Question

Express $\sin {89}^o+\tan {89}^{\mathrm{o}}$ in terms of angles between $0^{\mathrm{o}}\mathrm{and}{45}^{\mathrm{o}}$.

Answer 1

Express the trigonometric expression in terms of angles between $0^{\mathrm{o}}\mathrm{and}{45}^{\mathrm{o}}$:

Use the following trigonometric identity:

$$\begin{gathered} \left(i\right)\cos \left({90}^{\mathrm{o}}-\theta \right)=\sin \theta \\ \left(ii\right)\cot \left({90}^{\mathrm{o}}-\theta \right)=\tan \theta \end{gathered}$$

Now,

$$\begin{gathered} \sin {89}^o+\tan {89}^{\mathrm{o}} \\ =\sin \left({90}^{\mathrm{o}}-1^o\right)+\tan \left({90}^{\mathrm{o}}-1^o\right)\left[\because \sin {89}^o=\sin \left({90}^{\mathrm{o}}-1^o\right)\&\tan {89}^{\mathrm{o}}=\tan \left({90}^{\mathrm{o}}-1^o\right)\right] \\ =\cos 1^o+\cot 1^o \end{gathered}$$

Thus, $\sin {89}^o+\tan {89}^{\mathrm{o}}$ can be expressed in terms of angles between $0^{\mathrm{o}}\mathrm{and}{45}^{\mathrm{o}}$ as $\cos 1^o+\cot 1^o$.