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Trigonometry

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

Note by Rishit Joshi
4 years, 4 months ago

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

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

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

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

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

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

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

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

Pi Han Goh - 4 years, 4 months ago

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

Kushagraa Aggarwal - 4 years, 4 months ago

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can u explain me

Rishit Joshi - 4 years, 4 months ago

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

Rishit Joshi - 4 years, 4 months ago

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Let \(cos^2 x = t\) \(\Rightarrow t^2 + t - 1 = 0\) \(\Rightarrow t = \frac{\sqrt{5} - 1}{2}\) and \(\frac{-\sqrt{5} - 1}{2}\)

Since ,\(0 \leq t \leq 1\) \(\Rightarrow cos^2x = \frac{\sqrt{5} - 1}{2}\) , \(\Rightarrow sec^2 x = \frac{2}{\sqrt{5} - 1}\) = \(\frac{\sqrt{5} + 1}{2}\) ,\( tan^2 x = sec^2x - 1 = \frac{\sqrt{5} - 1}{2} = cos^2x\)

\(\Rightarrow tan^2x + tan^4x = cos^2x + cos^4x =1 \)

Varun Kaushik - 4 years, 4 months ago

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