Proving this identity: $\frac{\sin^2\theta}{\cos^2\theta} + \frac{\cos^2\theta}{\sin^2\theta} = \sec^2\theta \csc^2\theta - 2$

I have tried solving this trig. identity, but I get stuck when it comes to the $-2$ part. Any suggestions?

$$\frac{\sin^2\theta}{\cos^2\theta} + \frac{\cos^2\theta}{\sin^2\theta} = \sec^2\theta \csc^2\theta - 2$$

• Hint: expand $(\sin^2 \theta+\cos^2 \theta)^2$ – polfosol Sep 12 '16 at 6:45

\begin{align*} \frac{\sin^2\theta}{\cos^2\theta} + \frac{\cos^2\theta}{\sin^2\theta} &= \tan^2\theta + \cot^2\theta \\ &= \sec^2\theta - 1+ \csc^2\theta - 1\\ &= \frac{1}{\cos^2\theta} + \frac{1}{\sin^2\theta} - 2\\ &= \frac{\sin^2\theta + \cos^2\theta}{\sin^2\theta \cos^2\theta} -2\\ &= \sec^2\theta \csc^2\theta - 2 \end{align*}

• Man we're all doing it the hard way. – Dan Uznanski Sep 12 '16 at 6:48
• This is so easy when you do it this way! Thank you so much @Muralidharan – Aman Bhargava Sep 12 '16 at 6:52
• @DanUznanski Ikr! I don't know how I couldn't think of this before. – Aman Bhargava Sep 12 '16 at 6:57

hint: $\dfrac{a^2}{b^2} + \dfrac{b^2}{a^2} = \dfrac{a^4+b^4}{a^2b^2}= \dfrac{(a^2+b^2)^2}{a^2b^2}-2$. Apply this identity for $a = \cos \theta, b = \sin \theta$ to get the desired identity.

hint: $a^4+b^4 = a^4+2a^2b^2+b^4-2a^2b^2=(a^2+b^2)^2 - 2a^2b^2$

$$\frac{s^2}{c^2}+\frac{c^2}{s^2}=\frac{s^4+c^4}{s^2c^2}=\frac{(s^2+c^2)^2-2s^2c^2}{s^2c^2}.$$

This is straightforward when you have another identity in your pocket, namely $$\sec\theta \cdot \csc\theta = \tan\theta + \cot\theta \tag{\star}$$ This implies $$\sec^2\theta \csc^2\theta = ( \tan\theta + \cot\theta )^2 = \tan^2\theta + 2 \tan\theta\cot\theta + \cot^2\theta = \tan^2\theta + 2 + \cot^2\theta$$ which gets you the equivalent of the identity in question. $\square$

By the way, proof of $(\star)$ arises from this trigonograph: Computing the area of the big triangle in two ways gives $$\frac{1}{2}\;\sec\theta\cdot\csc\theta = \frac{1}{2}\cdot 1 \cdot\left(\;\tan\theta + \cot\theta\;\right)$$ and the identity follows. $\square$