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What is \(\frac{1-\cos 2x}{\sin 2x}\) equal to?

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Answer and explanation

Correct answer: \(\tan x\)

Use the identities \(1-\cos 2x=2\sin^2x\) and \(\sin 2x=2\sin x\cos x\). Therefore, \(\frac{1-\cos 2x}{\sin 2x}=\frac{2\sin^2x}{2\sin x\cos x}=\frac{\sin x}{\cos x}=\tan x\). \(\cot x\) is the reciprocal of \(\tan x\), so it is not correct. Exam tip: first rewrite both numerator and denominator using double-angle identities.

Tags

trigonometric functionsdouble angle identitiestangenttrigonometric simplificationclass 11 mathematics

Frequently asked questions

What is the correct answer to this question?

\(\tan x\)

Why is this the correct answer?

Use the identities \(1-\cos 2x=2\sin^2x\) and \(\sin 2x=2\sin x\cos x\). Therefore, \(\frac{1-\cos 2x}{\sin 2x}=\frac{2\sin^2x}{2\sin x\cos x}=\frac{\sin x}{\cos x}=\tan x\). \(\cot x\) is the reciprocal of \(\tan x\), so it is not correct. Exam tip: first rewrite both numerator and denominator using double-angle identities.

Which subject and chapter does this question cover?

This is a Class 12 Mathematics question. Chapter: Trigonometric Functions. Topic: Trigonometric functions and their properties.

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