Question

If \[\sin A = \dfrac{1}{3}\] ,then find the value of \[\cos A,\csc A + \tan A\] and \[\sec A\].

Answer 1

Hint: We are given the value of $\sin A$ therefore we can use the ratio of sine and then use the Pythagoras theorem of triangles to get the unknown values and then substitute the values in the formulas of trigonometric ratios to get the desired answers.

\[\sin \theta = \dfrac{{{\text{perpendicular}}}}{{{\text{hypotenuse}}}}\]
\[\cos A = \dfrac{{{\text{base}}}}{{{\text{hypotenuse}}}}\]
\[\csc A = \dfrac{1}{{\sin A}}\]
\[\tan A = \dfrac{{{\text{perpendicular}}}}{{{\text{base}}}}\]

Complete Step-by-step Solution
Given:
$\sin A= \dfrac{1}{3}$
Now, as we know that:
\[\sin \theta = \dfrac{{{\text{perpendicular}}}}{{{\text{hypotenuse}}}}\]
Comparing the values we get:
\[
  {\text{perpendicular}} = 1 \\
  {\text{hypotenuse}} = 3 \\
 \]
According to Pythagoras theorem of right angled triangle:
\[{\left( {{\text{perpendicular}}} \right)^2} + {\left( {{\text{base}}} \right)^2} = {\left( {{\text{hypotenuse}}} \right)^2}\]
Hence substituting the known values we get:
\[
  {\left( 1 \right)^2} + {\left( {{\text{base}}} \right)^2} = {\left( 3 \right)^2} \\
  {\left( {{\text{base}}} \right)^2} = 9 - 1 \\
  {\left( {{\text{base}}} \right)^2} = 8 \\
  {\text{base}} = 2\sqrt 2 \\
 \]
Now since we know that:
\[\cos A = \dfrac{{{\text{base}}}}{{{\text{hypotenuse}}}}\]
Therefore, substituting the values we get:
\[\cos A = \dfrac{{2\sqrt 2 }}{3}\]
Now we know that:
\[\csc A = \dfrac{1}{{\sin A}}\] and \[\tan A = \dfrac{{{\text{perpendicular}}}}{{{\text{base}}}}\]
Hence substituting the values we get:
\[
  \csc A = 3 \\
  \tan A = \dfrac{1}{{2\sqrt 2 }} \\
 \]
Now adding both the terms we get:
\[
  \csc A + \tan A = 3 + \dfrac{1}{{2\sqrt 2 }} \\
  \csc A + \tan A = \dfrac{{3\left( {2\sqrt 2 } \right) + 1\left( 1 \right)}}{{2\sqrt 2 }} \\
  \csc A + \tan A = \dfrac{{6\sqrt 2 + 1}}{{2\sqrt 2 }} \\
 \]
Now, as we know that:
\[\sec A = \dfrac{1}{{\cos A}}\]
Substituting the values we get:
\[\sec A = \dfrac{3}{{2\sqrt 2 }}\]

$\therefore $ The required values are \[\cos A = \dfrac{{2\sqrt 2 }}{3}\], $\csc A + \tan A = \dfrac{{6\sqrt 2 + 1}}{{2\sqrt 2 }}$ and $\sec A = \dfrac{3}{{2\sqrt 2 }}$

Note:
The trigonometric functions are the ratios of the sides of the right angled triangle.
Also, cosec A is the reciprocal of sin A and sec A is the reciprocal of cos A.
The values of cosec A and sec A can also be calculated by the formulas:
\[
  \sec A = \dfrac{{{\text{hypotenuse}}}}{{{\text{base}}}} \\
  \csc \theta = \dfrac{{{\text{hypotenuse}}}}{{{\text{perpendicular}}}} \\
 \]