Question

$sec\left(50°\right)+\tan \left(50°\right)$ is equal to

• A. $\tan (20°)+\tan (50°)$
• B. $2\tan (20°)+\tan (50°)$
• C. $\tan (20°)+2\tan (50°)$
• D. $2\tan (20°)+2\tan (50°)$

Answer 1

The correct option is C

$\tan (20°)+2\tan (50°)$

Explanation for the correct option:

Simplify the given expression.

An expression $sec\left(50°\right)+\tan \left(50°\right)$ is given.

Simplify the given expression as follows:

$$\begin{gathered} sec\left(50°\right)+\tan \left(50°\right)=\frac{1}{\cos (50°)}+\tan (50°) \\ \Rightarrow sec\left(50°\right)+\tan \left(50°\right)=\frac{1}{\cos (50°)}·\frac{\cos (20°)}{\cos (20°)}+\tan (50°) \\ \Rightarrow sec\left(50°\right)+\tan \left(50°\right)=\frac{\sin (70°)}{\cos (50°)·\cos (20°)}+\tan (50°) \\ \Rightarrow sec\left(50°\right)+\tan \left(50°\right)=\frac{\sin (50°+20°)}{\cos (50°)·\cos (20°)}+\tan (50°) \\ \Rightarrow sec\left(50°\right)+\tan \left(50°\right)=\frac{\sin (50°)·\cos (20°)+\cos (50°)·\sin (20°)}{\cos (50°)·\cos (20°)}+\tan (50°) \\ \Rightarrow sec\left(50°\right)+\tan \left(50°\right)=\frac{\sin (50°)·\cos (20°)}{\cos (50°)·\cos (20°)}+\frac{\cos (50°)·\sin (20°)}{\cos (50°)·\cos (20°)}+\tan (50°) \\ \Rightarrow sec\left(50°\right)+\tan \left(50°\right)=\frac{\sin (50°)}{\cos (50°)}+\frac{\sin (20°)}{\cos (20°)}+\tan (50°) \\ \Rightarrow sec\left(50°\right)+\tan \left(50°\right)=\tan (50°)+\tan (20°)+\tan (50°) \\ \Rightarrow sec\left(50°\right)+\tan \left(50°\right)=\tan (20°)+2\tan (50°) \end{gathered}$${Since, $sec\theta =\frac{1}{\cos \theta },\sin (A+B)=\sin \left(A\right)\cos \left(B\right)+\cos \left(A\right)\sin \left(B\right)$}

Therefore, the given expression $sec\left(50°\right)+\tan \left(50°\right)$ is equal to $\tan \left(20°\right)+2\tan \left(50°\right)$.

Hence, the option $C$ is the correct .