Solve $\frac{\tan^3\theta}{1+\tan^2\theta}+\frac{\cot^3\theta}{1+\cot^2\theta} = \sec\theta\csc\theta - 2\sin\theta\cos\theta$ [closed]

$$\frac{\tan^3\theta}{1+\tan^2\theta}+\frac{\cot^3\theta}{1+\cot^2\theta} = \sec\theta\csc\theta - 2\sin\theta\cos\theta$$

closed as off-topic by Claude Leibovici, N. F. Taussig, Najib Idrissi, user147263, Fly by NightJul 11 '15 at 22:41

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• Is it clear now? – Akshansh Bhura Jul 11 '15 at 8:03
• @AkshanshBhura You can use the pythagorean identites $\tan^2 \theta +1 = \sec^2 \theta$, etc, to simplify it. – Kartik Jul 11 '15 at 8:08
• The equation is actually an identity (i.e. true for all $\theta$ such that all terms are defined), so "How to prove .... ?" would be a better title than "Solve ....". – JimmyK4542 Jul 11 '15 at 8:11

Hint: $$\frac{\tan^3\theta}{1+\tan^2\theta}+\frac{\cot^3\theta}{1+\cot^2\theta}$$

$$=\frac{\tan^3\theta}{\sec^2\theta}+\frac{\cot^3\theta}{\csc^2\theta}$$

$$=\frac{\sin^3\theta}{\cos\theta}+\frac{\cos^3\theta}{\sin\theta}$$

$$=\frac{\sin^4\theta+\cos^4\theta}{\sin\theta \cos\theta}$$ Now use $(a^4+b^4)=(a^2+b^2)^2-2a^2b^2$ and solve by yourself.

• Yess!! Thank for help – Akshansh Bhura Jul 11 '15 at 8:22
• You are welcome :) , You are new here so I wanna inform you that next time if you want to ask any question, you have to show your work about your question. – Chiranjeev_Kumar Jul 11 '15 at 8:29

HINT:

$$\frac{\tan^3\theta}{1+\tan^2\theta}=\frac{\tan\theta(1+\tan^2\theta-1)}{1+\tan^2\theta}=\tan\theta-\frac{\tan\theta}{1+\tan^2\theta}$$

Multiplying the numerator & the denominator by $\cos^2\theta,$ $$\frac{\tan\theta}{1+\tan^2\theta}=\cdots=\sin\theta\cos\theta$$

Can you take it from here?

$$\frac{\tan^3(x)}{1+\tan^2(x)}+\frac{\cot^3(x)}{1+\cot^2(x)} = \sec(x)\csc(x) - 2\sin(x)\cos(x)\Longleftrightarrow$$

$$\frac{\tan^3(x)}{1+\tan^2(x)}+\frac{\cot^3(x)}{1+\cot^2(x)} = 2\csc(2x)-\sin(2x)\Longleftrightarrow$$

$$\sin^2(x)\tan(x)+\frac{\cot^3(x)}{1+\cot^2(x)} = 2\csc(2x)-\sin(2x)\Longleftrightarrow$$

$$\sin^2(x)\tan(x)+\cos^2(x)\cot(x) = 2\csc(2x)-\sin(2x)\Longleftrightarrow$$

$$-\sin(2x)+\tan(x)+\cot(x) = 2\csc(2x)-\sin(2x)\Longleftrightarrow$$

$$\frac{\cos(x)}{\sin(x)}+\frac{\cos(x)}{\sin(x)} = 2\left(\frac{1}{\sin(2x)}\right)\Longleftrightarrow$$

$$\frac{\cos^2(x)+\sin^2(x)}{\cos(x)\sin(x)} = \frac{2}{\sin(2x)}\Longleftrightarrow$$

$$\sin(2x)\left(\cos^2(x)+\sin^2(x)\right)=2\cos(x)\sin(x)\Longleftrightarrow$$

$$\sin(2x)\left(\cos^2(x)+1-\cos^2(x)\right)=2\cos(x)\sin(x)\Longleftrightarrow$$

$$\sin(2x)\left(1\right)=2\cos(x)\sin(x)\Longleftrightarrow$$

$$\sin(2x)=2\cos(x)\sin(x)\Longleftrightarrow$$

$$2\cos(x)\sin(x)=2\cos(x)\sin(x)$$

So it is equal!!!