Solve from 1 to 12

Mathematics

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Here are the solutions for problems 1 to 12.

: Prove sin⁡x1+cos⁡x+1+cos⁡xsin⁡x=2sin⁡x\frac{\sin x}{1+\cos x} + \frac{1+\cos x}{\sin x} = \frac{2}{\sin x}

  1. Combine the fractions on the left-hand side (LHS) using a common denominator.

    LHS=sin⁡2x+(1+cos⁡x)2(1+cos⁡x)sin⁡x\text{LHS} = \frac{\sin^2 x + (1+\cos x)^2}{(1+\cos x)\sin x}

  2. Expand the numerator.

    LHS=sin⁡2x+1+2cos⁡x+cos⁡2x(1+cos⁡x)sin⁡x\text{LHS} = \frac{\sin^2 x + 1 + 2\cos x + \cos^2 x}{(1+\cos x)\sin x}

  3. Use the identity sin⁡2x+cos⁡2x=1\sin^2 x + \cos^2 x = 1.

    LHS=1+1+2cos⁡x(1+cos⁡x)sin⁡x=2+2cos⁡x(1+cos⁡x)sin⁡x\text{LHS} = \frac{1 + 1 + 2\cos x}{(1+\cos x)\sin x} = \frac{2 + 2\cos x}{(1+\cos x)\sin x}

  4. Factor out 2 from the numerator and cancel the common term.

    LHS=2(1+cos⁡x)(1+cos⁡x)sin⁡x=2sin⁡x\text{LHS} = \frac{2(1+\cos x)}{(1+\cos x)\sin x} = \frac{2}{\sin x}

    This matches the right-hand side (RHS).

    2sin⁡x\boxed{\frac{2}{\sin x}}

    : Prove sin⁡x1−cos⁡x=1+cos⁡xsin⁡x\frac{\sin x}{1-\cos x} = \frac{1+\cos x}{\sin x}

  5. Start with the left-hand side (LHS) and multiply the numerator and denominator by the conjugate of the denominator, (1+cos⁡x)(1+\cos x).

    LHS=sin⁡x1−cos⁡x×1+cos⁡x1+cos⁡x\text{LHS} = \frac{\sin x}{1-\cos x} \times \frac{1+\cos x}{1+\cos x}

  6. Multiply the terms in the numerator and denominator.

    LHS=sin⁡x(1+cos⁡x)12−cos⁡2x\text{LHS} = \frac{\sin x (1+\cos x)}{1^2 - \cos^2 x}

  7. Use the identity 1−cos⁡2x=sin⁡2x1 - \cos^2 x = \sin^2 x.

    LHS=sin⁡x(1+cos⁡x)sin⁡2x\text{LHS} = \frac{\sin x (1+\cos x)}{\sin^2 x}

  8. Cancel out sin⁡x\sin x from the numerator and denominator.

    LHS=1+cos⁡xsin⁡x\text{LHS} = \frac{1+\cos x}{\sin x}

    This matches the right-hand side (RHS).

    1+cos⁡xsin⁡x\boxed{\frac{1+\cos x}{\sin x}}

    : Prove sin⁡x1−cos⁡x−1−cos⁡xsin⁡x=2cot⁡x\frac{\sin x}{1-\cos x} - \frac{1-\cos x}{\sin x} = 2\cot x

  9. Combine the fractions on the left-hand side (LHS) using a common denominator.

    LHS=sin⁡2x−(1−cos⁡x)2(1−cos⁡x)sin⁡x\text{LHS} = \frac{\sin^2 x - (1-\cos x)^2}{(1-\cos x)\sin x}

  10. Expand the numerator.

    LHS=sin⁡2x−(1−2cos⁡x+cos⁡2x)(1−cos⁡x)sin⁡x=sin⁡2x−1+2cos⁡x−cos⁡2x(1−cos⁡x)sin⁡x\text{LHS} = \frac{\sin^2 x - (1 - 2\cos x + \cos^2 x)}{(1-\cos x)\sin x} = \frac{\sin^2 x - 1 + 2\cos x - \cos^2 x}{(1-\cos x)\sin x}

  11. Use the identity sin⁡2x−1=−cos⁡2x\sin^2 x - 1 = -\cos^2 x.

    LHS=−cos⁡2x+2cos⁡x−cos⁡2x(1−cos⁡x)sin⁡x=2cos⁡x−2cos⁡2x(1−cos⁡x)sin⁡x\text{LHS} = \frac{-\cos^2 x + 2\cos x - \cos^2 x}{(1-\cos x)\sin x} = \frac{2\cos x - 2\cos^2 x}{(1-\cos x)\sin x}

  12. Factor out 2cos⁡x2\cos x from the numerator.

    LHS=2cos⁡x(1−cos⁡x)(1−cos⁡x)sin⁡x\text{LHS} = \frac{2\cos x (1-\cos x)}{(1-\cos x)\sin x}

  13. Cancel out the common term (1−cos⁡x)(1-\cos x).

    LHS=2cos⁡xsin⁡x=2cot⁡x\text{LHS} = \frac{2\cos x}{\sin x} = 2\cot x

    This matches the right-hand side (RHS).

    2cot⁡x\boxed{2\cot x}

    : Prove 1+sin⁡xcos⁡x+cos⁡x1+sin⁡x=2cos⁡x\frac{1+\sin x}{\cos x} + \frac{\cos x}{1+\sin x} = \frac{2}{\cos x}

  14. Combine the fractions on the left-hand side (LHS) using a common denominator.

    LHS=(1+sin⁡x)2+cos⁡2xcos⁡x(1+sin⁡x)\text{LHS} = \frac{(1+\sin x)^2 + \cos^2 x}{\cos x (1+\sin x)}

  15. Expand the numerator.

    LHS=1+2sin⁡x+sin⁡2x+cos⁡2xcos⁡x(1+sin⁡x)\text{LHS} = \frac{1 + 2\sin x + \sin^2 x + \cos^2 x}{\cos x (1+\sin x)}

  16. Use the identity sin⁡2x+cos⁡2x=1\sin^2 x + \cos^2 x = 1.

    LHS=1+2sin⁡x+1cos⁡x(1+sin⁡x)=2+2sin⁡xcos⁡x(1+sin⁡x)\text{LHS} = \frac{1 + 2\sin x + 1}{\cos x (1+\sin x)} = \frac{2 + 2\sin x}{\cos x (1+\sin x)}

  17. Factor out 2 from the numerator and cancel the common term.

    LHS=2(1+sin⁡x)cos⁡x(1+sin⁡x)=2cos⁡x\text{LHS} = \frac{2(1+\sin x)}{\cos x (1+\sin x)} = \frac{2}{\cos x}

    This matches the right-hand side (RHS).

    2cos⁡x\boxed{\frac{2}{\cos x}}

    : Prove 1+sin⁡xcos⁡x−cos⁡x1+sin⁡x=2tan⁡x\frac{1+\sin x}{\cos x} - \frac{\cos x}{1+\sin x} = 2\tan x

  18. Combine the fractions on the left-hand side (LHS) using a common denominator.

    LHS=(1+sin⁡x)2−cos⁡2xcos⁡x(1+sin⁡x)\text{LHS} = \frac{(1+\sin x)^2 - \cos^2 x}{\cos x (1+\sin x)}

  19. Expand the numerator.

    LHS=1+2sin⁡x+sin⁡2x−cos⁡2xcos⁡x(1+sin⁡x)\text{LHS} = \frac{1 + 2\sin x + \sin^2 x - \cos^2 x}{\cos x (1+\sin x)}

  20. Use the identity cos⁡2x=1−sin⁡2x\cos^2 x = 1 - \sin^2 x.

    LHS=1+2sin⁡x+sin⁡2x−(1−sin⁡2x)cos⁡x(1+sin⁡x)=1+2sin⁡x+sin⁡2x−1+sin⁡2xcos⁡x(1+sin⁡x)\text{LHS} = \frac{1 + 2\sin x + \sin^2 x - (1 - \sin^2 x)}{\cos x (1+\sin x)} = \frac{1 + 2\sin x + \sin^2 x - 1 + \sin^2 x}{\cos x (1+\sin x)}

  21. Simplify the numerator.

    LHS=2sin⁡x+2sin⁡2xcos⁡x(1+sin⁡x)\text{LHS} = \frac{2\sin x + 2\sin^2 x}{\cos x (1+\sin x)}

  22. Factor out 2sin⁡x2\sin x from the numerator and cancel the common term.

    LHS=2sin⁡x(1+sin⁡x)cos⁡x(1+sin⁡x)=2sin⁡xcos⁡x\text{LHS} = \frac{2\sin x (1+\sin x)}{\cos x (1+\sin x)} = \frac{2\sin x}{\cos x}

  23. Express the result in terms of tan⁡x\tan x.

    LHS=2tan⁡x\text{LHS} = 2\tan x

    This matches the right-hand side (RHS).

    2tan⁡x\boxed{2\tan x}

    : Prove cos⁡x1+sin⁡x+tan⁡x=sec⁡x\frac{\cos x}{1+\sin x} + \tan x = \sec x

  24. Start with the left-hand side (LHS) and express tan⁡x\tan x as sin⁡xcos⁡x\frac{\sin x}{\cos x}.

    LHS=cos⁡x1+sin⁡x+sin⁡xcos⁡x\text{LHS} = \frac{\cos x}{1+\sin x} + \frac{\sin x}{\cos x}

  25. Combine the fractions using a common denominator.

    LHS=cos⁡2x+sin⁡x(1+sin⁡x)(1+sin⁡x)cos⁡x\text{LHS} = \frac{\cos^2 x + \sin x (1+\sin x)}{(1+\sin x)\cos x}

  26. Expand the numerator.

    LHS=cos⁡2x+sin⁡x+sin⁡2x(1+sin⁡x)cos⁡x\text{LHS} = \frac{\cos^2 x + \sin x + \sin^2 x}{(1+\sin x)\cos x}

  27. Use the identity sin⁡2x+cos⁡2x=1\sin^2 x + \cos^2 x = 1.

    LHS=1+sin⁡x(1+sin⁡x)cos⁡x\text{LHS} = \frac{1 + \sin x}{(1+\sin x)\cos x}

  28. Cancel out the common term (1+sin⁡x)(1+\sin x).

    LHS=1cos⁡x=sec⁡x\text{LHS} = \frac{1}{\cos x} = \sec x

    This matches the right-hand side (RHS).

    sec⁡x\boxed{\sec x}

    : Prove cos⁡x1−sin⁡x−tan⁡x=sec⁡x\frac{\cos x}{1-\sin x} - \tan x = \sec x

  29. Start with the left-hand side (LHS) and express tan⁡x\tan x as sin⁡xcos⁡x\frac{\sin x}{\cos x}.

    LHS=cos⁡x1−sin⁡x−sin⁡xcos⁡x\text{LHS} = \frac{\cos x}{1-\sin x} - \frac{\sin x}{\cos x}

  30. Combine the fractions using a common denominator.

    LHS=cos⁡2x−sin⁡x(1−sin⁡x)(1−sin⁡x)cos⁡x\text{LHS} = \frac{\cos^2 x - \sin x (1-\sin x)}{(1-\sin x)\cos x}

  31. Expand the numerator.

    LHS=cos⁡2x−sin⁡x+sin⁡2x(1−sin⁡x)cos⁡x\text{LHS} = \frac{\cos^2 x - \sin x + \sin^2 x}{(1-\sin x)\cos x}

  32. Use the identity sin⁡2x+cos⁡2x=1\sin^2 x + \cos^2 x = 1.

    LHS=1−sin⁡x(1−sin⁡x)cos⁡x\text{LHS} = \frac{1 - \sin x}{(1-\sin x)\cos x}

  33. Cancel out the common term (1−sin⁡x)(1-\sin x).

    LHS=1cos⁡x=sec⁡x\text{LHS} = \frac{1}{\cos x} = \sec x

    This matches the right-hand side (RHS).

    sec⁡x\boxed{\sec x}

    : Prove sin⁡x1+cos⁡x+cot⁡x=csc⁡x\frac{\sin x}{1+\cos x} + \cot x = \csc x

  34. Start with the left-hand side (LHS) and express cot⁡x\cot x as cos⁡xsin⁡x\frac{\cos x}{\sin x}.

    LHS=sin⁡x1+cos⁡x+cos⁡xsin⁡x\text{LHS} = \frac{\sin x}{1+\cos x} + \frac{\cos x}{\sin x}

  35. Combine the fractions using a common denominator.

    LHS=sin⁡2x+cos⁡x(1+cos⁡x)(1+cos⁡x)sin⁡x\text{LHS} = \frac{\sin^2 x + \cos x (1+\cos x)}{(1+\cos x)\sin x}

  36. Expand the numerator.

    LHS=sin⁡2x+cos⁡x+cos⁡2x(1+cos⁡x)sin⁡x\text{LHS} = \frac{\sin^2 x + \cos x + \cos^2 x}{(1+\cos x)\sin x}

  37. Use the identity sin⁡2x+cos⁡2x=1\sin^2 x + \cos^2 x = 1.

    LHS=1+cos⁡x(1+cos⁡x)sin⁡x\text{LHS} = \frac{1 + \cos x}{(1+\cos x)\sin x}

  38. Cancel out the common term (1+cos⁡x)(1+\cos x).

    LHS=1sin⁡x=csc⁡x\text{LHS} = \frac{1}{\sin x} = \csc x

    This matches the right-hand side (RHS).

    csc⁡x\boxed{\csc x}

    : Prove sin⁡x1−cos⁡x−cot⁡x=csc⁡x\frac{\sin x}{1-\cos x} - \cot x = \csc x

  39. Start with the left-hand side (LHS) and express cot⁡x\cot x as cos⁡xsin⁡x\frac{\cos x}{\sin x}.

    LHS=sin⁡x1−cos⁡x−cos⁡xsin⁡x\text{LHS} = \frac{\sin x}{1-\cos x} - \frac{\cos x}{\sin x}

  40. Combine the fractions using a common denominator.

    LHS=sin⁡2x−cos⁡x(1−cos⁡x)(1−cos⁡x)sin⁡x\text{LHS} = \frac{\sin^2 x - \cos x (1-\cos x)}{(1-\cos x)\sin x}

  41. Expand the numerator.

    LHS=sin⁡2x−cos⁡x+cos⁡2x(1−cos⁡x)sin⁡x\text{LHS} = \frac{\sin^2 x - \cos x + \cos^2 x}{(1-\cos x)\sin x}

  42. Use the identity sin⁡2x+cos⁡2x=1\sin^2 x + \cos^2 x = 1.

    LHS=1−cos⁡x(1−cos⁡x)sin⁡x\text{LHS} = \frac{1 - \cos x}{(1-\cos x)\sin x}

  43. Cancel out the common term (1−cos⁡x)(1-\cos x).

    LHS=1sin⁡x=csc⁡x\text{LHS} = \frac{1}{\sin x} = \csc x

    This matches the right-hand side (RHS).

    csc⁡x\boxed{\csc x}

    : Prove 1+sin⁡x1−sin⁡x=(sec⁡x+tan⁡x)2\frac{1+\sin x}{1-\sin x} = (\sec x + \tan x)^2

  44. Start with the right-hand side (RHS) and express sec⁡x\sec x and tan⁡x\tan x in terms of sin⁡x\sin x and cos⁡x\cos x.

    RHS=(1cos⁡x+sin⁡xcos⁡x)2\text{RHS} = \left(\frac{1}{\cos x} + \frac{\sin x}{\cos x}\right)^2

  45. Combine the terms inside the parenthesis.

    RHS=(1+sin⁡xcos⁡x)2\text{RHS} = \left(\frac{1+\sin x}{\cos x}\right)^2

  46. Square the numerator and the denominator.

    RHS=(1+sin⁡x)2cos⁡2x\text{RHS} = \frac{(1+\sin x)^2}{\cos^2 x}

  47. Use the identity cos⁡2x=1−sin⁡2x\cos^2 x = 1 - \sin^2 x.

    RHS=(1+sin⁡x)21−sin⁡2x\text{RHS} = \frac{(1+\sin x)^2}{1 - \sin^2 x}

  48. Factor the denominator using the difference of squares formula, a2−b2=(a−b)(a+b)a^2 - b^2 = (a-b)(a+b).

    RHS=(1+sin⁡x)2(1−sin⁡x)(1+sin⁡x)\text{RHS} = \frac{(1+\sin x)^2}{(1-\sin x)(1+\sin x)}

  49. Cancel out one term of (1+sin⁡x)(1+\sin x) from the numerator and denominator.

    RHS=1+sin⁡x1−sin⁡x\text{RHS} = \frac{1+\sin x}{1-\sin x}

    This matches the left-hand side (LHS).

    1+sin⁡x1−sin⁡x\boxed{\frac{1+\sin x}{1-\sin x}}

    : Prove 1−cos⁡x1+cos⁡x=(csc⁡x−cot⁡x)2\frac{1-\cos x}{1+\cos x} = (\csc x - \cot x)^2

  50. Start with the right-hand side (RHS) and express csc⁡x\csc x and cot⁡x\cot x in terms of sin⁡x\sin x and cos⁡x\cos x.

    RHS=(1sin⁡x−cos⁡xsin⁡x)2\text{RHS} = \left(\frac{1}{\sin x} - \frac{\cos x}{\sin x}\right)^2

  51. Combine the terms inside the parenthesis.

    RHS=(1−cos⁡xsin⁡x)2\text{RHS} = \left(\frac{1-\cos x}{\sin x}\right)^2

  52. Square the numerator and the denominator.

    RHS=(1−cos⁡x)2sin⁡2x\text{RHS} = \frac{(1-\cos x)^2}{\sin^2 x}

  53. Use the identity sin⁡2x=1−cos⁡2x\sin^2 x = 1 - \cos^2 x.

    RHS=(1−cos⁡x)21−cos⁡2x\text{RHS} = \frac{(1-\cos x)^2}{1 - \cos^2 x}

  54. Factor the denominator using the difference of squares formula, a2−b2=(a−b)(a+b)a^2 - b^2 = (a-b)(a+b).

    RHS=(1−cos⁡x)2(1−cos⁡x)(1+cos⁡x)\text{RHS} = \frac{(1-\cos x)^2}{(1-\cos x)(1+\cos x)}

  55. Cancel out one term of (1−cos⁡x)(1-\cos x) from the numerator and denominator.

    RHS=1−cos⁡x1+cos⁡x\text{RHS} = \frac{1-\cos x}{1+\cos x}

    This matches the left-hand side (LHS).

    1−cos⁡x1+cos⁡x\boxed{\frac{1-\cos x}{1+\cos x}}

    : Prove sin⁡x1+cos⁡x=1−cos⁡xsin⁡x\frac{\sin x}{1+\cos x} = \frac{1-\cos x}{\sin x}

  56. Start with the left-hand side (LHS) and multiply the numerator and denominator by the conjugate of the denominator, (1−cos⁡x)(1-\cos x).

    LHS=sin⁡x1+cos⁡x×1−cos⁡x1−cos⁡x\text{LHS} = \frac{\sin x}{1+\cos x} \times \frac{1-\cos x}{1-\cos x}

  57. Multiply the terms in the numerator and denominator.

    LHS=sin⁡x(1−cos⁡x)12−cos⁡2x\text{LHS} = \frac{\sin x (1-\cos x)}{1^2 - \cos^2 x}

  58. Use the identity 1−cos⁡2x=sin⁡2x1 - \cos^2 x = \sin^2 x.

    LHS=sin⁡x(1−cos⁡x)sin⁡2x\text{LHS} = \frac{\sin x (1-\cos x)}{\sin^2 x}

  59. Cancel out sin⁡x\sin x from the numerator and denominator.

    LHS=1−cos⁡xsin⁡x\text{LHS} = \frac{1-\cos x}{\sin x}

    This matches the right-hand side (RHS).

    1−cos⁡xsin⁡x\boxed{\frac{1-\cos x}{\sin x}}
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