Question

The value of $\sec {\text{ }}{50^ \circ } + \tan {\text{ }}{50^ \circ }$ is equal to
A) $\tan {\text{ }}{20^ \circ } + \tan {\text{ }}{50^ \circ }$
B) $2\tan {\text{ }}{20^ \circ } + \tan {\text{ }}{50^ \circ }$
C) $\tan {\text{ }}{20^ \circ } + 2\tan {\text{ }}{50^ \circ }$
D) $2\tan {\text{ }}{20^ \circ } + 2\tan {\text{ }}{50^ \circ }$

Answer 1

Hint: We have to solve the given trigonometric expression. The expression is the sum of sec and tan functions, although we know the standard values of trigonometric functions, the degree which is asked is not a standard degree which means we have to use another method to find the solution. We will use the formulae:
$\sin \theta = \cos (90 - \theta )$
$\cos \theta = \sin (90 - \theta )$
We will first express the first term in terms of cos and sin and then solve it further.

Complete step by step answer:
We are given the expression:
$\sec {\text{ }}{50^ \circ } + \tan {\text{ }}{50^ \circ }$
 As $\sec \theta = \dfrac{1}{{\cos \theta }}$
$ = \dfrac{1}{{\cos {{50}^ \circ }}} + \tan {\text{ }}{50^ \circ }$
Multiplying numerator and denominator of the first term by $\cos {20^o}$ we get,
\[ \Rightarrow \left[ {\dfrac{{cos{\text{ }}20^\circ }}{{cos{\text{ }}20^\circ {\text{ }}cos{\text{ }}50^\circ }}} \right]{\text{ }} + {\text{ }}tan{\text{ }}50^\circ \]
$\cos {20^o}$ can be written as $\sin {70^o}$ from the formula $\cos \theta = \sin (90 - \theta )$, we can thus write,
\[ \Rightarrow \left[ {\dfrac{{sin{\text{ }}70^\circ }}{{cos{\text{ }}20^\circ {\text{ }}cos{\text{ }}50^\circ }}} \right]{\text{ }} + {\text{ }}tan{\text{ }}50^\circ \]
\[ \Rightarrow \left[ {\dfrac{{sin\left( {50^\circ {\text{ }} + {\text{ }}20^\circ } \right)}}{{\left( {cos{\text{ }}20^\circ {\text{ }}cos{\text{ }}50^\circ } \right)}}{\text{ }}} \right]{\text{ }} + {\text{ }}tan{\text{ }}50^\circ \]
Using the compound formula : $\sin (A + B) = \sin A\cos B + \cos A\sin B$
\[ \Rightarrow \left[ {\dfrac{{\left( {sin{\text{ }}50^\circ {\text{ }}cos{\text{ }}20^\circ {\text{ }} + {\text{ }}cos{\text{ }}50^\circ {\text{ }}sin{\text{ }}20^\circ } \right)}}{{\left( {cos{\text{ }}20^\circ {\text{ }}cos{\text{ }}50^\circ } \right)}}} \right]{\text{ }} + {\text{ }}tan{\text{ }}50^\circ \]
\[ \Rightarrow {\text{ }}\dfrac{{\left( {sin{\text{ }}50^\circ {\text{ }}cos{\text{ }}20^\circ } \right)}}{{\left( {cos{\text{ }}20^\circ {\text{ }}cos{\text{ }}50^\circ } \right)}} + {\text{ }}\dfrac{{\left( {cos{\text{ }}50^\circ {\text{ }}sin{\text{ }}20^\circ } \right)}}{{\left( {cos{\text{ }}20^\circ {\text{ }}cos{\text{ }}50^\circ } \right)}} + {\text{ }}tan{\text{ }}50^\circ \]
The first two expression will now be reduced first by cancelling the common factors present in the numerator and the denominator and then it will be written in the terms of tan using the formula for converting sin and cos into the tan function:
\[\tan \theta = \dfrac{{\sin \theta }}{{\cos \theta }}\]
First upon reduction we can write,
\[ \Rightarrow {\text{ }}\dfrac{{sin{\text{ }}50^\circ }}{{cos{\text{ }}50^\circ }} + {\text{ }}\dfrac{{sin{\text{ }}20^\circ }}{{cos{\text{ }}20^\circ }} + {\text{ }}tan{\text{ }}50^\circ \]
Then upon using the formula we will write,
$ \Rightarrow tan{\text{ }}50^\circ {\text{ }} + {\text{ }}tan{\text{ }}20^\circ {\text{ }} + {\text{ }}tan{\text{ }}50^\circ $
\[\; \Rightarrow tan{\text{ }}20^\circ {\text{ }} + {\text{ }}2{\text{ }}tan{\text{ }}50^\circ \]
Therefore, option (C) is correct.

Note:
It is important to remember the complementary angle formula if we want to change the trigonometric ratios like sin and cos to each other.
The formulae are given below: (also used in the question)
\[sin(\theta ) = cos({90^o} - \theta )\]
\[cos(\theta ) = sin({90^o} - \theta )\]
Similar formulae can be written for tan and cot , and sec and cosec , they are also mentioned below:
\[
  tan(\theta ) = cot({90^o} - \theta ) \\
  cot(\theta ) = tan({90^o} - \theta ) \\
\]
\[ sec(\theta ) = csc({90^o} - \theta ) \\
  csc(\theta ) = sec({90^o} - \theta ) \\
 \]