Question

3
(Cos^430°- Sin^445') - 3(Sin^260°-
Sec^2-45°) + 1/4Cot^230°​

Answer 1

Question :-  ↓

$3/2(Cos^430°- Sin^445') - 3(Sin^260°-Sec^2-45°) + 1/4Cot^230°​$

Answer :-    ↓

                  $= 3/2(Cos^430°- Sin^445') - 3(Sin^260°-Sec^2-45°) + 1/4Cot^230°​$

$= \frac{2}{3} [(\frac{\sqrt{3} }{2} ^4 - (\frac{1}{\sqrt{2} } )^4] - 3[(\frac{\sqrt{3} }{2} ^2 - (\sqrt{2} )^2] + \frac{1}{4} (\sqrt{3} )^2$​

$= \frac{2}{3} (\frac{9}{16} - \frac{1}{4} ) - 3(\frac{3}{4} -2) + \frac{3}{4}$

$= \frac{2}{3} * \frac{5}{16} + \frac{15}{4} + \frac{3}{4}$

$= \frac{5}{24} + \frac{9}{2}$

$= \frac{113}{24}$

THANK YOU   ↑

Answer 2

Answer:

113/24

Step-by-step explanation:

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