Question Number 111002 by bemath last updated on 01/Sep/20
$$\sqrt{\mathrm{bemath}} \\ $$$$\Rightarrow\:\mathrm{sin}\:\mathrm{14}°+\mathrm{cos}\:\mathrm{14}°\mathrm{tan}\:\mathrm{38}°−\mathrm{1}=? \\ $$
Commented by Khanacademy last updated on 01/Sep/20
$$\alpha\boldsymbol{{beMath\_uz}}\:\:\:\boldsymbol{{sizning}}\:\:\boldsymbol{{kanalingizmi}} \\ $$
Answered by som(math1967) last updated on 01/Sep/20
$$\:\frac{\mathrm{sin14cos38}+\mathrm{cos14sin38}}{\mathrm{cos38}}−\mathrm{1} \\ $$$$=\frac{\mathrm{sin}\left(\mathrm{38}+\mathrm{14}\right)}{\mathrm{cos38}}\:−\mathrm{1} \\ $$$$=\frac{\mathrm{sin52}}{\mathrm{sin}\left(\mathrm{90}−\mathrm{38}\right)}\:−\mathrm{1} \\ $$$$=\frac{\mathrm{sin52}}{\mathrm{sin52}}\:−\mathrm{1}=\mathrm{1}−\mathrm{1}=\mathrm{0ans} \\ $$$$ \\ $$