Question

If \[\tan A = \dfrac{b}{a}\] where a and b are real numbers find the value of \[{\sin ^2}A\] and \[{\cos ^2}A\] also verify that \[{\sin ^2}A + {\cos ^2}A = 1\]

Answer 1

Hint: we are going to use the properties of trigonometric functions to get the solution.
1) \[{\text{tan}}\;{\text{A = }}\dfrac{{{\text{perpendicular}}}}{{{\text{base}}}}\]
2) \[{\text{sin A = }}\dfrac{{{\text{Perpendicular}}}}{{{\text{hypotenuse}}}}\]
3) \[{\text{cos A = }}\dfrac{{{\text{base}}}}{{{\text{hypotenuse}}}}\]
Where $A$ is an angle between the hypotenuse and base sides on a right angled triangle.

Complete step by step solution:
It is given that \[\tan A = \dfrac{b}{a}\] where ‘a’ and ‘b’ are real numbers.
In a right angle triangle, we know that
\[{\text{tan A = }}\dfrac{{{\text{perpendicular}}}}{{{\text{base}}}}\]
Where
\[
  {\text{Perpendicular = b}} \\
  {\text{Base = a}} \\
 \]

Now our aim is to find the values of \[{\sin ^2}A\] and \[{\cos ^2}A\].
Now we are going to use the Pythagoras theorem. Pythagoras theorem states that in a right triangle the square of the hypotenuse side is the sum of the square of the other two sides.
So we have,
$\text{Hypotenuse}^2=\text{Perpendicular}^2 + \text{base}^2$
$\text{Hypotenuse}^2=\text{b}^2 + \text{a}^2$
Now we are going to take square root on both sides. Then we get,
\[{\text{hypotenuse = }}\sqrt {{{\text{b}}^{\text{2}}}{\text{ + }}{{\text{a}}^{\text{2}}}} \]
And we already know that, in a right angled triangle
\[{\text{sin A = }}\dfrac{{{\text{Perpendicular}}}}{{{\text{hypotenuse}}}}\]
Where \[Perpendicular = b\] and we already found the value of hypotenuse that \[{\text{hypotenuse = }}\sqrt {{{\text{b}}^{\text{2}}}{\text{ + }}{{\text{a}}^{\text{2}}}} \]
Now we are going to substitute these values on \[{\text{sin}}\;{\text{A = }}\dfrac{{{\text{Perpendicular}}}}{{{\text{hypotenuse}}}}\].
\[ \Rightarrow \sin A = \dfrac{b}{{\sqrt {{b^2} + {a^2}} }}\]
Now we are going to take the square on both sides.
\[ \Rightarrow {\sin ^2}A = \dfrac{{{b^2}}}{{{b^2} + {a^2}}} \ldots \ldots \left( 1 \right)\]
And we already know that, in a right angled triangle,
\[{\text{cos A = }}\dfrac{{{\text{base}}}}{{{\text{hypotenuse}}}}\]
Where \[base = a\] and we already found the value of hypotenuse that \[{\text{hypotenuse = }}\sqrt {{{\text{b}}^{\text{2}}}{\text{ + }}{{\text{a}}^{\text{2}}}} \]
Now we are going to substitute these values on \[{\text{cos A = }}\dfrac{{{\text{base}}}}{{{\text{hypotenuse}}}}\].
\[ \Rightarrow \cos A = \dfrac{a}{{\sqrt {{b^2} + {a^2}} }}\]
Now we are going to take the square on both sides.
\[ \Rightarrow {\cos ^2}A = \dfrac{{{a^2}}}{{{b^2} + {a^2}}} \ldots \ldots ..\left( 2 \right)\]
Now we found the values of \[{\sin ^2}A\] and \[{\cos ^2}A\] .
Now we are going to verify that \[{\sin ^2}A + {\cos ^2}A = 1\]
So we are going to add equation (1) and (2),
\[ \Rightarrow {\sin ^2}A + {\cos ^2}A = \dfrac{{{b^2}}}{{{b^2} + {a^2}}} + \dfrac{{{a^2}}}{{{b^2} + {a^2}}}\]
Now we are going to take LCM
\[ \Rightarrow {\sin ^2}A + {\cos ^2}A = \dfrac{{{b^2} + {a^2}}}{{{b^2} + {a^2}}}\]
Numerator and denominator have the same term so we can cancel them.
\[ \Rightarrow {\sin ^2}A + {\cos ^2}A = 1\]

Hence we can conclude that \[{\sin ^2}A + {\cos ^2}A = 1\]

Note:
The trigonometric identities or properties are equations that are true and valid for right-angled triangles. We have to connect the formula of trigonometric identities that are sin, cos, and tan to solve this problem.