# 26,27,28 pls answer ASAP

$28\phantom{\rule{0ex}{0ex}}\left(se{c}^{2}\theta -1\right)(\mathrm{cos}e{c}^{2}\theta -1)\phantom{\rule{0ex}{0ex}}Weknowthat1+{\mathrm{tan}}^{2}\theta =se{c}^{2}\theta and1+co{t}^{2}\theta =\mathrm{cos}e{c}^{2}\theta \phantom{\rule{0ex}{0ex}}So,weget\phantom{\rule{0ex}{0ex}}{\mathrm{tan}}^{2}\theta \times co{t}^{2}\theta \phantom{\rule{0ex}{0ex}}=\frac{1}{co{t}^{2}\theta}\times co{t}^{2}\theta \left[as\mathrm{tan}x=\frac{1}{cotx}\right]\phantom{\rule{0ex}{0ex}}=1\phantom{\rule{0ex}{0ex}}=RHS$

For remaining queries we request you to post them in separate threads to have rapid assistance from our experts.

Regards

**
**