Question

Plz do this
Q no 21 plzzzzz

Answer 1

4sec2 (90 - 21 )- cot2theta -2/3 + 3 tan2(90-26) / cot2 20 =x/3

4(cosec2 21-cot2 21)-2/3 +3 = x/3

4-2/3+3=x/ 3

X=19

Answer 2

Solution:

$4\bigg( \frac{ {sec}^{2}69° - {cot}^{2}21°}{3}\bigg) - \frac{2}{3} sin \: 90° + \frac{3 {tan}^{2}64° }{ {cot}^{2} 26°} = \frac{x}{3} \\ \\ as \: we \: know \: that$
$sec(90° - \theta) = cosec \: \theta \\ \\ tan(90° - \theta) = cot \: \theta \\ \\ {cosec}^{2} \theta - {cot}^{2} \theta = 1 \\ \\ so \\ \\ 4( \frac{ {sec}^{2}(90° - 21°) - {cot}^{2}21°}{3}) - \frac{2}{3} sin \: 90° + \frac{3 {tan}^{2}(90° - 26°) }{ {cot}^{2} 26°} = \frac{x}{3}$
$= 4( \frac{ {cosec}^{2}21° - {cot}^{2}21°}{3}) - \frac{2}{3} sin \: 90° + \frac{3 {cot}^{2}26° }{ {cot}^{2} 26°} = \frac{x}{3} \\\\apply\: identity\:and\:sin 90°=1\\ \\ \frac{4(1)}{3} - \frac{2(1)}{3} + 3(1) = \frac{x}{3} \\ \\ \frac{4}{3} - \frac{2}{3} + 3 = \frac{x}{3} \\ \\ \frac{4 - 2 + 9}{3} = \frac{x}{3} \\ \\ x = 11$
Hope it helps you