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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
3 years, 6 months ago

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

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

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

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

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

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

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

$$\tan^4 (x) + \tan^2 (x) = 1$$ · 3 years, 6 months ago

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. · 3 years, 6 months ago

can u explain me · 3 years, 6 months ago

IN THE FIRST STEP I DIDN 'T GOT HOW 1+COS^2X CAME CAN U EXPLAIN ME AGAIN PLSS · 3 years, 6 months ago

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

$$\Rightarrow tan^2x + tan^4x = cos^2x + cos^4x =1$$ · 3 years, 6 months ago