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

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

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

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

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

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

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

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

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

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

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

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

2. \displaystyle \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.