# Solve identity : $\frac{1+\sin x\cos x}{\cos^3 x - \sin^3 x}+\frac{1}{\sin x+\cos x}+\frac{\sin^2x-2\cos x-1}{\cos^2x-\sin^2x}=\frac{1}{\tan^2x-1}$

$$\frac{1+\sin x\cos x}{\cos^3 x - \sin^3 x}+\frac{1}{\sin x+\cos x}+\frac{\sin^2x-2\cos x-1}{\cos^2x-\sin^2x}=\frac{1}{\tan^2x-1}$$

I am doing this identity for about an hour and I can't get to the result. I use a common denominator but that just makes a mess that takes forever to do and gets me nowhere. Surely there is an easier way that I am not seeing? Some hint or help please?

• Can you factor $\cos^3 x - \sin^3{x}$? – Bob Sep 18 '17 at 21:18
• Got it. I definitely need some sleep. Can't believe I didn't see that one! :) Thanks! – CevapMathMaster Sep 18 '17 at 21:26

Use $a^3-b^3=(a-b)(a^2+ab+b^2)$

Therefore we have,

$$\cos^3x-\sin^3x=(\cos x-\sin x)(1+\sin x \cos x)$$

Your expression in LHS

$$\frac{1+\sin x\cos x}{\cos^3 x - \sin^3 x}+\frac{1}{\sin x+\cos x}+\frac{\sin^2x-2\cos x-1}{\cos^2x-\sin^2x}$$

Simplifies to

\begin{align} \text{LHS} &=\frac{1}{\cos x - \sin x}+\frac{1}{\sin x+\cos x}+\frac{\sin^2x-2\cos x-1}{\cos^2x-\sin^2x}\\ &= \frac{2\cos x}{\cos^2 x - \sin^2 x}+\frac{\sin^2x-2\cos x-1}{\cos^2x-\sin^2x}\\ &= \frac{2 \cos x+\sin^2x-2\cos x-1}{\cos^2x-\sin^2x}\\ &= \frac{\sin^2 x-1}{\cos^2x-\sin^2x}\\ &= \frac{cos^2 x}{\sin ^2x-\cos^2x}\\ &= \frac{1}{\tan^2x-1}= \text{RHS}\\ \end{align}

• Wow... I can't believe I didn't see that one. Haha! Really simple actually. I must be tired. Thank you! :) – CevapMathMaster Sep 18 '17 at 21:24
• @CevapMathMaster Numerator in the LCM of first two terms will be $$(\cos x+\sin x)+(\cos x-\sin x)$$ – Jaideep Khare Sep 18 '17 at 22:19
• Thanks @Jaideep Khare! Finished the problem successfully :) – CevapMathMaster Sep 18 '17 at 22:24