Trigonometry

If cos2(x)+cos4(x)=1\cos^2(x) + \cos^4(x) = 1, prove that tan2(x)+tan4(x)=1\tan^2(x) + \tan^4(x) = 1.

Note by Rishit Joshi
7 years, 9 months ago

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Comments

Divide both sides by cos2(x) \cos^2 (x)

1+cos2(x)=sec2(x)1 + \cos^2 (x) = \sec^2 (x)

1+cos2(x)=tan2(x)+11 + \cos^2 (x) = \tan^2 (x) + 1

cos2(x)=tan2(x) \cos^2 (x) = \tan^2 (x)

1sec2(x)=tan2(x) \frac {1}{ \sec^2 (x) } = \tan^2 (x)

1tan2(x)+1=tan2(x) \frac {1}{ \tan^2 (x) + 1 } = \tan^2 (x)

1=tan2(x)(tan2(x)+1)1 = \tan^2 (x) (\tan^2 (x) + 1 )

tan4(x)+tan2(x)=1 \tan^4 (x) + \tan^2 (x) = 1

Pi Han Goh - 7 years, 9 months ago

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I was complete at the step cos2(x)=tan2(x) cos^2(x) = tan^2(x) . The next steps weren't required. You just had to replace cos2(x) cos^2(x) by tan2(x)tan^2(x) in the given equation.

kushagraa aggarwal - 7 years, 9 months ago

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can u explain me

Rishit Joshi - 7 years, 9 months ago

IN THE FIRST STEP I DIDN 'T GOT HOW 1+COS^2X CAME CAN U EXPLAIN ME AGAIN PLSS

Rishit Joshi - 7 years, 9 months ago

Let cos2x=tcos^2 x = t t2+t1=0\Rightarrow t^2 + t - 1 = 0 t=512\Rightarrow t = \frac{\sqrt{5} - 1}{2} and 512\frac{-\sqrt{5} - 1}{2}

Since ,0t10 \leq t \leq 1 cos2x=512\Rightarrow cos^2x = \frac{\sqrt{5} - 1}{2} , sec2x=251\Rightarrow sec^2 x = \frac{2}{\sqrt{5} - 1} = 5+12\frac{\sqrt{5} + 1}{2} ,tan2x=sec2x1=512=cos2x tan^2 x = sec^2x - 1 = \frac{\sqrt{5} - 1}{2} = cos^2x

tan2x+tan4x=cos2x+cos4x=1\Rightarrow tan^2x + tan^4x = cos^2x + cos^4x =1

varun kaushik - 7 years, 9 months ago
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