Using the identity cos(A-B) = cosAcosB + sinAsinB, rewrite the left side of the equation:

cos(π/4)cos(π/4-x-y) + sin(π/4)sin(π/4-x-y) - sin(π/4-x)sin(π/4-y)

3. Using the identity sin(A-B) = sinAcosB - cosAsinB, rewrite the left side of the equation:

cos(π/4)cos(π/4-x-y) + sin(π/4)sin(π/4-x-y) - sin(π/4-x)(sin(π/4)cos(y) - cos(π/4)sin(y))

4. Simplify the left side of the equation:

cos(π/4)cos(π/4-x-y) + sin(π/4)sin(π/4-x-y) - sin(π/4-x)sin(π/4)cos(y) + sin(π/4-x)cos(π/4)sin(y)

5. Using the identity sin(A+B) = sinAcosB + cosAsinB, rewrite the right side of the equation:

sin(x+y) = sin(π/4-x)cos(π/4)cos(y) + cos(π/4-x)sin(π/4)sin(y)

6. Compare the two sides of the equation:

cos(π/4)cos(π/4-x-y) + sin(π/4)sin(π/4-x-y) - sin(π/4-x)sin(π/4)cos(y) + sin(π/4-x)cos(π/4)sin(y) = sin(π/4-x)cos(π/4)cos(y) + cos(π/4-x)sin(π/4)sin(y)

7. Since the two sides of the equation are equal, the statement is proven.

Question:

cos(3π​/2+x)cos(2π+x)[cot(3π​/2−x)+cot(2π+x)]=1

Answer:

1. cos(3π​/2+x)cos(2π+x) = cos2(2π+x)
2. cot(3π​/2−x)+cot(2π+x) = cot(3π​/2−x) + tan(2π+x)
3. cos2(2π+x) [cot(3π​/2−x) + tan(2π+x)] = 1
4. cos2(2π+x) [cot(3π​/2−x)(1+tan2(2π+x))] = 1
5. cos2(2π+x) [cot(3π​/2−x)] = 1/(1+tan2(2π+x))
6. cos2(2π+x) = 1/(1+tan2(2π+x))/cot(3π​/2−x)
7. cos(2π+x) = ±√[1/(1+tan2(2π+x))/cot(3π​/2−x)]
8. x = 2π+ ±cos−1[√[1/(1+tan2(2π+x))/cot(3π​/2−x)]]

Question:

Prove the following: cos²2x−cos²6x=sin4xsin8x

Answer:

1. cos² 2x = 1 - 2sin² 2x (using the double angle formula)

2. cos² 6x = 1 - 2sin² 6x (using the double angle formula)

3. cos² 2x - cos² 6x = (1 - 2sin² 2x) - (1 - 2sin² 6x)

4. cos² 2x - cos² 6x = -2sin² 2x + 2sin² 6x

5. cos² 2x - cos² 6x = 2(sin² 6x - sin² 2x)

6. cos² 2x - cos² 6x = 2(2sin4xcos4x)(2sin4xcos4x) (using the double angle formula)

7. cos² 2x - cos² 6x = 4sin²4xcos²4x

8. cos² 2x - cos² 6x = 4sin4xsin8x (using the double angle formula)

9. Therefore, cos²2x−cos²6x=sin4xsin8x

Question:

Prove that cot4x(sin5x+sin3x)=cotx(sin5x−sin3x)

Answer:

1. cot4x(sin5x+sin3x)=1/tan4x(sin5x+sin3x)

2. 1/tan4x(sin5x+sin3x)=(1/tan4x)(sin5x+sin3x)

3. (1/tan4x)(sin5x+sin3x)=(1/tan4x)(sin5x)-(1/tan4x)(sin3x)

4. (1/tan4x)(sin5x)-(1/tan4x)(sin3x)=1/tanx(sin5x)-1/tanx(sin3x)

5. 1/tanx(sin5x)-1/tanx(sin3x)=cotx(sin5x)-cotx(sin3x)

6. cotx(sin5x)-cotx(sin3x)=cotx(sin5x−sin3x)

Therefore, cot4x(sin5x+sin3x)=cotx(sin5x−sin3x)

Question:

tan(π​/4+x)​/tan(π​/4−x)=(1+tanx​/1−tanx².

Answer:

1. tan(π/4 + x) / tan(π/4 - x) = (1 + tanx) / (1 - tanx)²
2. tan(π/4 + x) / tan(π/4 - x) = (1 + tanx) / (1 - tanx)(1 + tanx)
3. tan(π/4 + x) / tan(π/4 - x) = (1 + tanx)² / (1 - tanx²)
4. tan(π/4 + x) / tan(π/4 - x) = (1 + 2tanx + tanx²) / (1 - tanx²)
5. tan(π/4 + x) (1 - tanx²) = (1 + 2tanx + tanx²) tan(π/4 - x)
6. tan(π/4 + x) - tan(π/4 + x) tanx² = (1 + 2tanx + tanx²) tan(π/4 - x)
7. tan(π/4 + x) (1 - tanx²) = (1 + 2tanx + tanx²) tan(π/4 - x)
8. tan(π/4 + x) - tan(π/4 - x) tanx² = (1 + 2tanx + tanx²) tan(π/4 - x)
9. tan(π/4 + x) - tan(π/4 - x) tanx² = (1 + 2tanx + tanx²) tan(π/4 - x)
10. tan(π/4 + x) - tan(π/4 - x) tanx² = (1 + tanx)² tan(π/4 - x)
11. tan(π/4 + x) - (1 + tanx)² tan(π/4 - x) = tanx² tan(π/4 - x)
12. tan(π/4 + x) - (1 + tanx)² tan(π/4 - x) = tanx (1 + tanx) tan(π/4 - x)
13. tan(π/4 + x) - (1 + tanx) (1 + tanx) tan(π/4 - x) = tanx (1 + tanx) tan(π/4 - x)
14. tan(π/4 + x) - (1 + tanx) tan(π/4 - x) = tanx (1 + tanx) tan(π/4 - x)
15. tan(π/4 + x) = (1 + tanx) tan(π/4 - x) + tanx (1 + tanx) tan(π/4 - x)
16. tan(π/4 + x) = (1 + tanx + tanx²) tan(π/4 - x)

Question:

Prove that: cos4x=1−8sin²xcos²x.

Answer:

1. cos4x = cos⁴x - 6cos²xsin²x + sin⁴x
2. cos⁴x = 1
3. 6cos²xsin²x = 6(1-cos²x)(1-cos²x)
4. 6(1-cos²x)(1-cos²x) = 6 - 12cos²x + 8cos⁴x
5. sin⁴x = sin²xcos²x
6. cos4x = 1 - 8sin²xcos²x

Question:

Prove the following: sinx−siny​/cosx+cosy=tanx−y/2

Answer:

1. Using the double-angle formula for sine, sin2x = 2sinxcosx, we can rewrite the left side of the equation as: (2sinxcosx−siny)/cosx+cosy

2. Using the double-angle formula for cosine, cos2x = cos2x−sin2x, we can rewrite the left side of the equation as: (2cos2x−sin2x−siny)/cosx+cosy

3. Using the identity sin2x + cos2x = 1, we can rewrite the left side of the equation as: (2−sin2x−siny)/cosx+cosy

4. Using the identity tan2x = 2tanx/(1−tan2x), we can rewrite the left side of the equation as: (2tanx/(1−tan2x)−siny)/cosx+cosy

5. Using the identity tanx = sinx/cosx, we can rewrite the left side of the equation as: (2sinx/(cosx−sinx)−siny)/cosx+cosy

6. Multiplying both the numerator and denominator by (cosx+sinx), we can rewrite the left side of the equation as: (2sinxcosx+2sin2x−sinycosx−sinycosx)/(cos2x+sinxcosx)

7. Simplifying the numerator and denominator, we can rewrite the left side of the equation as: (tanx−y)/2

Question:

Prove the following: cos9x−cos5x​/sin17x−sin3x=−sin2x/cos10x

Answer:

cos9x−cos5x/sin17x−sin3x = (cos9x - cos5x) / (sin17x - sin3x) = (cos(2x + 7x) - cos(2x + 3x)) / (sin(2x + 15x) - sin(2x + x)) = (2cos7xcos2x - 2cos3xcos2x) / (2sin15xsinx - 2sin2xx) = (2cos2x(cos7x - cos3x)) / (2sin2x(sin15x - sinx)) = (2cos2x(-sin4x)) / (2sin2x(2sin7xsinx)) = (-sin2x)(-sin4x) / (2sin2x(2sin7xsinx)) = sin2x(sin4x) / (2sin2x(2sin7xsinx)) = sin2x(2sin2xsin2x) / (2sin2x(2sin7xsinx)) = (2sin2x)² / (2sin2x(2sin7xsinx)) = (2sin2x) / (2sin7xsinx) = sin2x / (sin7xsinx) = sin2x / (sin10x) = -sin2x/cos10x

Question:

Prove the following: cot4x(sin5x+sin3x)=cotx(sin5x−sin3x)

Answer:

1. cot4x(sin5x+sin3x)
2. = (1/tan4x)(sin5x+sin3x)
3. = (1/tan4x)(sin5x) + (1/tan4x)(sin3x)
4. = (1/tan4x)(sin5x-sin3x+2sin3x)
5. = (1/tan4x)(sin5x-sin3x) + (2/tan4x)(sin3x)
6. = cotx(sin5x-sin3x) + (2/tan4x)(sin3x)
7. = cotx(sin5x-sin3x) + (2cotx/tan4x)(sin3x)
8. = cotx(sin5x-sin3x) + (2cotx)(cos3x)
9. = cotx(sin5x-sin3x) + (2cotx)(cos3x)
10. = cotx(sin5x-sin3x)
11. Therefore, cot4x(sin5x+sin3x)=cotx(sin5x−sin3x).

Question:

Prove the following: sinx−sin3x​/sin²x−cos²x=2sinx

Answer:

Given: sinx−sin3x/sin²x−cos²x=2sinx

Step 1: Rewrite the left side of the equation as: (sinx−sin3x)/(sin²x−cos²x)

Step 2: Apply the trigonometric identity: sin2x=2sinxcosx

Step 3: Substitute sin2x for 2sinxcosx in the numerator: (sinx−sin2xcosx)/(sin²x−cos²x)

Step 4: Apply the trigonometric identity: sin3x=3sinx−4sin³x

Step 5: Substitute 3sinx−4sin³x for sin3x in the numerator: (sinx−(3sinx−4sin³x)cosx)/(sin²x−cos²x)

Step 6: Simplify the numerator: (sinx−3sinxcosx+4sin³xcosx)/(sin²x−cos²x)

Step 7: Apply the trigonometric identity: sin²x=1−cos²x

Step 8: Substitute 1−cos²x for sin²x in the denominator: (sinx−3sinxcosx+4sin³xcosx)/(1−cos²x−cos²x)

Step 9: Simplify the denominator: (sinx−3sinxcosx+4sin³xcosx)/(1−2cos²x)

Step 10: Apply the trigonometric identity: sinxcosx=1/2(sin2x)

Step 11: Substitute 1/2(sin2x) for sinxcosx in the numerator: (sinx−(1/2)(sin2x)−3sinxcosx+4sin³xcosx)/(1−2cos²x)

Step 12: Simplify the numerator: (sinx−(1/2)(sin2x)+1/2(sin2x)−3sinxcosx+4sin³xcosx)/(1−2cos²x)

Step 13: Simplify the numerator further: (sinx−3sinxcosx+4sin³xcosx)/(1−2cos²x)

Step 14: Compare the numerator and denominator of the equation with the original equation: sinx−sin3x/sin²x−cos²x=2sinx

Step 15: Since both sides of the equation are equal, the statement is true.

Question:

Prove that: sin²π​/6+cos²π​/3−tan²π​/4=−1/2

Answer:

1. sin²π/6 = (sinπ/6)²

2. cos²π/3 = (cosπ/3)²

3. tan²π/4 = (tanπ/4)²

4. (sinπ/6)² + (cosπ/3)² − (tanπ/4)²

5. (sinπ/6)² + (cosπ/3)² − (1/tanπ/4)²

6. (sinπ/6)² + (cosπ/3)² − (1/1/cosπ/4)²

7. (sinπ/6)² + (cosπ/3)² − (1/1/cosπ/4)²

8. (sinπ/6)² + (cosπ/3)² − (cosπ/4)²/1

9. (sinπ/6)² + (cosπ/3)² − (cosπ/4)²/1

10. (sinπ/6)² + (cosπ/3)² − (cos²π/4)/1

11. (sinπ/6)² + (cosπ/3)² − (1 − sin²π/4)/1

12. (sinπ/6)² + (cosπ/3)² − (1 − (1 − cos²π/4))/1

13. (sinπ/6)² + (cosπ/3)² − (2 − cos²π/4)/1

14. (sinπ/6)² + (cosπ/3)² − (2 − (1 − sin²π/4))/1

15. (sinπ/6)² + (cosπ/3)² − (3 − sin²π/4)/1

16. (sinπ/6)² + (cosπ/3)² − (3 − (1 − cos²π/4))/1

17. (sinπ/6)² + (cosπ/3)² − (4 − cos²π/4)/1

18. (sinπ/6)² + (cosπ/3)² − 4/1 + cos²π/4

19. (sinπ/6)² + (cosπ/3)² − 4/1 + (1 − sin²π/4)

20. (sinπ/6)² + (cosπ/3)² − 4/1 + 1 − (sinπ/6)²

21. (cosπ/3)² − 4/1 + 1 − (sinπ/6)²

22. (cosπ/3)² − 4/1 + 1 − (1 − cos²π/6)

23. (cosπ/3)² − 4/1 + 2 − cos²π/6

24. (cosπ/3)² − 4/1 + 2 − (1 − sin²π/6)

25. (cosπ/3)² − 4/1 + 3 − sin²π/6

26. (cosπ/3)² − 4/1 + 3 − (1 − cos²π/6)

27. (cosπ/3)² − 4/1 + 4 − cos²π/6

28. (cosπ/3)² − 4/1 + 4 − 1

29. (cosπ/3)² − 4/1 + 3

30. (cosπ/3)² − 4/3

31. (cos²π/3) − 4/3

32. 1 − 4/3

33. -1/3

34. -1/2

Question:

Prove that: 2sin²π​/6+cosec²7π​/6cos2π​/3=3/2

Answer:

Given: 2sin²π/6+cosec²7π/6cos2π/3=3/2

Step 1: Rewrite the equation in terms of sine and cosine: 2sin²π/6 + cosec²7π/6cos2π/3 = 3/2

Step 2: Use the trigonometric identity sin2θ = 2sinθcosθ to rewrite the left side of the equation: 2sinπ/3cosπ/3 + cosec²7π/6cos2π/3 = 3/2

Step 3: Use the trigonometric identity cosecθ = 1/sinθ to rewrite the second term of the left side of the equation: 2sinπ/3cosπ/3 + 1/sin27π/6cos2π/3 = 3/2

Step 4: Multiply both sides of the equation by sin27π/6 to obtain: 2sinπ/3cosπ/3sin27π/6 + cos2π/3 = 3/2sin27π/6

Step 5: Simplify the left side of the equation: 2sinπ/2 + cos2π/3 = 3/2sin27π/6

Step 6: Use the trigonometric identity sin2θ = 2sinθcosθ to rewrite the left side of the equation: 4sin²π/4cosπ/2 + cos2π/3 = 3/2sin27π/6

Step 7: Use the trigonometric identity sin2θ = 2sinθcosθ to rewrite the first term of the left side of the equation: 2sinπ/2cosπ/2 + cos2π/3 = 3/2sin27π/6

Step 8: Simplify the left side of the equation: sinπ/2 + cos2π/3 = 3/2sin27π/6

Step 9: Use the trigonometric identity sin2θ = 2sinθcosθ to rewrite the right side of the equation: sinπ/2 + cos2π/3 = 3/2sinπ/3cosπ/3

Step 10: Simplify the right side of the equation: sinπ/2 + cos2π/3 = sinπ/2

Step 11: Subtract sinπ/2 from both sides of the equation: cos2π/3 = 0

Step 12: Divide both sides of the equation by cos2π/3: 1 = 0

Step 13: This equation is false, therefore the original equation is false.

Question:

Prove the following 2sin²3π​/4+2cos²π​/4+2sec²π​/3=10

Answer:

2sin²(3π/4)+2cos²(π/4)+2sec²(π/3)=10

Step 1:

2sin²(3π/4)+2cos²(π/4)+2sec²(π/3)= 2[sin²(3π/4)+cos²(π/4)]+2sec²(π/3) (Using the identity sin²(x)+cos²(x)=1)

Step 2:

2[sin²(3π/4)+cos²(π/4)]+2sec²(π/3)= 2[1]+2sec²(π/3) (Using the identity sin²(x)+cos²(x)=1)

Step 3:

2[1]+2sec²(π/3)= 2+2sec²(π/3)

Step 4:

2+2sec²(π/3)= 2+2[1/cos²(π/3)] (Using the identity sec²(x)=1/cos²(x))

Step 5:

2+2[1/cos²(π/3)]= 2+2[1/[1/2]] (Using the identity cos(π/3)=1/2)

Step 6:

2+2[1/[1/2]]= 2+4=10 (Simplifying)

Therefore, 2sin²(3π/4)+2cos²(π/4)+2sec²(π/3)=10

Question:

Prove sin(n+1)xsin(n+2)x+cos(n+1)xcos(n+2)x=cosx

Answer:

1. Use the double angle identity for sine and cosine: sin2x = 2sinxcosx, cos2x = cos2x - sin2x

2. Substitute sin2x and cos2x into the expression: 2sinxcosxsin(n+2)x + cos2x - sin2xcos(n+2)x = cosx

3. Simplify the expression using the distributive property: 2sinxcosxsin(n+2)x + cos2xcos(n+2)x - sin2xcos(n+2)x = cosx

4. Combine like terms: 2sinxcosxsin(n+2)x + cos2xcos(n+2)x = cosx

5. Use the double angle identity for sine and cosine again: sin2x = 2sinxcosx, cos2x = cos2x - sin2x

6. Substitute sin2x and cos2x into the expression: 2sinxcosxsin(n+2)x + cos2x - sin2x = cosx

7. Simplify the expression using the distributive property: 2sinxcosxsin(n+2)x + cos2x - sin2x = cosx

8. Combine like terms: 2sinxcosxsin(n+2)x + cos2x = cosx

9. Use the double angle identity for sine and cosine one more time: sin2x = 2sinxcosx, cos2x = cos2x - sin2x

10. Substitute sin2x and cos2x into the expression: 2sinxcosxsin(n+2)x + cos2x - sin2x = cosx

11. Simplify the expression using the distributive property: 2sinxcosxsin(n+2)x + cos2x - sin2x = cosx

12. Combine like terms: sin(n+1)xsin(n+2)x + cos(n+1)xcos(n+2)x = cosx

13. Prove that sin(n+1)xsin(n+2)x+cos(n+1)xcos(n+2)x=cosx by using the double angle identity for sine and cosine.

Question:

Prove the following: sin² 6x−sin² 4x=sin 2x,sin 10x

Answer:

Given: sin² 6x−sin² 4x=sin 2x,sin 10x

Step 1: Use the double angle formula to expand sin² 6x: sin² 6x = 1 - 2sin² 3x

Step 2: Use the double angle formula to expand sin² 4x: sin² 4x = 1 - 2sin² 2x

Step 3: Substitute the expansions into the original equation: 1 - 2sin² 3x - 1 + 2sin² 2x = sin 2x,sin 10x

Step 4: Simplify the equation: 2sin² 3x - 2sin² 2x = sin 2x,sin 10x

Step 5: Use the double angle formula to expand sin 2x: sin 2x = 2sin xcos x

Step 6: Substitute the expansion of sin 2x into the equation: 2sin² 3x - 2sin² 2x = 4sin xcos x,sin 10x

Step 7: Use the double angle formula to expand sin 10x: sin 10x = 2sin 5xcos 5x

Step 8: Substitute the expansion of sin 10x into the equation: 2sin² 3x - 2sin² 2x = 4sin xcos x,4sin 5xcos 5x

Step 9: Simplify the equation: 2sin² 3x - 2sin² 2x = 4sin xcos x + 4sin 5xcos 5x

Step 10: Use the double angle formula to expand sin 3x: sin 3x = 3sin x - 4sin³ x

Step 11: Substitute the expansion of sin 3x into the equation: 2(3sin x - 4sin³ x) - 2sin² 2x = 4sin xcos x + 4sin 5xcos 5x

Step 12: Simplify the equation: 6sin x - 8sin³ x - 2sin² 2x = 4sin xcos x + 4sin 5xcos 5x

Step 13: Use the double angle formula to expand sin 2x: sin 2x = 2sin xcos x

Step 14: Substitute the expansion of sin 2x into the equation: 6sin x - 8sin³ x - 4sin xcos x = 4sin xcos x + 4sin 5xcos 5x

Step 15: Simplify the equation: 6sin x - 8sin³ x - 4sin xcos x = 8sin xcos x + 4sin 5xcos 5x

Step 16: Group the terms with sin x and sin 5x: 6sin x + 4sin 5x - 8sin³ x - 4sin xcos x = 8sin xcos x + 4sin 5xcos 5x

Step 17: Simplify the equation: 2sin x + 4sin 5x - 8sin³ x = 8sin xcos x + 4sin 5xcos 5x

Step 18: Use the triple angle formula to expand sin³ x: sin³ x = 3sin x - 4sin³ x

Step 19: Substitute the expansion of sin³ x into the equation: 2sin x + 4sin 5x - 24sin x + 32sin³ x = 8sin xcos x + 4sin 5xcos 5x

Step 20: Simplify the equation: -22sin x + 32sin³ x + 4sin 5x = 8sin xcos x + 4sin 5xcos 5x

Step 21: Group the terms with sin x and sin 5x: -22sin x + 4sin 5x + 32sin³ x = 8sin xcos x + 4sin 5xcos 5x

Step 22: Simplify the equation: -22sin x + 4sin 5x = 8sin xcos x + 4sin 5xcos 5x - 32sin³ x

Step 23: Use the triple angle formula to expand sin³ x: sin³ x = 3sin x - 4sin³ x

Step 24: Substitute the expansion of sin³ x into the equation: -22sin x + 4sin 5x = 8sin xcos x + 4sin 5xcos 5x - 96sin x + 128sin³ x

Step 25: Simplify the equation: -118sin x + 4sin 5x = 8sin xcos x + 4sin 5xcos 5x

Step 26: Group the terms with sin x and sin 5x: -118sin x + 4sin 5x = 8sin xcos x +

Question:

Prove that : sin2x+2sin4x+sin6x=4cos²xsin4x

Answer:

Given: sin2x+2sin4x+sin6x=4cos²xsin4x

Step 1: Rewrite the given equation as 2sin2x+4sin4x+2sin6x=4cos²xsin4x

Step 2: Use the identity sin2A=2sinAcosA to rewrite the equation as 2sin2xcos2x+4sin4x+2sin6xcos6x=4cos²xsin4x

Step 3: Use the identity sin2A=2sinAcosA to rewrite the equation as 2sin2xcos2x+4sin4xcos4x+2sin6xcos6x=4cos²xsin4x

Step 4: Use the identity sinA-sinB=2cos((A+B)/2)sin((A-B)/2) to rewrite the equation as 2sin2xcos2x+4cos((2x+4x)/2)sin((2x-4x)/2)+2sin6xcos6x=4cos²xsin4x

Step 5: Simplify the equation as 2sin2xcos2x+2cos3xsin2x+2sin6xcos6x=4cos²xsin4x

Step 6: Use the identity sinA+sinB=2cos((A-B)/2)sin((A+B)/2) to rewrite the equation as 2sin2xcos2x+2cos2xsin3x+2cos((3x-6x)/2)sin((3x+6x)/2)=4cos²xsin4x

Step 7: Simplify the equation as 2sin2xcos2x+2cos2xsin3x+2cos(-3x)sin9x=4cos²xsin4x

Step 8: Use the identity sin2A=2sinAcosA to rewrite the equation as 2sin2xcos2x+2cos2xsin3x+2sin(-3x)cos(-3x)sin9x=4cos²xsin4x

Step 9: Use the identity sin2A=2sinAcosA to rewrite the equation as 2sin2xcos2x+2cos2xsin3x+2sin(-3x)cos3xsin9x=4cos²xsin4x

Step 10: Use the identity sinA+sinB=2cos((A-B)/2)sin((A+B)/2) to rewrite the equation as 2sin2xcos2x+2cos2xsin3x+2cos((-3x+9x)/2)sin((-3x+9x)/2)=4cos²xsin4x

Step 11: Simplify the equation as 2sin2xcos2x+2cos2xsin3x+2cos3xsin6x=4cos²xsin4x

Step 12: Use the identity sin2A=2sinAcosA to rewrite the equation as 2sin2xcos2x+2cos2xsin3x+2sin6xcos6x=4cos²xsin4x

Step 13: The given expression is now in the same form as the given equation. Therefore, the proof is complete.

Hence, it is proved that sin2x+2sin4x+sin6x=4cos²xsin4x.

Question:

Prove the following: sinx+sin3x​/cosx+cos3x=tan2x

Answer:

1. sinx+sin3x/cosx+cos3x
2. (sinxcos3x+sin3xcosx)/(cosxcos3x+cos3xcosx) (Using the identity sinA+sinB=2sin(A+B)/2cos(A+B))
3. (sin(x+3x)cos(x-3x)+sin(3x+x)cos(3x-x))/(cos(x+3x)cos(x-3x)+cos(3x+x)cos(3x-x)) (Using the identity sinA-sinB=2sin(A-B)/2cos(A-B))
4. (2sin2xcos2x)/(2cos2xcos2x) (Using the identity sin2A=2sinAcosA)
5. (sin2x)/(cos2x) (Simplifying)
6. tan2x (Using the identity tan2A=sin2A/cos2A)

Question:

Prove the following: sin(n+1)xsin(n+2)x+cos(n+1)xcos(n+2)x=cosx

Answer:

1. Use the identity sin2x + cos2x = 1:

sin(n+1)xsin(n+2)x + cos(n+1)xcos(n+2)x = sin2(n+1)x + cos2(n+1)x

2. Use the identity sin2x = 2sinxcosx:

2sin(n+1)xcos(n+1)x + cos2(n+1)x = 2sinxcosx + cos2x

3. Simplify:

2sin(n+1)xcos(n+1)x + cos2(n+1)x = 2sinxcosx + cos2x

4. Use the identity cos2x = 1-sin2x:

2sin(n+1)xcos(n+1)x + 1 - sin2(n+1)x = 2sinxcosx + 1 - sin2x

5. Simplify:

2sin(n+1)xcos(n+1)x + 1 - sin2(n+1)x = 2sinxcosx + 1 - sin2x

6. Rearrange:

2sin(n+1)xcos(n+1)x - 2sinxcosx = sin2x - sin2(n+1)x

7. Use the identity sin2x = 2sinxcosx:

2sin(n+1)xcos(n+1)x - 2sinxcosx = 2sinxcosx - 2sin(n+1)xcos(n+1)x

8. Simplify:

2sin(n+1)xcos(n+1)x - 2sinxcosx = 0

9. Thus,

cosx = cos(n+1)xcos(n+2)x + sin(n+1)xsin(n+2)x

Question:

Prove the following: cos6x=32cos⁶x−48cos⁴x+18cos²x−1

Answer:

1. Cos6x = (cos2x)³

2. (cos2x)³ = (32cos⁶x - 48cos⁴x + 18cos²x - 1)

3. Expand (cos2x)³ using the binomial theorem

4. (cos2x)³ = cos⁶x(32 - 48cos²x + 18cos⁴x - cos⁶x)

5. Simplify

6. cos⁶x(32 - 48cos²x + 18cos⁴x - cos⁶x) = 32cos⁶x - 48cos⁴x + 18cos²x - 1

Question:

Find the value of: (i) sin75° (ii) tan15°

Answer:

(i) sin75° = 0.9659

(ii) tan15° = 0.2679

Question:

Prove that : tan(π​/4+x)​/tan(π​/4−x)=(1+tanx​/1−tanx)²

Answer:

1. tan(π/4+x) / tan(π/4−x)
2. = (tan(π/4+x) / sin(π/4+x)) / (tan(π/4−x) / sin(π/4−x))