Question

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Answer 1

Question :

Complete the equation :

sin²θ + cos²θ = $\sf \boxed{ \: \: \: \: \: \: \: \: \: }$

Answer :

sin²θ + cos²θ = $\sf \boxed{1}$

Solution :

We know that, In a right angled triangle :

• $\sf sin\theta = \dfrac{Side \: opposite \: to \: \theta}{Hypotenuse}$
• $\sf cos\theta = \dfrac{Side \: adjacent \: to \: \theta}{Hypotenuse}$

Let's consider :

• Side opposite to θ = P
• Side adjacent to θ = B
• Hypotenuse = H

Now, we have to find sin²θ + cos²θ,

So, let's find sin²θ first :

We have :

$\sf sin\theta = \dfrac{P}{H}$

By squaring both sides :

$\sf : \implies sin^{2} \theta = \Bigg(\dfrac{P}{H}\Bigg)^{2}$

$\sf : \implies sin^{2} \theta = \dfrac{(P)^{2}}{(H)^{2}}$

Now, let's find cos²θ :

We have :

$\sf cos\theta = \dfrac{B}{H}$

By squaring both sides :

$\sf : \implies cos^{2}\theta = \Bigg(\dfrac{B}{H}\Bigg)^{2}$

$\sf : \implies cos^{2} \theta = \dfrac{(B)^{2}}{(H)^{2}}$

Now, we have sin²θ and cos²θ,

Let's find sin²θ + cos²θ :

$\sf : \implies sin^{2} \theta + cos^{2} \theta = \dfrac{(P)^{2}}{(H)^{2}} + \dfrac{(B)^{2}}{(H)^{2}}$

$\sf : \implies sin^{2} \theta + cos^{2} \theta = \dfrac{(P)^{2} + (B)^{2}}{(H)^{2}}$

Now, by Pythagoras' theorem :

• $\Large \underline{\boxed{\bf{H^{2} = B^{2} + P^{2}}}}$

$\sf : \implies sin^{2} \theta + cos^{2} \theta = \dfrac{(H)^{2}}{(H)^{2}}$

$\sf : \implies sin^{2} \theta + cos^{2} \theta = \cancel{\dfrac{(H)^{2}}{(H)^{2}}}$

$\sf : \implies sin^{2} \theta + cos^{2} \theta = 1$

Hence, value of sin²θ + cos²θ = 1.

Answer 2

Answer:

Answer :

sin²θ + cos²θ = \sf \boxed{1}

1

Solution :

We know that, In a right angled triangle :

\sf sin\theta = \dfrac{Side \: opposite \: to \: \theta}{Hypotenuse}sinθ=

Hypotenuse

Sideoppositetoθ

\sf cos\theta = \dfrac{Side \: adjacent \: to \: \theta}{Hypotenuse}cosθ=

Hypotenuse

Sideadjacenttoθ

Let's consider :

Side opposite to θ = P

Side adjacent to θ = B

Hypotenuse = H

Now, we have to find sin²θ + cos²θ,

So, let's find sin²θ first :

We have :

\sf sin\theta = \dfrac{P}{H}sinθ=

H

P

By squaring both sides :

\sf : \implies sin^{2} \theta = \Bigg(\dfrac{P}{H}\Bigg)^{2}:⟹sin

2

θ=(

H

P

)

2

\sf : \implies sin^{2} \theta = \dfrac{(P)^{2}}{(H)^{2}}:⟹sin

2

θ=

(H)

2

(P)

2

Now, let's find cos²θ :

We have :

\sf cos\theta = \dfrac{B}{H}cosθ=

H

B

By squaring both sides :

\sf : \implies cos^{2}\theta = \Bigg(\dfrac{B}{H}\Bigg)^{2}:⟹cos

2

θ=(

H

B

)

2

\sf : \implies cos^{2} \theta = \dfrac{(B)^{2}}{(H)^{2}}:⟹cos

2

θ=

(H)

2

(B)

2

Now, we have sin²θ and cos²θ,

Let's find sin²θ + cos²θ :

\sf : \implies sin^{2} \theta + cos^{2} \theta = \dfrac{(P)^{2}}{(H)^{2}} + \dfrac{(B)^{2}}{(H)^{2}}:⟹sin

2

θ+cos

2

θ=

(H)

2

(P)

2

+

(H)

2

(B)

2

\sf : \implies sin^{2} \theta + cos^{2} \theta = \dfrac{(P)^{2} + (B)^{2}}{(H)^{2}}:⟹sin

2

θ+cos

2

θ=

(H)

2

(P)

2

+(B)

2

Now, by Pythagoras' theorem :

\Large \underline{\boxed{\bf{H^{2} = B^{2} + P^{2}}}}

H

2

=B

2

+P

2

\sf : \implies sin^{2} \theta + cos^{2} \theta = \dfrac{(H)^{2}}{(H)^{2}}:⟹sin

2

θ+cos

2

θ=

(H)

2

(H)

2

\sf : \implies sin^{2} \theta + cos^{2} \theta = \cancel{\dfrac{(H)^{2}}{(H)^{2}}}:⟹sin

2

θ+cos

2

θ=

(H)

2

(H)

2

\sf : \implies sin^{2} \theta + cos^{2} \theta = 1:⟹sin

2

θ+cos

2

θ=1

Hence, value of sin²θ + cos²θ = 1.