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Question Number 170737 by naka3546 last updated on 30/May/22

(3/(sin^2 40°)) − (1/(cos^2 40°)) + 64 sin^2 40° = ?

$$\frac{\mathrm{3}}{\mathrm{sin}^{\mathrm{2}} \mathrm{40}°}\:−\:\frac{\mathrm{1}}{\mathrm{cos}^{\mathrm{2}} \mathrm{40}°}\:+\:\mathrm{64}\:\mathrm{sin}^{\mathrm{2}} \mathrm{40}°\:\:=\:\:? \\ $$

Answered by som(math1967) last updated on 30/May/22

(3/(sin^2 40)) +((64sin^2 40cos^2 40−1)/(cos^2 40)) ★ =(3/(sin^2 40)) +((16sin^2 80−1)/(cos^2 40)) =((3cos^2 40+16sin^2 80sin^2 40−sin^2 40)/(sin^2 40cos^2 40)) =((4cos^2 40−1+4(2sin80sin40)^2 )/(sin^2 40cos^2 40)) =((4cos^2 40−1+(2cos40−2cos120)^2 )/(sin^2 40cos^2 40)) =((8cos^2 40−1+4cos40+1)/(sin^2 40cos^2 40)) =((4cos40(2c0s40+1))/(sin^2 40cos^2 40)) =((4(2cos40+1))/(sin40sin40cos40)) =((8(2cos40+1))/(2sin40cos40sin40)) =((16(2cos40+1))/(2sin80sin40)) =((16(2cos40+1))/((cos40−cos120))) =((32(2cos40+1))/((2cos40+1)))=32 ans ★64sin^2 40cos^2 40 =16(2sin40cos40)^2 =16sin^2 80

$$\frac{\mathrm{3}}{{sin}^{\mathrm{2}} \mathrm{40}}\:+\frac{\mathrm{64}{sin}^{\mathrm{2}} \mathrm{40}{cos}^{\mathrm{2}} \mathrm{40}−\mathrm{1}}{{cos}^{\mathrm{2}} \mathrm{40}}\:\bigstar \\ $$$$=\frac{\mathrm{3}}{{sin}^{\mathrm{2}} \mathrm{40}}\:+\frac{\mathrm{16}{sin}^{\mathrm{2}} \mathrm{80}−\mathrm{1}}{{cos}^{\mathrm{2}} \mathrm{40}} \\ $$$$=\frac{\mathrm{3}{cos}^{\mathrm{2}} \mathrm{40}+\mathrm{16}{sin}^{\mathrm{2}} \mathrm{80}{sin}^{\mathrm{2}} \mathrm{40}−{sin}^{\mathrm{2}} \mathrm{40}}{{sin}^{\mathrm{2}} \mathrm{40}{cos}^{\mathrm{2}} \mathrm{40}} \\ $$$$=\frac{\mathrm{4}{cos}^{\mathrm{2}} \mathrm{40}−\mathrm{1}+\mathrm{4}\left(\mathrm{2}{sin}\mathrm{80}{sin}\mathrm{40}\right)^{\mathrm{2}} }{{sin}^{\mathrm{2}} \mathrm{40}{cos}^{\mathrm{2}} \mathrm{40}} \\ $$$$=\frac{\mathrm{4}{cos}^{\mathrm{2}} \mathrm{40}−\mathrm{1}+\left(\mathrm{2}{cos}\mathrm{40}−\mathrm{2}{cos}\mathrm{120}\right)^{\mathrm{2}} }{{sin}^{\mathrm{2}} \mathrm{40}{cos}^{\mathrm{2}} \mathrm{40}} \\ $$$$=\frac{\mathrm{8}{cos}^{\mathrm{2}} \mathrm{40}−\mathrm{1}+\mathrm{4}{cos}\mathrm{40}+\mathrm{1}}{{sin}^{\mathrm{2}} \mathrm{40}{cos}^{\mathrm{2}} \mathrm{40}} \\ $$$$=\frac{\mathrm{4}{cos}\mathrm{40}\left(\mathrm{2}{c}\mathrm{0}{s}\mathrm{40}+\mathrm{1}\right)}{{sin}^{\mathrm{2}} \mathrm{40}{cos}^{\mathrm{2}} \mathrm{40}} \\ $$$$=\frac{\mathrm{4}\left(\mathrm{2}{cos}\mathrm{40}+\mathrm{1}\right)}{{sin}\mathrm{40}{sin}\mathrm{40}{cos}\mathrm{40}} \\ $$$$=\frac{\mathrm{8}\left(\mathrm{2}{cos}\mathrm{40}+\mathrm{1}\right)}{\mathrm{2}{sin}\mathrm{40}{cos}\mathrm{40}{sin}\mathrm{40}} \\ $$$$=\frac{\mathrm{16}\left(\mathrm{2}{cos}\mathrm{40}+\mathrm{1}\right)}{\mathrm{2}{sin}\mathrm{80}{sin}\mathrm{40}} \\ $$$$=\frac{\mathrm{16}\left(\mathrm{2}{cos}\mathrm{40}+\mathrm{1}\right)}{\left({cos}\mathrm{40}−{cos}\mathrm{120}\right)} \\ $$$$=\frac{\mathrm{32}\left(\mathrm{2}{cos}\mathrm{40}+\mathrm{1}\right)}{\left(\mathrm{2}{cos}\mathrm{40}+\mathrm{1}\right)}=\mathrm{32}\:{ans} \\ $$$$\bigstar\mathrm{64}{sin}^{\mathrm{2}} \mathrm{40}{cos}^{\mathrm{2}} \mathrm{40} \\ $$$$=\mathrm{16}\left(\mathrm{2}{sin}\mathrm{40}{cos}\mathrm{40}\right)^{\mathrm{2}} \\ $$$$=\mathrm{16}{sin}^{\mathrm{2}} \mathrm{80} \\ $$

Commented by naka3546 last updated on 30/May/22

Thank you, sir.

$$\mathrm{Thank}\:\mathrm{you},\:\mathrm{sir}. \\ $$

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