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Question Number 111002 by bemath last updated on 01/Sep/20

$$\sqrt{\mathrm{bemath}}$$$$\Rightarrow \sin 14°+\cos 14°\tan 38°-1=?$$

Commented by Khanacademy last updated on 01/Sep/20

$$\alpha \mathit{b}\mathit{e}\mathit{M}\mathit{a}\mathit{t}\mathit{h}\mathbf{\_}\mathit{u}\mathit{z}\mathit{s}\mathit{i}\mathit{z}\mathit{n}\mathit{i}\mathit{n}\mathit{g}\mathit{k}\mathit{a}\mathit{n}\mathit{a}\mathit{l}\mathit{i}\mathit{n}\mathit{g}\mathit{i}\mathit{z}\mathit{m}\mathit{i}$$

Answered by som(math1967) last updated on 01/Sep/20

$$\frac{\sin 14\cos 38+\cos 14\sin 38}{\cos 38}-1$$$$=\frac{\sin (38+14)}{\cos 38}-1$$$$=\frac{\sin 52}{\sin (90-38)}-1$$$$=\frac{\sin 52}{\sin 52}-1=1-1=0\mathrm{ans}$$$$$$

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