Question

Express $\cos \left(68\right)°+\tan \left(76\right)°$ in terms of the angles between $0°\mathrm{and}45°$.

Answer 1

Solution:

$\cos \left(68\right)°+\tan \left(76\right)°$

$$\begin{gathered} =\cos \left(90°-22°\right)+\tan \left(90°-14°\right) \\ =\sin \left(22°\right)+\cot \left(14°\right)\left[\mathbf{\because }\mathbf{}\mathbf{cos}\mathbf{(}\mathbf{90}\mathbf{°}\mathbf{-}\mathbf{\theta }\mathbf{)}\mathbf{=}\mathbf{sin}\mathbf{\left(}\mathbf{\theta }\mathbf{\right)}\mathbf{}\mathbf{and}\mathbf{}\mathbf{tan}\mathbf{(}\mathbf{90}\mathbf{°}\mathbf{-}\mathbf{\theta }\mathbf{)}\mathbf{=}\mathbf{cot}\mathbf{\left(}\mathbf{\theta }\mathbf{\right)}\right] \end{gathered}$$

Final Answer: $\mathbf{cos}\left(68\right)\mathbf{°}\mathbf{+}\mathbf{tan}\left(76\right)\mathbf{°}$ in terms of angles between $0°$ and $45°$ is $\mathbf{sin}\left(22\right)\mathbf{°}\mathbf{+}\mathbf{cot}\left(14\right)\mathbf{°}$.