Answer:
Solution given:
Right angled triangle ABC is drawn where <C=[tex]\theta[/tex]
we know that
[tex]\displaystyle Sin\theta=\frac{opposite}{hypotenuse} =\frac{AB}{AC}[/tex]
[tex]\displaystyle Cos\theta=\frac{adjacent}{hypotenuse}=\frac{BC}{AC} [/tex]
Now
left hand side
[tex] \displaystyle {sin}^{2} \theta + {cos}^{2} \:\theta[/tex]
Substituting value
[tex](\frac{AB}{AC})²+(\frac{BC}{AC})²[/tex]
distributing power
[tex]\frac{AB²}{AC²}+\frac{BC²}{AC²}[/tex]
Taking L.C.M
[tex]\displaystyle \frac{AB²+BC²}{AC²}[/tex]....[I]
In ∆ABC By using Pythagoras law we get
[tex]\boxed{\green{\bold{Opposite²+adjacent²=hypotenuse²}}}[/tex]
AB²+BC²=AC²
Substituting value of AB²+BC² in equation [I]
we get
[tex]\displaystyle \frac{AC²}{AC²}[/tex]
=1
Right hand side
proved
