Step-by-step explanation:
Let ABC be a right angled triangle where there's a right angle in B . Let the measure of the angle BAC be ∅. So,
[tex] \sin(∅) = \frac{BC}{AC} [/tex]
[tex] = > {\sin}^{2}∅ = \frac{ {BC}^{2} }{ {AC}^{2} } [/tex]
Also,
[tex] \cos(∅) = \frac{AB}{AC} [/tex]
[tex] = > { \cos}^{2}∅ = \frac{ {AB}^{2} }{ {AC}^{2} } [/tex]
Now adding the values of sin^2∅& cos^2∅,
[tex] { \sin}^{2} ∅ + { \cos}^{2} ∅ = \frac{ {BC}^{2} }{ {AC}^{2} } + \frac{ {AB}^{2} }{ {AC}^{2} } [/tex]
[tex] = > { \sin}^{2} ∅ + { \cos}^{2} ∅ = \frac{ {AB}^{2} + {BC}^{2} }{ {AC}^{2} } [/tex]
But we know that
[tex] {AC}^{2} = {AB}^{2} + {BC}^{2} [/tex] by applying Pythagorean Theorem
So ,
[tex] = > { \sin}^{2} ∅ + { \cos}^{2} ∅ = \frac{ {AC}^{2} }{ {AC}^{2} } = 1[/tex]
