Question

Proof that :eta%3D%201" id="TexFormula1" title=" {sin}^{2} \theta + {cos}^{2} \:\theta= 1" alt=" {sin}^{2} \theta + {cos}^{2} \:\theta= 1" align="absmiddle" class="latex-formula">
Thx.
​

Answer 1

Answer:

Solution given:

Right angled triangle ABC is drawn where <C=$\theta$

we know that

$\displaystyle Sin\theta=\frac{opposite}{hypotenuse} =\frac{AB}{AC}$

$\displaystyle Cos\theta=\frac{adjacent}{hypotenuse}=\frac{BC}{AC}$

Now

left hand side

$\displaystyle {sin}^{2} \theta + {cos}^{2} \:\theta$

Substituting value

$(\frac{AB}{AC})²+(\frac{BC}{AC})²$

distributing power

$\frac{AB²}{AC²}+\frac{BC²}{AC²}$

Taking L.C.M

$\displaystyle \frac{AB²+BC²}{AC²}$....[I]

In ∆ABC By using Pythagoras law we get

$\boxed{\green{\bold{Opposite²+adjacent²=hypotenuse²}}}$

AB²+BC²=AC²

Substituting value of AB²+BC² in equation [I]

we get

$\displaystyle \frac{AC²}{AC²}$

=1

Right hand side

proved

Answer 2

Step-by-step explanation:

HOPE IT HELPS YOU

MARK ME DOWN BRAINLIEST