Question

$ cosec {69^0} + \cot {69^0}$ When expressed in terms of angles between ${{\text{0}}^0}$ and ${45^0}$ becomes.
$\begin{gathered}
  A.{\text{ }}\sec {21^ \circ } + \tan {21^ \circ } \\
  B.{\text{ }}\sin {21^ \circ } + \cot {21^ \circ } \\
  C.{\text{ }}\sin {21^ \circ } + \cos {21^ \circ } \\
  D.{\text{ }}\sec {21^ \circ } + \cot {21^ \circ } \\
\end{gathered} $

Answer 1

Hint:-Here, we go through the conversion of trigonometric angles.

For this type of question, we have to proceed according to options given, and just we have to change trigonometric functions to get a required option.
We can write ${\text{6}}{{\text{9}}^0}{\text{ = 9}}{{\text{0}}^0} - {21^0}$
And we know that all trigonometric functions are positive in the first quadrant.
Now, we can write question as,

$\cos ec\left( {{{90}^0} - {{21}^0}} \right) + \cot \left( {{{90}^0} - {{21}^0}} \right)$
= $\sec {21^0} + \tan {21^0}$
Here, we know that in first quadrant $\cos ec\left( {{{90}^0} - \theta } \right) = \cot \theta $ and $\cot \left( {{{90}^0} - \theta } \right) = \tan \theta $
Hence option $A$ is the correct answer.

Note:-Whenever we face such a type of question we have to proceed according to the option and remember how to change trigonometric functions.