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1-Find-sin52-sin8-cos22-2-If-a-2-1-a-2-6-find-a-3-1-a-3-3-Find-tan32-tan13-1-tan32-tan13-

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Question Number 190091 by Shrinava last updated on 27/Mar/23
1. Find sin52° + sin8° − cos22° 2. If a^2 + (1/a^2 ) = 6 find a^3 + (1/a^3 ) 3. Find ((tan32° + tan13°)/(1 − tan32° ∙ tan13°))
$$\mathrm{1}.\:\mathrm{Find}\:\:\:\mathrm{sin52}°\:+\:\mathrm{sin8}°\:−\:\mathrm{cos22}° \\ $$$$\mathrm{2}.\:\mathrm{If}\:\:\:\mathrm{a}^{\mathrm{2}} \:+\:\frac{\mathrm{1}}{\mathrm{a}^{\mathrm{2}} }\:=\:\mathrm{6}\:\:\:\mathrm{find}\:\:\:\mathrm{a}^{\mathrm{3}} \:+\:\frac{\mathrm{1}}{\mathrm{a}^{\mathrm{3}} } \\ $$$$\mathrm{3}.\:\mathrm{Find}\:\:\:\frac{\mathrm{tan32}°\:+\:\mathrm{tan13}°}{\mathrm{1}\:−\:\mathrm{tan32}°\:\centerdot\:\mathrm{tan13}°} \\ $$
Answered by BaliramKumar last updated on 27/Mar/23
1. (sin52° + sin8°) − cos22° 2sin(((52° + 8°)/2))cos(((52° − 8°)/2)) − cos22° 2sin30°cos22° − cos22° 2∙(1/2)cos22° − cos22° cos22° − cos22° = 0 2. a^2 + (1/a^2 ) +2 = 6 + 2 (a + (1/a))^2 = 8 (a + (1/a)) = ± 2(√(2 )) ............(i) a^3 + (1/a^3 ) = (a + (1/a))^3 − 3(a + (1/a)) a^3 + (1/a^3 ) = (±2(√2))^3 − 3(±2(√2)) a^3 + (1/a^3 ) = ±16(√2) ∓ 6(√2) = ± 10(√2) 3. ((tan32° + tan13°)/(1 − tan32° ∙ tan13°)) = tan(32° +13°) ((tan32° + tan13°)/(1 − tan32° ∙ tan13°)) = tan45° = 1
$$\mathrm{1}.\:\left({sin}\mathrm{52}°\:+\:{sin}\mathrm{8}°\right)\:−\:{cos}\mathrm{22}° \\ $$$$\:\:\:\:\:\:\mathrm{2}{sin}\left(\frac{\mathrm{52}°\:+\:\mathrm{8}°}{\mathrm{2}}\right){cos}\left(\frac{\mathrm{52}°\:−\:\mathrm{8}°}{\mathrm{2}}\right)\:−\:{cos}\mathrm{22}° \\ $$$$\:\:\:\:\:\mathrm{2}{sin}\mathrm{30}°{cos}\mathrm{22}°\:−\:{cos}\mathrm{22}° \\ $$$$\:\:\:\:\:\mathrm{2}\centerdot\frac{\mathrm{1}}{\mathrm{2}}{cos}\mathrm{22}°\:−\:{cos}\mathrm{22}° \\ $$$$\:\:\:\:\:{cos}\mathrm{22}°\:−\:{cos}\mathrm{22}°\:=\:\mathrm{0} \\ $$$$ \\ $$$$\mathrm{2}.\:{a}^{\mathrm{2}} \:+\:\frac{\mathrm{1}}{{a}^{\mathrm{2}} }\:+\mathrm{2}\:=\:\mathrm{6}\:+\:\mathrm{2} \\ $$$$\:\:\:\:\left({a}\:+\:\frac{\mathrm{1}}{{a}}\right)^{\mathrm{2}} \:=\:\mathrm{8} \\ $$$$\:\:\:\:\left({a}\:+\:\frac{\mathrm{1}}{{a}}\right)\:=\:\pm\:\mathrm{2}\sqrt{\mathrm{2}\:}\:…………\left({i}\right) \\ $$$${a}^{\mathrm{3}} \:+\:\frac{\mathrm{1}}{{a}^{\mathrm{3}} }\:=\:\left({a}\:+\:\frac{\mathrm{1}}{{a}}\right)^{\mathrm{3}} \:−\:\mathrm{3}\left({a}\:+\:\frac{\mathrm{1}}{{a}}\right) \\ $$$${a}^{\mathrm{3}} \:+\:\frac{\mathrm{1}}{{a}^{\mathrm{3}} }\:=\:\left(\pm\mathrm{2}\sqrt{\mathrm{2}}\right)^{\mathrm{3}} \:−\:\mathrm{3}\left(\pm\mathrm{2}\sqrt{\mathrm{2}}\right) \\ $$$${a}^{\mathrm{3}} +\:\frac{\mathrm{1}}{{a}^{\mathrm{3}} }\:\:=\:\pm\mathrm{16}\sqrt{\mathrm{2}}\:\mp\:\mathrm{6}\sqrt{\mathrm{2}}\:=\:\pm\:\mathrm{10}\sqrt{\mathrm{2}} \\ $$$$\: \\ $$$$\mathrm{3}.\:\:\:\:\frac{\mathrm{tan32}°\:+\:\mathrm{tan13}°}{\mathrm{1}\:−\:\mathrm{tan32}°\:\centerdot\:\mathrm{tan13}°}\:=\:{tan}\left(\mathrm{32}°\:+\mathrm{13}°\right) \\ $$$$\:\:\:\:\:\:\frac{\mathrm{tan32}°\:+\:\mathrm{tan13}°}{\mathrm{1}\:−\:\mathrm{tan32}°\:\centerdot\:\mathrm{tan13}°}\:=\:{tan}\mathrm{45}°\:=\:\mathrm{1} \\ $$$$ \\ $$$$ \\ $$$$ \\ $$
Commented by Shrinava last updated on 01/Apr/23
thank you professor cool
$$\mathrm{thank}\:\mathrm{you}\:\mathrm{professor}\:\mathrm{cool} \\ $$

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