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Combine trigonometric identities and radical manipulation to rewrite the expression cos⁡(θ)1−sin⁡(θ)−tan⁡(θ)\frac{\cos(\theta)}{1 - \sin(\theta)} - \tan(\theta) in an equivalent form containing only a single basic trigonometric function, valid for all θ≠π2+kπ\theta \neq \frac{\pi}{2} + k\pi where kk is an integer.

Answer and steps

Final answer

sec⁡(θ)\sec(\theta)

Steps

01

Express all terms with sine and cosine

To simplify the expression, we can rewrite the tangent function in terms of sine and cosine, and then combine the terms over a common denominator.

Recall the quotient identity for tangent: tan⁡(θ)=sin⁡(θ)cos⁡(θ)\tan(\theta) = \frac{\sin(\theta)}{\cos(\theta)}

Substitute this into the original expression: cos⁡(θ)1−sin⁡(θ)−tan⁡(θ)=cos⁡(θ)1−sin⁡(θ)−sin⁡(θ)cos⁡(θ)\frac{\cos(\theta)}{1 - \sin(\theta)} - \tan(\theta) = \frac{\cos(\theta)}{1 - \sin(\theta)} - \frac{\sin(\theta)}{\cos(\theta)}

02

Combine the fractions over a common denominator

The common denominator for the two fractions is cos⁡(θ)(1−sin⁡(θ))\cos(\theta)(1 - \sin(\theta)). Multiply the numerator and denominator of each fraction accordingly: cos⁡(θ)⋅cos⁡(θ)(1−sin⁡(θ))cos⁡(θ)−sin⁡(θ)(1−sin⁡(θ))cos⁡(θ)(1−sin⁡(θ))=cos⁡2(θ)−sin⁡(θ)(1−sin⁡(θ))cos⁡(θ)(1−sin⁡(θ))\frac{\cos(\theta)\cdot \cos(\theta)}{(1 - \sin(\theta))\cos(\theta)} - \frac{\sin(\theta)(1 - \sin(\theta))}{\cos(\theta)(1 - \sin(\theta))} = \frac{\cos^2(\theta) - \sin(\theta)(1 - \sin(\theta))}{\cos(\theta)(1 - \sin(\theta))}

03

Expand and apply the Pythagorean identity

Expand the numerator: cos⁡2(θ)−sin⁡(θ)+sin⁡2(θ)\cos^2(\theta) - \sin(\theta) + \sin^2(\theta)

Group the squared terms using the Pythagorean identity sin⁡2(θ)+cos⁡2(θ)=1\sin^2(\theta) + \cos^2(\theta) = 1: (sin⁡2(θ)+cos⁡2(θ))−sin⁡(θ)=1−sin⁡(θ)(\sin^2(\theta) + \cos^2(\theta)) - \sin(\theta) = 1 - \sin(\theta)

04

Simplify the rational expression

Substitute the simplified numerator back into the fraction: 1−sin⁡(θ)cos⁡(θ)(1−sin⁡(θ))\frac{1 - \sin(\theta)}{\cos(\theta)(1 - \sin(\theta))}

For all valid θ≠π2+kπ\theta \neq \frac{\pi}{2} + k\pi, the factor 1−sin⁡(θ)1 - \sin(\theta) is non-zero. Dividing numerator and denominator by (1−sin⁡(θ))(1 - \sin(\theta)) gives: 1cos⁡(θ)=sec⁡(θ)\frac{1}{\cos(\theta)} = \sec(\theta)

Alternative method (Multiplying by the conjugate): You can also simplify the first term directly by multiplying the numerator and denominator by the conjugate (1+sin⁡(θ))(1 + \sin(\theta)): cos⁡(θ)(1+sin⁡(θ))(1−sin⁡(θ))(1+sin⁡(θ))=cos⁡(θ)(1+sin⁡(θ))1−sin⁡2(θ)=cos⁡(θ)(1+sin⁡(θ))cos⁡2(θ)=1+sin⁡(θ)cos⁡(θ)\frac{\cos(\theta)(1 + \sin(\theta))}{(1 - \sin(\theta))(1 + \sin(\theta))} = \frac{\cos(\theta)(1 + \sin(\theta))}{1 - \sin^2(\theta)} = \frac{\cos(\theta)(1 + \sin(\theta))}{\cos^2(\theta)} = \frac{1 + \sin(\theta)}{\cos(\theta)} Splitting this into two fractions gives 1cos⁡(θ)+sin⁡(θ)cos⁡(θ)=sec⁡(θ)+tan⁡(θ)\frac{1}{\cos(\theta)} + \frac{\sin(\theta)}{\cos(\theta)} = \sec(\theta) + \tan(\theta). Subtracting tan⁡(θ)\tan(\theta) yields sec⁡(θ)\sec(\theta).

Educational illustration for this question

Typical errors

OverviewACEpath Teacher's Diagnosis

01 A student incorrectly distributes the negative sign when multiplying -sin(θ) by (1 - sin(θ)), yielding -sin²(θ) instead of +sin²(θ).

Sample Student Work

Reviewed by ACEpath Teacher
Step-by-Step Review
Walk Through It with Your Teacher
Incorrect
  1. Step 1

    cos⁡(θ)1−sin⁡(θ)−sin⁡(θ)cos⁡(θ)\frac{\cos(\theta)}{1 - \sin(\theta)} - \frac{\sin(\theta)}{\cos(\theta)}

  2. Step 2

    = cos⁡2(θ)−sin⁡(θ)−sin⁡2(θ)(1−sin⁡(θ))cos⁡(θ)\frac{\cos^2(\theta) - \sin(\theta) - \sin^2(\theta)}{(1 - \sin(\theta))\cos(\theta)}

    Correct

    = cos⁡2(θ)−sin⁡(θ)(1−sin⁡(θ))(1−sin⁡(θ))cos⁡(θ)\frac{\cos^2(\theta) - \sin(\theta)(1 - \sin(\theta))}{(1 - \sin(\theta))\cos(\theta)} = cos⁡2(θ)−sin⁡(θ)+sin⁡2(θ)(1−sin⁡(θ))cos⁡(θ)\frac{\cos^2(\theta) - \sin(\theta) + \sin^2(\theta)}{(1 - \sin(\theta))\cos(\theta)}

    When combining over the common denominator, distributing the negative sign gives −sin⁡(θ)⋅(1−sin⁡(θ))=−sin⁡(θ)+sin⁡2(θ)-\sin(\theta) \cdot (1 - \sin(\theta)) = -\sin(\theta) + \sin^2(\theta), not −sin⁡(θ)−sin⁡2(θ)-\sin(\theta) - \sin^2(\theta).

  3. Step 3

    = cos⁡(2θ)−sin⁡(θ)(1−sin⁡(θ))cos⁡(θ)\frac{\cos(2\theta) - \sin(\theta)}{(1 - \sin(\theta))\cos(\theta)}

    Correct

    = 1−sin⁡(θ)(1−sin⁡(θ))cos⁡(θ)\frac{1 - \sin(\theta)}{(1 - \sin(\theta))\cos(\theta)} = 1cos⁡(θ)\frac{1}{\cos(\theta)} = \sec(θ\theta)

    Because of the sign error in the previous step, you grouped cos⁡2(θ)−sin⁡2(θ)\cos^2(\theta) - \sin^2(\theta) into cos⁡(2θ)\cos(2\theta) instead of combining sin⁡2(θ)+cos⁡2(θ)=1\sin^2(\theta) + \cos^2(\theta) = 1 to cancel the common factor.

WHAT YOU DID WELL
Compliments from Your Teacher

You correctly substituted tan⁡(θ)=sin⁡(θ)cos⁡(θ)\tan(\theta) = \frac{\sin(\theta)}{\cos(\theta)}.

You correctly identified the common denominator (1−sin⁡(θ))cos⁡(θ)(1 - \sin(\theta))\cos(\theta).

You correctly recognized the double-angle identity cos⁡(2θ)=cos⁡2(θ)−sin⁡2(θ)\cos(2\theta) = \cos^2(\theta) - \sin^2(\theta) on your resulting terms.

THE ERROR
Where Your ACEpath Teacher Spotted the Mistake
Sign Distribution Error

You wrote −sin⁡2(θ)-\sin^2(\theta) instead of +sin⁡2(θ)+\sin^2(\theta) when expanding −sin⁡(θ)(1−sin⁡(θ))-\sin(\theta)(1 - \sin(\theta)).

WHY IT HAPPENED
ACEpath Teacher's Diagnosis

Dropping the negative sign when multiplying a negative factor by a subtracted term inside parentheses: (−a)(b−c)=−ab+ac(-a)(b - c) = -ab + ac.

REASONING PATH
Your ACEpath teacher’s suggested approach
Parentheses Protection

  • Always write out the numerator subtraction with explicit grouping parentheses before expanding: A−B(C−D)=A−(BC−BD)A - B(C - D) = A - (BC - BD).

  • Write −sin⁡(θ)(1−sin⁡(θ))-\sin(\theta)(1 - \sin(\theta)) as a separate intermediate grouping step before distributing.

CORRECTED PATH
ACEpath Teacher's Demonstration

  1. Expand the numerator with correct signs: cos⁡2(θ)−sin⁡(θ)(1−sin⁡(θ))=cos⁡2(θ)−sin⁡(θ)+sin⁡2(θ)\cos^2(\theta) - \sin(\theta)(1 - \sin(\theta)) = \cos^2(\theta) - \sin(\theta) + \sin^2(\theta).

  2. Apply the Pythagorean identity sin⁡2(θ)+cos⁡2(θ)=1\sin^2(\theta) + \cos^2(\theta) = 1: numerator becomes 1−sin⁡(θ)1 - \sin(\theta).

  3. Simplify the fraction: 1−sin⁡(θ)(1−sin⁡(θ))cos⁡(θ)=1cos⁡(θ)\frac{1 - \sin(\theta)}{(1 - \sin(\theta))\cos(\theta)} = \frac{1}{\cos(\theta)}.

  4. Use the reciprocal identity: 1cos⁡(θ)=sec⁡(θ)\frac{1}{\cos(\theta)} = \sec(\theta).

Final answer

sec⁡(θ)\sec(\theta)

NEXT STEPS
Final Takeaways from Your Teacher

  • Practice keeping explicit grouping parentheses around numerators before expanding subtraction.

  • Look for opportunities to apply the Pythagorean identity sin⁡2(θ)+cos⁡2(θ)=1\sin^2(\theta) + \cos^2(\theta) = 1 whenever sums of squared trigonometric terms appear.

OverviewACEpath Teacher's Diagnosis

02 A student mistakenly equates 1 - sin(θ) to cos(θ) by misremembering the Pythagorean identity as a linear relation.

Sample Student Work

Reviewed by ACEpath Teacher
Step-by-Step Review
Walk Through It with Your Teacher
Incorrect
  1. Step 1

    cos⁡(θ)1−sin⁡(θ)−tan⁡(θ)⟹cos⁡(θ)cos⁡(θ)−tan⁡(θ)\frac{\cos(\theta)}{1 - \sin(\theta)} - \tan(\theta) \Longrightarrow \frac{\cos(\theta)}{\cos(\theta)} - \tan(\theta)

    Correct

    cos⁡(θ)(1+sin⁡(θ))(1−sin⁡(θ))(1+sin⁡(θ))\frac{\cos(\theta)(1 + \sin(\theta))}{(1 - \sin(\theta))(1 + \sin(\theta))} - \tan(θ\theta) = cos⁡(θ)(1+sin⁡(θ))1−sin⁡2(θ)\frac{\cos(\theta)(1 + \sin(\theta))}{1 - \sin^2(\theta)} - \tan(θ\theta)

    You replaced 1−sin⁡(θ)1 - \sin(\theta) in the denominator with cos⁡(θ)\cos(\theta). The Pythagorean identity is cos⁡2(θ)=1−sin⁡2(θ)\cos^2(\theta) = 1 - \sin^2(\theta), not a linear identity. Therefore, 1−sin⁡(θ)≠cos⁡(θ)1 - \sin(\theta) \neq \cos(\theta).

  2. Step 2

    cos⁡(θ)cos⁡(θ)−tan⁡(θ)=1−tan⁡(θ)\frac{\cos(\theta)}{\cos(\theta)} - \tan(\theta) = 1 - \tan(\theta)

    Correct

    sec⁡(θ)\sec(\theta)

    You simplified the ratio correctly from your previous line, but because the denominator substitution was invalid, the final expression of 1−tan⁡(θ)1 - \tan(\theta) is incorrect.

WHAT YOU DID WELL
Compliments from Your Teacher

You transcribed the initial expression accurately.

You correctly evaluated the quotient cos⁡(θ)cos⁡(θ)\frac{\cos(\theta)}{\cos(\theta)} as 11 in your final step.

THE ERROR
Where Your ACEpath Teacher Spotted the Mistake
Linear misapplication of the Pythagorean identity

Replacing the linear factor 1−sin⁡(θ)1 - \sin(\theta) directly with cos⁡(θ)\cos(\theta).

WHY IT HAPPENED
ACEpath Teacher's Diagnosis

You treated the Pythagorean relationship as linear (1−sin⁡(θ)=cos⁡(θ)1 - \sin(\theta) = \cos(\theta)) instead of quadratic (sin⁡2(θ)+cos⁡2(θ)=1\sin^2(\theta) + \cos^2(\theta) = 1). To create a squared term from 1−sin⁡(θ)1 - \sin(\theta), you must multiply by its conjugate 1+sin⁡(θ)1 + \sin(\theta) or convert tan⁡(θ)\tan(\theta) to sine and cosine and find a common denominator.

REASONING PATH
Your ACEpath teacher’s suggested approach
Check exponent on Pythagorean terms

  • sin⁡2(θ)+cos⁡2(θ)=1  ⟹  1−sin⁡2(θ)=cos⁡2(θ)\sin^2(\theta) + \cos^2(\theta) = 1 \implies 1 - \sin^2(\theta) = \cos^2(\theta), but 1−sin⁡(θ)≠cos⁡(θ)1 - \sin(\theta) \neq \cos(\theta) in general.

  • Whenever you see 1±sin⁡(θ)1 \pm \sin(\theta) or 1±cos⁡(θ)1 \pm \cos(\theta), do not replace it with a single function unless the term is already squared.

Conjugate multiplication strategy

  • (1−sin⁡(θ))(1+sin⁡(θ))=1−sin⁡2(θ)=cos⁡2(θ)(1 - \sin(\theta))(1 + \sin(\theta)) = 1 - \sin^2(\theta) = \cos^2(\theta).

  • Multiply the numerator and denominator by the conjugate (1+sin⁡(θ))(1 + \sin(\theta)) to turn the denominator into a single squared term.

CORRECTED PATH
ACEpath Teacher's Demonstration

  1. Multiply numerator and denominator of the first fraction by (1+sin⁡(θ))(1 + \sin(\theta)): cos⁡(θ)(1+sin⁡(θ))(1−sin⁡(θ))(1+sin⁡(θ))=cos⁡(θ)(1+sin⁡(θ))1−sin⁡2(θ)\frac{\cos(\theta)(1 + \sin(\theta))}{(1 - \sin(\theta))(1 + \sin(\theta))} = \frac{\cos(\theta)(1 + \sin(\theta))}{1 - \sin^2(\theta)}.

  2. Apply the true Pythagorean identity 1−sin⁡2(θ)=cos⁡2(θ)1 - \sin^2(\theta) = \cos^2(\theta): cos⁡(θ)(1+sin⁡(θ))cos⁡2(θ)=1+sin⁡(θ)cos⁡(θ)\frac{\cos(\theta)(1 + \sin(\theta))}{\cos^2(\theta)} = \frac{1 + \sin(\theta)}{\cos(\theta)}.

  3. Split the fraction into two terms: 1cos⁡(θ)+sin⁡(θ)cos⁡(θ)=sec⁡(θ)+tan⁡(θ)\frac{1}{\cos(\theta)} + \frac{\sin(\theta)}{\cos(\theta)} = \sec(\theta) + \tan(\theta).

  4. Subtract the remaining term: (sec⁡(θ)+tan⁡(θ))−tan⁡(θ)=sec⁡(θ)(\sec(\theta) + \tan(\theta)) - \tan(\theta) = \sec(\theta).

Final answer

sec⁡(θ)\sec(\theta)

NEXT STEPS
Final Takeaways from Your Teacher

  • Test 1−sin⁡(θ)1 - \sin(\theta) with a test angle like θ=0\theta = 0 to verify: 1−sin⁡(0)=11 - \sin(0) = 1, whereas cos⁡(0)=1\cos(0) = 1; but at θ=π6\theta = \frac{\pi}{6}, 1−sin⁡(π6)=12≠32=cos⁡(π6)1 - \sin(\frac{\pi}{6}) = \frac{1}{2} \neq \frac{\sqrt{3}}{2} = \cos(\frac{\pi}{6}).

  • Practice using conjugate multiplication whenever a denominator has the binomial form 1±sin⁡(θ)1 \pm \sin(\theta) or 1±cos⁡(θ)1 \pm \cos(\theta).

OverviewACEpath Teacher's Diagnosis

03 A student incorrectly attempts to cancel terms between the numerator and denominator before factoring, subtracting 1 and sin(θ) invalidly.

Sample Student Work

Reviewed by ACEpath Teacher
Step-by-Step Review
Walk Through It with Your Teacher
Incorrect
  1. Step 1

    cos⁡(θ)1−sin⁡(θ)−sin⁡(θ)cos⁡(θ)\frac{\cos(\theta)}{1 - \sin(\theta)} - \frac{\sin(\theta)}{\cos(\theta)}

  2. Step 2

    = cos⁡2(θ)−sin⁡(θ)+sin⁡2(θ)(1−sin⁡(θ))cos⁡(θ)\frac{\cos^2(\theta) - \sin(\theta) + \sin^2(\theta)}{(1 - \sin(\theta))\cos(\theta)}   ⟹  \implies = 1−sin⁡(θ)(1−sin⁡(θ))cos⁡(θ)\frac{1 - \sin(\theta)}{(1 - \sin(\theta))\cos(\theta)}

  3. Step 3

    =0cos⁡(θ)= \frac{0}{\cos(\theta)}

    Correct

    1cos⁡(θ)\frac{1}{\cos(\theta)}

    Cancelling a non-zero factor from both numerator and denominator is division by that factor: 1−sin⁡(θ)1−sin⁡(θ)=1\frac{1 - \sin(\theta)}{1 - \sin(\theta)} = 1, not 00.

  4. Step 4

    = 0

    Correct

    sec⁡(θ)\sec(\theta)

    This result follows directly from the cancellation error in the previous step.

WHAT YOU DID WELL
Compliments from Your Teacher

You correctly rewrote the tangent function in terms of sine and cosine.

You accurately combined the fractions across a common denominator.

You correctly applied the Pythagorean identity sin⁡2(θ)+cos⁡2(θ)=1\sin^2(\theta) + \cos^2(\theta) = 1 to simplify the numerator.

THE ERROR
Where Your ACEpath Teacher Spotted the Mistake
Multiplicative Cancellation Replaced with Zero

You treated cancelling an entire factor from the numerator as subtraction rather than division, leaving 00 instead of the multiplicative identity 11.

WHY IT HAPPENED
ACEpath Teacher's Diagnosis

Confusing division cancellation with additive cancellation: when an entire expression divides out via AA\frac{A}{A}, the quotient is 11, not 00.

REASONING PATH
Your ACEpath teacher’s suggested approach
Cancellation is Division

  • For any non-zero expression AA, dividing numerator and denominator by AA yields A⋅1A⋅B=1B\frac{A \cdot 1}{A \cdot B} = \frac{1}{B}.

  • Always explicitly write a factor of 11 before crossing out an entire numerator: 1⋅(1−sin⁡(θ))(1−sin⁡(θ))cos⁡(θ)=1cos⁡(θ)\frac{1 \cdot (1 - \sin(\theta))}{(1 - \sin(\theta))\cos(\theta)} = \frac{1}{\cos(\theta)}.

CORRECTED PATH
ACEpath Teacher's Demonstration

  1. Divide numerator and denominator by the common factor (1−sin⁡(θ))(1 - \sin(\theta)):

  2. 1⋅(1−sin⁡(θ))(1−sin⁡(θ))cos⁡(θ)=1cos⁡(θ)\frac{1 \cdot (1 - \sin(\theta))}{(1 - \sin(\theta))\cos(\theta)} = \frac{1}{\cos(\theta)}

  3. Apply the reciprocal identity 1cos⁡(θ)=sec⁡(θ)\frac{1}{\cos(\theta)} = \sec(\theta).

Final answer

sec⁡(θ)\sec(\theta)

NEXT STEPS
Final Takeaways from Your Teacher

  • Practice simplifying fractions where the entire numerator factors out, reinforcing that xx⋅y=1y\frac{x}{x \cdot y} = \frac{1}{y}.

  • Review reciprocal identities: express 1cos⁡(θ)\frac{1}{\cos(\theta)} as sec⁡(θ)\sec(\theta).

Lock it in

Question

What is the expanded form of −x(1−x)-x(1 - x)?

Answer

−x+x2-x + x^2. A negative times a negative produces a positive term.

Similar question

01

Simplify the expression into a single basic trigonometric function: sin⁡(θ)1−cos⁡(θ)−cot⁡(θ)\frac{\sin(\theta)}{1 - \cos(\theta)} - \cot(\theta) for valid θ\theta.

Answer

csc⁡(θ)\csc(\theta). Key steps: Rewrite cot⁡(θ)=cos⁡(θ)sin⁡(θ)\cot(\theta) = \frac{\cos(\theta)}{\sin(\theta)}, combine over sin⁡(θ)(1−cos⁡(θ))\sin(\theta)(1 - \cos(\theta)) to get sin⁡2(θ)−cos⁡(θ)(1−cos⁡(θ))sin⁡(θ)(1−cos⁡(θ))=sin⁡2(θ)−cos⁡(θ)+cos⁡2(θ)sin⁡(θ)(1−cos⁡(θ))=1−cos⁡(θ)sin⁡(θ)(1−cos⁡(θ))=1sin⁡(θ)=csc⁡(θ)\frac{\sin^2(\theta) - \cos(\theta)(1 - \cos(\theta))}{\sin(\theta)(1 - \cos(\theta))} = \frac{\sin^2(\theta) - \cos(\theta) + \cos^2(\theta)}{\sin(\theta)(1 - \cos(\theta))} = \frac{1 - \cos(\theta)}{\sin(\theta)(1 - \cos(\theta))} = \frac{1}{\sin(\theta)} = \csc(\theta).

02

Simplify into a single term without fractions: 1+cos⁡(x)sin⁡(x)+sin⁡(x)1+cos⁡(x)\frac{1 + \cos(x)}{\sin(x)} + \frac{\sin(x)}{1 + \cos(x)}.

Answer

2csc⁡(x)2\csc(x). Key steps: Find common denominator sin⁡(x)(1+cos⁡(x))\sin(x)(1 + \cos(x)), numerator becomes (1+cos⁡(x))2+sin⁡2(x)=1+2cos⁡(x)+cos⁡2(x)+sin⁡2(x)=2+2cos⁡(x)=2(1+cos⁡(x))(1 + \cos(x))^2 + \sin^2(x) = 1 + 2\cos(x) + \cos^2(x) + \sin^2(x) = 2 + 2\cos(x) = 2(1 + \cos(x)). Cancel (1+cos⁡(x))(1 + \cos(x)) to obtain 2sin⁡(x)=2csc⁡(x)\frac{2}{\sin(x)} = 2\csc(x).

03

Rewrite the expression as a single trigonometric function: sec⁡(θ)−cos⁡(θ)tan⁡(θ)\frac{\sec(\theta) - \cos(\theta)}{\tan(\theta)}.

Answer

sin⁡(θ)\sin(\theta). Key steps: Replace sec⁡(θ)\sec(\theta) with 1cos⁡(θ)\frac{1}{\cos(\theta)} to get 1−cos⁡2(θ)cos⁡(θ)sin⁡(θ)cos⁡(θ)=sin⁡2(θ)cos⁡(θ)⋅cos⁡(θ)sin⁡(θ)=sin⁡(θ)\frac{\frac{1 - \cos^2(\theta)}{\cos(\theta)}}{\frac{\sin(\theta)}{\cos(\theta)}} = \frac{\sin^2(\theta)}{\cos(\theta)} \cdot \frac{\cos(\theta)}{\sin(\theta)} = \sin(\theta).

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