Proving the identity $\frac{\cos^2\theta+\tan^2\theta-1}{\sin^2\theta}=\tan^2\theta$

I am stuck with this trigonometric identity. It appeared in a question paper of mine, and I am wondering whether there is a print error or something, because I absolutely have no idea how to solve this.

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

I would really appreciate some inputs!

• The identity is invalid for $θ = 0$. – user21820 Sep 11 '16 at 13:53

$$\frac{\cos^2\theta + \frac{\sin^2 \theta}{\cos^2\theta}-1}{\sin^2\theta}=$$ $$= \frac{\cos^2 \theta +\frac{\sin^2\theta}{\cos^2 \theta}-\sin^2 \theta-\cos^2\theta}{\sin^2 \theta}$$ $$= \frac{\frac{\sin^2\theta-\cos^2\theta \cdot \sin^2\theta}{\cos^2\theta}}{\sin^2\theta}$$ $$=\frac{\sin^2\theta \cdot (1-\cos^2\theta)}{\sin^2\theta \cdot \cos^2\theta}$$ $$=\frac{\sin^2\theta \cdot \sin^2 \theta}{sin^2\theta \cdot \cos^2\theta}$$ $$=\frac{\sin^2\theta}{\cos^2\theta}=\tan^2\theta$$

• Thank you so much for a concise proof such as this! Helped me greatly :) – Aman Bhargava Sep 11 '16 at 15:58

Follow the usual process of manipulating only one side to make it look like the other side, and leave the other side completely alone. The LHS is much more complicated, so let's start there and try to make it look like the RHS. \begin{align*} \frac{\cos^2\theta + \tan^2\theta - 1}{\sin^2\theta} &= \frac{\cos^2\theta - 1 + \tan^2\theta}{\sin^2\theta}\\[0.3cm] &= \frac{\cos^2\theta - 1}{\sin^2\theta} + \frac{\tan^2\theta}{\sin^2\theta}\\[0.3cm] &= \frac{\cos^2\theta - 1}{1 - \cos^2\theta} + \tan^2\theta \cdot \frac1{\sin^2\theta}\\[0.3cm] &= \frac{-(1-\cos^2\theta)}{1 - \cos^2\theta} + \frac{\sin^2\theta}{\cos^2\theta} \cdot \frac1{\sin^2\theta}\\[0.3cm] &= -1 + \frac1{\cos^2\theta}\\[0.3cm] &= -1 + \sec^2\theta\\[0.3cm] &= \tan^2\theta \end{align*}

• Thank you so much! I can see what I did wrong. Kudos for the clear proof :) – Aman Bhargava Sep 11 '16 at 15:57

Hint:

Multiplying by $\sin^2 \theta$ and using the definition of $\tan$ and $1=\cos^2 \theta +\sin^2 \theta$, we have: $$\cos^2 \theta +\frac{\sin^2 \theta}{\cos^2 \theta}-\cos^2 \theta -\sin^2 \theta = \sin^2 \theta \tan^2 \theta$$ Now it is easy....

Multiply both sides with $\sin^2\theta$ and add the same to get

$$\cos^2\theta+\tan^2\theta-1+\sin^2\theta=\sin^2 \theta+ \sin^2 \theta\tan^2 \theta \implies\tan^2 \theta=\sin^2 \theta+ \sin^2 \theta\tan^2 \theta.$$

Now divide by $\tan^2 \theta$ to obtain $$1=\cos^2 \theta+\sin^2 \theta,$$

which is true.

$$\cos^2\theta=\dfrac1{1+\tan^2\theta}\text{ and }\sin^2\theta=\dfrac{\tan^2\theta}{1+\tan^2\theta}$$

Writing $\tan^2\theta=t$

$$\dfrac{\dfrac1{1+t}+t-1}{\dfrac t{1+t}}=\dfrac{(1+t^2-1)(1+t)}{(1+t)t}=t$$

Here, $$L.H.S=\frac {cos^2\theta+Tan^2\theta-1}{sin^2\theta}$$ $$=\frac {Tan^2\theta - sin^2\theta}{sin^2\theta}$$ $$=\frac {Tan^2\theta}{sin^2\theta} - 1$$ $$=sec^2\theta -1$$ $$=Tan^2\theta=R.H.S$$ Proved.