1. ## tan^2(x)-sin^2(x)

Question asks to find equivalent of $\tan ^2 (x) - \sin ^2 (x)$

with selection of A, B, C, or D below respectively
$
\begin{array}{l}
\sec ^2 (x) \\
2\cot (x)\csc (x) \\
\frac{{\sin ^3 (x)}}{{\cos ^2 (x)}} \\
\tan ^2 (x)\sin ^2 (x) \\
\end{array}
$

I've made several attempts at this and still draw blank, below is my latest.

$\frac{{\sin ^2 (x)}}{{\cos ^2 (x)}} - \sin ^2 (x) \\$

$\frac{{\sin ^2 (x)}}{{\cos ^2 (x)}} - (1 - \cos ^2 (x)) \\$

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

$\frac{{\sin ^2 (x)}}{{\cos ^2 (x)}} - \frac{{\cos ^2 (x)}}{{\cos ^2 (x)}} + \frac{{\cos ^4 (x)}}{{\cos ^2 (x)}} \\$

$\frac{{\sin ^2 (x) - \cos ^2 (x) + \cos ^4 (x)}}{{\cos ^2 (x)}} \\$

$\frac{{(1 - \cos ^2 (x)) - \cos ^2 (x) + \cos ^4 (x)}}{{\cos ^2 (x)}} \\$

$\frac{{1 - 2\cos ^2 (x) + \cos ^4 }}{{\cos ^2 (x)}} \\$

And here a come to what appear to be a dead end

2. \begin{aligned}
\tan ^{2}(x)-\sin ^{2}(x)&=\sin ^{2}(x)\left( \frac{1}{\cos ^{2}(x)}-1 \right) \\
& =\sin ^{2}(x)\cdot \frac{\overbrace{1-\cos ^{2}(x)}^{\sin ^{2}(x)}}{\cos ^{2}x} \\
& =\overbrace{\frac{\sin ^{2}(x)}{\cos ^{2}(x)}}^{\tan ^{2}(x)}\cdot \sin ^{2}(x) \\
& =\tan ^{2}(x)\sin ^{2}(x).
\end{aligned}

3. Thankyou for your reply. I'm not sure how you got to the first step though

4. I just factorised.

5. Oh okay now I see it. Thankyou.