Question

Find the values of other five trigonometric ratios if $\sin \theta =\dfrac{{{a}^{2}}-{{b}^{2}}}{{{a}^{2}}+{{b}^{2}}}$ .

Answer 1

Hint: First we consider a right angled triangle and draw a diagram. We use the Pythagoras theorem to find the length of the side of a right triangle. We use the trigonometric ratio property to find the values of five trigonometric ratios which are $cos\theta ,tan\theta ,cosec\theta ,sec\theta $ and $\cot \theta $.

Complete step-by-step answer:
We have been given the value of $\sin \theta =\dfrac{{{a}^{2}}-{{b}^{2}}}{{{a}^{2}}+{{b}^{2}}}$.
We have to find the values of the other five trigonometric ratios.
First let us consider a right triangle $\Delta ABC$, in which $\angle B=90{}^\circ $.

We know that $\sin \theta =\dfrac{\text{perpendicular}}{\text{hypotenuse}}$ and we have $\sin \theta =\dfrac{{{a}^{2}}-{{b}^{2}}}{{{a}^{2}}+{{b}^{2}}}$.
When we compare both, we get
Perpendicular $AB={{a}^{2}}-{{b}^{2}}$ and hypotenuse $AC={{a}^{2}}+{{b}^{2}}$
Now, we know that according to the Pythagoras theorem in a right angle triangle, the square of the hypotenuse is equal to the sum of squares of the other two sides.
So, for $\Delta ABC$, we have
$A{{C}^{2}}=A{{B}^{2}}+B{{C}^{2}}$
Now, substituting the values we have
\[\begin{align}
  & {{\left( {{a}^{2}}+{{b}^{2}} \right)}^{2}}={{\left( {{a}^{2}}-{{b}^{2}} \right)}^{2}}+B{{C}^{2}} \\
 & \Rightarrow B{{C}^{2}}={{\left( {{a}^{2}}+{{b}^{2}} \right)}^{2}}-{{\left( {{a}^{2}}-{{b}^{2}} \right)}^{2}} \\
\end{align}\]
Now, we know that
 \[\begin{align}
  & {{\left( a+b \right)}^{2}}={{a}^{2}}+{{b}^{2}}+2ab \\
 & {{\left( a-b \right)}^{2}}={{a}^{2}}+{{b}^{2}}-2ab \\
\end{align}\]
Now, substituting the values, we get
\[\begin{align}
  & \Rightarrow B{{C}^{2}}=\left( {{a}^{4}}+{{b}^{4}}+2{{a}^{2}}{{b}^{2}} \right)-\left( {{a}^{4}}+{{b}^{4}}-2{{a}^{2}}{{b}^{2}} \right) \\
 & \Rightarrow B{{C}^{2}}={{a}^{4}}+{{b}^{4}}+2{{a}^{2}}{{b}^{2}}-{{a}^{4}}-{{b}^{4}}+2{{a}^{2}}{{b}^{2}} \\
 & \Rightarrow B{{C}^{2}}=4{{a}^{2}}{{b}^{2}} \\
\end{align}\]
Or we can write
\[\begin{align}
  & B{{C}^{2}}={{\left( 2ab \right)}^{2}} \\
 & BC=2ab \\
\end{align}\]
So, we have base \[BC=2ab\]
Now, we know that the values of trigonometric ratios are as follows:
\[\begin{align}
  & \cos \theta =\dfrac{\text{Base}}{\text{Hypotenuse}} \\
 & \tan \theta =\dfrac{\text{Perpendicular}}{\text{Base}} \\
 & cosec\theta =\dfrac{\text{Hypotenuse}}{\text{Perpendicular}} \\
 & \sec \theta =\dfrac{\text{Hypotenuse}}{\text{Base}} \\
 & \cot \theta =\dfrac{\text{Base}}{\text{Perpendicular}} \\
\end{align}\]
Now, we have Perpendicular $AB={{a}^{2}}-{{b}^{2}}$, hypotenuse $AC={{a}^{2}}+{{b}^{2}}$ and base \[BC=2ab\]
Substituting the values, we get
\[\begin{align}
  & \cos \theta =\dfrac{2ab}{{{a}^{2}}+{{b}^{2}}} \\
 & \tan \theta =\dfrac{{{a}^{2}}-{{b}^{2}}}{\text{2ab}} \\
 & cosec\theta =\dfrac{{{a}^{2}}+{{b}^{2}}}{{{a}^{2}}-{{b}^{2}}} \\
 & \sec \theta =\dfrac{{{a}^{2}}+{{b}^{2}}}{\text{2ab}} \\
 & \cot \theta =\dfrac{\text{2ab}}{{{a}^{2}}-{{b}^{2}}} \\
\end{align}\]
So, we get the values of all trigonometric ratios.

Note: To solve such types of questions the key concept is to draw a right triangle and apply Pythagoras theorem. Also, it is necessary to remember the formulas of all trigonometric ratios. In a right triangle the side opposite to the right angle is hypotenuse and the other two sides are base and perpendicular.