Maths Olympiad Prep

Track / Stage 5 / 241 of 400 #841 of 1964

Problem 841

AIME late
Algebra Difficulty 5.5 Find the answer

1. Simplify the expression 1cos4xsin4x2tan2x\frac{1-\cos ^{4} x}{\sin ^{4} x}-2 \tan ^{-2} x.

A number or a short expression. Fractions can be typed as 3/2, and spacing doesn't matter.

Official solution

1. In the numerator of the first fraction, we use the formula for the difference of squares and write the product (1cos2x)(1+cos2x)\left(1-\cos ^{2} x\right)\left(1+\cos ^{2} x\right). We consider the following relationships: tanx=sinxcosx\tan x=\frac{\sin x}{\cos x} and 1cos2x=sin2x1-\cos ^{2} x=\sin ^{2} x. We simplify the first fraction by ssin2x\mathrm{s} \sin ^{2} x and raise the second fraction to the exponent -2. We add the fractions and get 1cos2xsin2x\frac{1-\cos ^{2} x}{\sin ^{2} x}. In the numerator, we again consider the relationship 1cos2x=sin2x1-\cos ^{2} x=\sin ^{2} x, so the result is 1.

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Source: NuminaMath-1.5, licensed Apache-2.0. Statement and solution reproduced as published; topic, difficulty and ordering added by this site.