Solve from 1 to 12

Mathematics

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Answer

2sin⁡x\frac{2}{\sin x}

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}

Step 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⁡xLHS = \frac{\sin^2 x + (1+\cos x)^2}{(1+\cos x)\sin x}

Step 2: Expand the numerator. LHS=sin⁡2x+1+2cos⁡x+cos⁡2x(1+cos⁡x)sin⁡xLHS = \frac{\sin^2 x + 1 + 2\cos x + \cos^2 x}{(1+\cos x)\sin x}

Step 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⁡xLHS = \frac{1 + 1 + 2\cos x}{(1+\cos x)\sin x} = \frac{2 + 2\cos x}{(1+\cos x)\sin x}

Step 4: Factor out 2 from the numerator and cancel the common term. LHS=2(1+cos⁡x)(1+cos⁡x)sin⁡x=2sin⁡xLHS = \frac{2(1+\cos x)}{(1+\cos x)\sin x} = \frac{2}{\sin x} This matches the right-hand side (RHS). \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}

Step 1: 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⁡xLHS = \frac{\sin x}{1-\cos x} \times \frac{1+\cos x}{1+\cos x}

Step 2: Multiply the terms in the numerator and denominator. LHS=sin⁡x(1+cos⁡x)12−cos⁡2xLHS = \frac{\sin x (1+\cos x)}{1^2 - \cos^2 x}

Step 3: Use the identity 1−cos⁡2x=sin⁡2x1 - \cos^2 x = \sin^2 x. LHS=sin⁡x(1+cos⁡x)sin⁡2xLHS = \frac{\sin x (1+\cos x)}{\sin^2 x}

Step 4: Cancel out sin⁡x\sin x from the numerator and denominator. LHS=1+cos⁡xsin⁡xLHS = \frac{1+\cos x}{\sin x} This matches the right-hand side (RHS). \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

Step 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⁡xLHS = \frac{\sin^2 x - (1-\cos x)^2}{(1-\cos x)\sin x}

Step 2: 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⁡xLHS = \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}

Step 3: 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⁡xLHS = \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}

Step 4: Factor out 2cos⁡x2\cos x from the numerator. LHS=2cos⁡x(1−cos⁡x)(1−cos⁡x)sin⁡xLHS = \frac{2\cos x (1-\cos x)}{(1-\cos x)\sin x}

Step 5: Cancel out the common term (1−cos⁡x)(1-\cos x). LHS=2cos⁡xsin⁡x=2cot⁡xLHS = \frac{2\cos x}{\sin x} = 2\cot x This matches the right-hand side (RHS). 2cot⁡x2\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}

Step 1: Combine the fractions on the left-hand side (LHS) using a common denominator. LHS=(1+sin⁡x)2+cos⁡2xcos⁡x(1+sin⁡x)LHS = \frac{(1+\sin x)^2 + \cos^2 x}{\cos x (1+\sin x)}

Step 2: Expand the numerator. LHS=1+2sin⁡x+sin⁡2x+cos⁡2xcos⁡x(1+sin⁡x)LHS = \frac{1 + 2\sin x + \sin^2 x + \cos^2 x}{\cos x (1+\sin x)}

Step 3: 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)LHS = \frac{1 + 2\sin x + 1}{\cos x (1+\sin x)} = \frac{2 + 2\sin x}{\cos x (1+\sin x)}

Step 4: Factor out 2 from the numerator and cancel the common term. LHS=2(1+sin⁡x)cos⁡x(1+sin⁡x)=2cos⁡xLHS = \frac{2(1+\sin x)}{\cos x (1+\sin x)} = \frac{2}{\cos x} This matches the right-hand side (RHS). \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

Step 1: Combine the fractions on the left-hand side (LHS) using a common denominator. LHS=(1+sin⁡x)2−cos⁡2xcos⁡x(1+sin⁡x)LHS = \frac{(1+\sin x)^2 - \cos^2 x}{\cos x (1+\sin x)}

Step 2: Expand the numerator. LHS=1+2sin⁡x+sin⁡2x−cos⁡2xcos⁡x(1+sin⁡x)LHS = \frac{1 + 2\sin x + \sin^2 x - \cos^2 x}{\cos x (1+\sin x)}

Step 3: 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)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)}

Step 4: Simplify the numerator. LHS=2sin⁡x+2sin⁡2xcos⁡x(1+sin⁡x)LHS = \frac{2\sin x + 2\sin^2 x}{\cos x (1+\sin x)}

Step 5: 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⁡xLHS = \frac{2\sin x (1+\sin x)}{\cos x (1+\sin x)} = \frac{2\sin x}{\cos x}

Step 6: Express the result in terms of tan⁡x\tan x. LHS=2tan⁡xLHS = 2\tan x This matches the right-hand side (RHS). 2tan⁡x2\tan x

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

Step 1: 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⁡xLHS = \frac{\cos x}{1+\sin x} + \frac{\sin x}{\cos x}

Step 2: Combine the fractions using a common denominator. LHS=cos⁡2x+sin⁡x(1+sin⁡x)(1+sin⁡x)cos⁡xLHS = \frac{\cos^2 x + \sin x (1+\sin x)}{(1+\sin x)\cos x}

Step 3: Expand the numerator. LHS=cos⁡2x+sin⁡x+sin⁡2x(1+sin⁡x)cos⁡xLHS = \frac{\cos^2 x + \sin x + \sin^2 x}{(1+\sin x)\cos x}

Step 4: Use the identity sin⁡2x+cos⁡2x=1\sin^2 x + \cos^2 x = 1. LHS=1+sin⁡x(1+sin⁡x)cos⁡xLHS = \frac{1 + \sin x}{(1+\sin x)\cos x}

Step 5: Cancel out the common term (1+sin⁡x)(1+\sin x). LHS=1cos⁡x=sec⁡xLHS = \frac{1}{\cos x} = \sec x This matches the right-hand side (RHS). sec⁡x\sec x

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

Step 1: 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⁡xLHS = \frac{\cos x}{1-\sin x} - \frac{\sin x}{\cos x}

Step 2: Combine the fractions using a common denominator. LHS=cos⁡2x−sin⁡x(1−sin⁡x)(1−sin⁡x)cos⁡xLHS = \frac{\cos^2 x - \sin x (1-\sin x)}{(1-\sin x)\cos x}

Step 3: Expand the numerator. LHS=cos⁡2x−sin⁡x+sin⁡2x(1−sin⁡x)cos⁡xLHS = \frac{\cos^2 x - \sin x + \sin^2 x}{(1-\sin x)\cos x}

Step 4: Use the identity sin⁡2x+cos⁡2x=1\sin^2 x + \cos^2 x = 1. LHS=1−sin⁡x(1−sin⁡x)cos⁡xLHS = \frac{1 - \sin x}{(1-\sin x)\cos x}

Step 5: Cancel out the common term (1−sin⁡x)(1-\sin x). LHS=1cos⁡x=sec⁡xLHS = \frac{1}{\cos x} = \sec x This matches the right-hand side (RHS). sec⁡x\sec x

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

Step 1: 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⁡xLHS = \frac{\sin x}{1+\cos x} + \frac{\cos x}{\sin x}

Step 2: Combine the fractions using a common denominator. LHS=sin⁡2x+cos⁡x(1+cos⁡x)(1+cos⁡x)sin⁡xLHS = \frac{\sin^2 x + \cos x (1+\cos x)}{(1+\cos x)\sin x}

Step 3: Expand the numerator. LHS=sin⁡2x+cos⁡x+cos⁡2x(1+cos⁡x)sin⁡xLHS = \frac{\sin^2 x + \cos x + \cos^2 x}{(1+\cos x)\sin x}

Step 4: Use the identity sin⁡2x+cos⁡2x=1\sin^2 x + \cos^2 x = 1. LHS=1+cos⁡x(1+cos⁡x)sin⁡xLHS = \frac{1 + \cos x}{(1+\cos x)\sin x}

Step 5: Cancel out the common term (1+cos⁡x)(1+\cos x). LHS=1sin⁡x=csc⁡xLHS = \frac{1}{\sin x} = \csc x This matches the right-hand side (RHS). csc⁡x\csc x

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

Step 1: 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⁡xLHS = \frac{\sin x}{1-\cos x} - \frac{\cos x}{\sin x}

Step 2: Combine the fractions using a common denominator. LHS=sin⁡2x−cos⁡x(1−cos⁡x)(1−cos⁡x)sin⁡xLHS = \frac{\sin^2 x - \cos x (1-\cos x)}{(1-\cos x)\sin x}

Step 3: Expand the numerator. LHS=sin⁡2x−cos⁡x+cos⁡2x(1−cos⁡x)sin⁡xLHS = \frac{\sin^2 x - \cos x + \cos^2 x}{(1-\cos x)\sin x}

Step 4: Use the identity sin⁡2x+cos⁡2x=1\sin^2 x + \cos^2 x = 1. LHS=1−cos⁡x(1−cos⁡x)sin⁡xLHS = \frac{1 - \cos x}{(1-\cos x)\sin x}

Step 5: Cancel out the common term (1−cos⁡x)(1-\cos x). LHS=1sin⁡x=csc⁡xLHS = \frac{1}{\sin x} = \csc x This matches the right-hand side (RHS). csc⁡x\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

Step 1: 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)2RHS = \left(\frac{1}{\cos x} + \frac{\sin x}{\cos x}\right)^2

Step 2: Combine the terms inside the parenthesis. RHS=(1+sin⁡xcos⁡x)2RHS = \left(\frac{1+\sin x}{\cos x}\right)^2

Step 3: Square the numerator and the denominator. RHS=(1+sin⁡x)2cos⁡2xRHS = \frac{(1+\sin x)^2}{\cos^2 x}

Step 4: Use the identity cos⁡2x=1−sin⁡2x\cos^2 x = 1 - \sin^2 x. RHS=(1+sin⁡x)21−sin⁡2xRHS = \frac{(1+\sin x)^2}{1 - \sin^2 x}

Step 5: 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)RHS = \frac{(1+\sin x)^2}{(1-\sin x)(1+\sin x)}

Step 6: Cancel out one term of (1+sin⁡x)(1+\sin x) from the numerator and denominator. RHS=1+sin⁡x1−sin⁡xRHS = \frac{1+\sin x}{1-\sin x} This matches the left-hand side (LHS). \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

Step 1: 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)2RHS = \left(\frac{1}{\sin x} - \frac{\cos x}{\sin x}\right)^2

Step 2: Combine the terms inside the parenthesis. RHS=(1−cos⁡xsin⁡x)2RHS = \left(\frac{1-\cos x}{\sin x}\right)^2

Step 3: Square the numerator and the denominator. RHS=(1−cos⁡x)2sin⁡2xRHS = \frac{(1-\cos x)^2}{\sin^2 x}

Step 4: Use the identity sin⁡2x=1−cos⁡2x\sin^2 x = 1 - \cos^2 x. RHS=(1−cos⁡x)21−cos⁡2xRHS = \frac{(1-\cos x)^2}{1 - \cos^2 x}

Step 5: 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)RHS = \frac{(1-\cos x)^2}{(1-\cos x)(1+\cos x)}

Step 6: Cancel out one term of (1−cos⁡x)(1-\cos x) from the numerator and denominator. RHS=1−cos⁡x1+cos⁡xRHS = \frac{1-\cos x}{1+\cos x} This matches the left-hand side (LHS). \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}

Step 1: 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⁡xLHS = \frac{\sin x}{1+\cos x} \times \frac{1-\cos x}{1-\cos x}

Step 2: Multiply the terms in the numerator and denominator. LHS=sin⁡x(1−cos⁡x)12−cos⁡2xLHS = \frac{\sin x (1-\cos x)}{1^2 - \cos^2 x}

Step 3: Use the identity 1−cos⁡2x=sin⁡2x1 - \cos^2 x = \sin^2 x. LHS=sin⁡x(1−cos⁡x)sin⁡2xLHS = \frac{\sin x (1-\cos x)}{\sin^2 x}

Step 4: Cancel out sin⁡x\sin x from the numerator and denominator. LHS=1−cos⁡xsin⁡xLHS = \frac{1-\cos x}{\sin x} This matches the right-hand side (RHS). \frac{1-\cos x{\sin x}}

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Quick Answer

Prove ( x)/(1+ x) + (1+ x)/( x) = (2)/( x) Step 1: Combine the fractions on the left-hand side (LHS) using a common denominator.

Solve from 1 to 12
Mathematics

This mathematics problem involves applying core mathematical principles and formulas. Below you will find a complete step-by-step solution with detailed explanations for each step, helping you understand not just the answer but the method behind it.

ScanSolve AI Answer
Here are the solutions for problems 1 to 12. Problem 1: Prove ( x)/(1+ x) + (1+ x)/( x) = (2)/( x) Step 1: Combine the fractions on the left-hand side (LHS) using a common denominator. LHS = (^2 x + (1+ x)^2)/((1+ x) x) Step 2: Expand the numerator. LHS = (^2 x + 1 + 2 x + ^2 x)/((1+ x) x) Step 3: Use the identity ^2 x + ^2 x = 1. LHS = (1 + 1 + 2 x)/((1+ x) x) = (2 + 2 x)/((1+ x) x) Step 4: Factor out 2 from the numerator and cancel the common term. LHS = (2(1+ x))/((1+ x) x) = (2)/( x) This matches the right-hand side (RHS). (2)/( x) Problem 2: Prove ( x)/(1- x) = (1+ x)/( x) Step 1: Start with the left-hand side (LHS) and multiply the numerator and denominator by the conjugate of the denominator, (1+ x). LHS = ( x)/(1- x) × (1+ x)/(1+ x) Step 2: Multiply the terms in the numerator and denominator. LHS = ( x (1+ x))/(1^2 - ^2 x) Step 3: Use the identity 1 - ^2 x = ^2 x. LHS = ( x (1+ x))/(^2 x) Step 4: Cancel out x from the numerator and denominator. LHS = (1+ x)/( x) This matches the right-hand side (RHS). (1+ x)/( x) Problem 3: Prove ( x)/(1- x) - (1- x)/( x) = 2 x Step 1: Combine the fractions on the left-hand side (LHS) using a common denominator. LHS = (^2 x - (1- x)^2)/((1- x) x) Step 2: Expand the numerator. LHS = (^2 x - (1 - 2 x + ^2 x))/((1- x) x) = (^2 x - 1 + 2 x - ^2 x)/((1- x) x) Step 3: Use the identity ^2 x - 1 = -^2 x. LHS = (-^2 x + 2 x - ^2 x)/((1- x) x) = (2 x - 2^2 x)/((1- x) x) Step 4: Factor out 2 x from the numerator. LHS = (2 x (1- x))/((1- x) x) Step 5: Cancel out the common term (1- x). LHS = (2 x)/( x) = 2 x This matches the right-hand side (RHS). 2 x Problem 4: Prove (1+ x)/( x) + ( x)/(1+ x) = (2)/( x) Step 1: Combine the fractions on the left-hand side (LHS) using a common denominator. LHS = ((1+ x)^2 + ^2 x)/( x (1+ x)) Step 2: Expand the numerator. LHS = (1 + 2 x + ^2 x + ^2 x)/( x (1+ x)) Step 3: Use the identity ^2 x + ^2 x = 1. LHS = (1 + 2 x + 1)/( x (1+ x)) = (2 + 2 x)/( x (1+ x)) Step 4: Factor out 2 from the numerator and cancel the common term. LHS = (2(1+ x))/( x (1+ x)) = (2)/( x) This matches the right-hand side (RHS). (2)/( x) Problem 5: Prove (1+ x)/( x) - ( x)/(1+ x) = 2 x Step 1: Combine the fractions on the left-hand side (LHS) using a common denominator. LHS = ((1+ x)^2 - ^2 x)/( x (1+ x)) Step 2: Expand the numerator. LHS = (1 + 2 x + ^2 x - ^2 x)/( x (1+ x)) Step 3: Use the identity ^2 x = 1 - ^2 x. LHS = (1 + 2 x + ^2 x - (1 - ^2 x))/( x (1+ x)) = (1 + 2 x + ^2 x - 1 + ^2 x)/( x (1+ x)) Step 4: Simplify the numerator. LHS = (2 x + 2^2 x)/( x (1+ x)) Step 5: Factor out 2 x from the numerator and cancel the common term. LHS = (2 x (1+ x))/( x (1+ x)) = (2 x)/( x) Step 6: Express the result in terms of x. LHS = 2 x This matches the right-hand side (RHS). 2 x Problem 6: Prove ( x)/(1+ x) + x = x Step 1: Start with the left-hand side (LHS) and express x as ( x)/( x). LHS = ( x)/(1+ x) + ( x)/( x) Step 2: Combine the fractions using a common denominator. LHS = (^2 x + x (1+ x))/((1+ x) x) Step 3: Expand the numerator. LHS = (^2 x + x + ^2 x)/((1+ x) x) Step 4: Use the identity ^2 x + ^2 x = 1. LHS = (1 + x)/((1+ x) x) Step 5: Cancel out the common term (1+ x). LHS = (1)/( x) = x This matches the right-hand side (RHS). x Problem 7: Prove ( x)/(1- x) - x = x Step 1: Start with the left-hand side (LHS) and express x as ( x)/( x). LHS = ( x)/(1- x) - ( x)/( x) Step 2: Combine the fractions using a common denominator. LHS = (^2 x - x (1- x))/((1- x) x) Step 3: Expand the numerator. LHS = (^2 x - x + ^2 x)/((1- x) x) Step 4: Use the identity ^2 x + ^2 x = 1. LHS = (1 - x)/((1- x) x) Step 5: Cancel out the common term (1- x). LHS = (1)/( x) = x This matches the right-hand side (RHS). x Problem 8: Prove ( x)/(1+ x) + x = x Step 1: Start with the left-hand side (LHS) and express x as ( x)/( x). LHS = ( x)/(1+ x) + ( x)/( x) Step 2: Combine the fractions using a common denominator. LHS = (^2 x + x (1+ x))/((1+ x) x) Step 3: Expand the numerator. LHS = (^2 x + x + ^2 x)/((1+ x) x) Step 4: Use the identity ^2 x + ^2 x = 1. LHS = (1 + x)/((1+ x) x) Step 5: Cancel out the common term (1+ x). LHS = (1)/( x) = x This matches the right-hand side (RHS). x Problem 9: Prove ( x)/(1- x) - x = x Step 1: Start with the left-hand side (LHS) and express x as ( x)/( x). LHS = ( x)/(1- x) - ( x)/( x) Step 2: Combine the fractions using a common denominator. LHS = (^2 x - x (1- x))/((1- x) x) Step 3: Expand the numerator. LHS = (^2 x - x + ^2 x)/((1- x) x) Step 4: Use the identity ^2 x + ^2 x = 1. LHS = (1 - x)/((1- x) x) Step 5: Cancel out the common term (1- x). LHS = (1)/( x) = x This matches the right-hand side (RHS). x Problem 10: Prove (1+ x)/(1- x) = ( x + x)^2 Step 1: Start with the right-hand side (RHS) and express x and x in terms of x and x. RHS = ((1)/( x) + ( x)/( x))^2 Step 2: Combine the terms inside the parenthesis. RHS = ((1+ x)/( x))^2 Step 3: Square the numerator and the denominator. RHS = ((1+ x)^2)/(^2 x) Step 4: Use the identity ^2 x = 1 - ^2 x. RHS = ((1+ x)^2)/(1 - ^2 x) Step 5: Factor the denominator using the difference of squares formula, a^2 - b^2 = (a-b)(a+b). RHS = ((1+ x)^2)/((1- x)(1+ x)) Step 6: Cancel out one term of (1+ x) from the numerator and denominator. RHS = (1+ x)/(1- x) This matches the left-hand side (LHS). (1+ x)/(1- x) Problem 11: Prove (1- x)/(1+ x) = ( x - x)^2 Step 1: Start with the right-hand side (RHS) and express x and x in terms of x and x. RHS = ((1)/( x) - ( x)/( x))^2 Step 2: Combine the terms inside the parenthesis. RHS = ((1- x)/( x))^2 Step 3: Square the numerator and the denominator. RHS = ((1- x)^2)/(^2 x) Step 4: Use the identity ^2 x = 1 - ^2 x. RHS = ((1- x)^2)/(1 - ^2 x) Step 5: Factor the denominator using the difference of squares formula, a^2 - b^2 = (a-b)(a+b). RHS = ((1- x)^2)/((1- x)(1+ x)) Step 6: Cancel out one term of (1- x) from the numerator and denominator. RHS = (1- x)/(1+ x) This matches the left-hand side (LHS). (1- x)/(1+ x) Problem 12: Prove ( x)/(1+ x) = (1- x)/( x) Step 1: Start with the left-hand side (LHS) and multiply the numerator and denominator by the conjugate of the denominator, (1- x). LHS = ( x)/(1+ x) × (1- x)/(1- x) Step 2: Multiply the terms in the numerator and denominator. LHS = ( x (1- x))/(1^2 - ^2 x) Step 3: Use the identity 1 - ^2 x = ^2 x. LHS = ( x (1- x))/(^2 x) Step 4: Cancel out x from the numerator and denominator. LHS = (1- x)/( x) This matches the right-hand side (RHS). (1- x)/( x) Send me the next one 📸