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Question Number 190091 by Shrinava last updated on 27/Mar/23

$$1.\mathrm{Find}\sin 52°+\sin 8°-\cos 22°$$$$2.\mathrm{If}{\mathrm{a}}^2+\frac{1}{{\mathrm{a}}^2}=6\mathrm{find}{\mathrm{a}}^3+\frac{1}{{\mathrm{a}}^3}$$$$3.\mathrm{Find}\frac{\tan 32°+\tan 13°}{1-\tan 32°\cdot \tan 13°}$$

Answered by BaliramKumar last updated on 27/Mar/23

$$1.(sin52°+sin8°)-cos22°$$$$2sin\left(\frac{52°+8°}{2}\right)cos\left(\frac{52°-8°}{2}\right)-cos22°$$$$2sin30°cos22°-cos22°$$$$2\cdot \frac{1}{2}cos22°-cos22°$$$$cos22°-cos22°=0$$$$$$$$2.a^2+\frac{1}{a^2}+2=6+2$$$${(a+\frac{1}{a})}^2=8$$$$(a+\frac{1}{a})=\pm 2\sqrt{2}............\left(i\right)$$$$a^3+\frac{1}{a^3}={(a+\frac{1}{a})}^3-3(a+\frac{1}{a})$$$$a^3+\frac{1}{a^3}={(\pm 2\sqrt{2})}^3-3(\pm 2\sqrt{2})$$$$a^3+\frac{1}{a^3}=\pm 16\sqrt{2}\mp 6\sqrt{2}=\pm 10\sqrt{2}$$$$$$$$3.\frac{\tan 32°+\tan 13°}{1-\tan 32°\cdot \tan 13°}=tan(32°+13°)$$$$\frac{\tan 32°+\tan 13°}{1-\tan 32°\cdot \tan 13°}=tan45°=1$$$$$$$$$$$$$$

Commented by Shrinava last updated on 01/Apr/23

$$\mathrm{thank}\mathrm{you}\mathrm{professor}\mathrm{cool}$$

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