which of the following is equivalent to $\frac{1-\cos(2x)}{\sin(2x)}$ in terms of $\tan(x)$?

(a) $\tan(x)$
(b) $\cot(x)$
(c) $\tan^2(x)$
(d) $\frac{1}{\tan(x)}$

Question

which of the following is equivalent to $\frac{1-\cos(2x)}{\sin(2x)}$ in terms of $\tan(x)$?

(a) $\tan(x)$
(b) $\cot(x)$
(c) $\tan^2(x)$
(d) $\frac{1}{\tan(x)}$

Accepted Answer

{ "quality": "clear", "question_count": 1, "skills_matched": [ "step_cot", "multiple_choice_formatter", "explain_then_answer", "tutor", "mastery_adaptive_brevity" ], "needs_drawing": false, "concepts_used": [ "Trigonometric Identities" ], "new_concepts": [], "current_concepts": [ "Trigonometric Identities", "Double Angle Formulas" ] } ## Identify the given expression and target We need to simplify the trigonometric expression: $$ \frac{1-\cos(2x)}{\sin(2x)} $$ ## Apply double-angle identities Using the *Trigonometric Identities* knowledge point $$ \begin{aligned} \cos(2x) &= 1 - 2\sin^2(x) \ \sin(2x) &= 2\sin(x)\cos(x) \end{aligned} $$ ## Substitute and simplify the expression Using the *Trigonometric Identities* knowledge point $$ \begin{aligned} \frac{1 - (1 - 2\sin^2(x))}{2\sin(x)\cos(x)} &= \frac{2\sin^2(x)}{2\sin(x)\cos(x)} \ &= \frac{\sin(x)}{\cos(x)} \ &= \tan(x) \end{aligned} $$

(A) $\tan(x)$

(B) $\cot(x)$

(C) $\tan^2(x)$

(D) $\frac{1}{\tan(x)}$

{ "subject": "Mathematics", "question_type": "Multiple Choice", "knowledge_point": [ "Mathematics", "Geometry", "Trigonometric Identities" ] }

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QUESTION IMAGE

Question

Explanation:

Identify the given expression and target

Apply double-angle identities

Substitute and simplify the expression

Answer: