Question

Why Sin²¤ + Cos²¤ = 1​

Answer 1

Here Is Your Ans

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BY Pythagoras Theorem

H² = P² + B² ----- 1

Divide H² In Equation 1

$= > \frac{ {h}^{2} }{ {h}^{2} } = \frac{ {p}^{2} }{ {h}^{2} } + \frac{ {b}^{2} }{ {h}^{2} } \\ \\ = > 1 = ( { \frac{p}{h} )}^{2} + {( \frac{b}{h} )}^{2}$

= > 1 = Sin²¤ + Cos²¤

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$<marquee > hope it helps u$

Answer 2

$\underline{\underline{\mathfrak{\green{Answer:-}}}}$

Sin²Φ + cos²Φ = 1

$\underline{\underline{\mathfrak{\green{Explanation:-}}}}$

$\underline{\pink{To \: Prove}}$

sin²Φ + cos²Φ = 1

$\underline{\pink{ </strong><strong>Proof</strong><strong>}}$

let us consider LHS,

= sin²Φ + cos²Φ

$\boxed{\red{sinΦ = \dfrac{opp. side}{Hypoteneuse} }}$

$\boxed{\red{cosΦ = \dfrac{Adj.side}{Hypoteneuse}}}$

On putting the values,

= ${(\dfrac{Opp.side}{Hyp})}^{2} +{(\dfrac{Adj.side}{Hyp})}^{2}$

= $\dfrac{{(Opp.side)}^{2}}{{(Hyp)}^{2}}+\dfrac{{(Adj.side)}^{2}}{{(Hyp)}^{2}}$

= $\dfrac{{(opp.side)}^{2}+{(Adj.side)}^{2}}{({Hyp})^{2}}$

From pythagoras theorem,

(opp.side)²+(Adj.side)²= (Hyp)²

= $\dfrac{{(Hyp)}^{2}}{({Hyp})^{2}}$

= 1

Therefore, LHS = 1

RHS = 1

Here,

LHS = RHS

$\blue{HENCE \: PROVED}$