Question

If a trigonometric ratio is given as \[\cos \theta =\dfrac{4}{5}\] . Find all the other trigonometric ratios of angle ‘\[\theta \]'.

Answer 1

Hint: As we know that cosine of any angle is the ratio of base to hypotenuse of the triangle of the triangle. Now we will use Pythagoras theorem to calculate the perpendicular of the triangle and thus we can find all other trigonometric ratios of the given angle.

Complete step-by-step solution -
We have been given \[\cos \theta =\dfrac{4}{5}\].
Let us suppose a \[\Delta ABC\], whose base is 4k and hypotenuse is 5k where k is any constant t.

We know that according to Pythagoras theorem for right angle triangles the sum of squares of perpendicular and base of the triangle is equal to the square of hypotenuse of that triangle.
So in \[\Delta ABC\],
\[\begin{align}
  & A{{B}^{2}}+B{{C}^{2}}=A{{C}^{2}} \\
 & A{{B}^{2}}+{{\left( 4K \right)}^{2}}={{\left( SK \right)}^{2}} \\
 & A{{B}^{2}}+16{{K}^{2}}=25{{K}^{2}} \\
 & A{{B}^{2}}=25{{K}^{2}}-16{{K}^{2}} \\
 & A{{B}^{2}}=9{{K}^{2}} \\
\end{align}\]
On taking squares on both sides, we get as follows:
\[\begin{align}
  & {{\left( A{{B}^{2}} \right)}^{{}^{1}/{}_{2}}}={{\left( 9{{K}^{2}} \right)}^{{}^{1}/{}_{2}}} \\
 & AB=3K \\
\end{align}\]
Now in \[\Delta ABC\] we have as follows:
\[\cos \theta =\dfrac{base}{hypotenuse}=\dfrac{4k}{5k}=\dfrac{4}{5}\] which is given to us.
\[\sin \theta =\dfrac{perpendicular}{hypotenuse}=\dfrac{AB}{AC}=\dfrac{3k}{5k}=\dfrac{3}{5}\]
\[\tan \theta =\dfrac{perpendicular}{base}=\dfrac{AB}{BC}=\dfrac{3k}{4k}=\dfrac{3}{4}\]
\[\cot \theta =\dfrac{base}{perpendicular}=\dfrac{BC}{AB}=\dfrac{4k}{3k}=\dfrac{4}{3}\]
\[cosec\theta =\dfrac{hypotenuse}{perpendicular}=\dfrac{AC}{AB}=\dfrac{5k}{3k}=\dfrac{5}{3}\]
\[cosec\theta =\dfrac{hypotenuse}{base}=\dfrac{AC}{BC}=\dfrac{5k}{4k}=\dfrac{5}{4}\]
Hence, all the trigonometric ratios are \[\sin \theta =\dfrac{3}{5},\tan \theta =\dfrac{3}{4},\cot \theta =\dfrac{4}{3},cosec\theta =\dfrac{5}{3},\sec \theta =\dfrac{5}{4}\].

Note: Be careful while calculating the other trigonometric ratios and don’t get confused among perpendicular, base and hypotenuse of the triangle or else you would get the wrong answer. Don’t take the sides of the triangle as ‘4’ and ‘5’because it can be in the multiple of any fixed number. So, it’s easier to assume them as 4k and 5k.