If \(\tan x=2\), what is the value of \(\frac{1-\tan^2 x}{1+\tan^2 x}\)?
Answer and explanation
Correct answer: \(-\frac{3}{5}\)
Given \(\tan x=2\), we get \(\tan^2x=4\). Hence, \(\frac{1-\tan^2 x}{1+\tan^2 x}=\frac{1-4}{1+4}=\frac{-3}{5}=-\frac{3}{5}\). The value \(\frac{3}{5}\) results from a sign error in the numerator. Exam tip: square the given trigonometric value first, then substitute carefully in both numerator and denominator.
Frequently asked questions
What is the correct answer to this question?
\(-\frac{3}{5}\)
Why is this the correct answer?
Given \(\tan x=2\), we get \(\tan^2x=4\). Hence, \(\frac{1-\tan^2 x}{1+\tan^2 x}=\frac{1-4}{1+4}=\frac{-3}{5}=-\frac{3}{5}\). The value \(\frac{3}{5}\) results from a sign error in the numerator. Exam tip: square the given trigonometric value first, then substitute carefully in both numerator and denominator.
Which subject and chapter does this question cover?
This is a Class 12 Mathematics question. Chapter: Trigonometric Functions. Topic: Trigonometric functions and their properties.
