Math, asked by prarthanatank999, 7 months ago

sin²theta + cos² theta = 1
prove the trigonometric identities


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Answers

Answered by Anonymous
6

Answer:

Let ABC be a right angled triangle right angled at B.

Let angle C ie angle ACB = theeta.

Now by Pythagoras theorem

AC^2= AB^2+BC^2 ……(1)

We know that sin theta = opposite side/ hypotenuse.

So sin theta = AB/AC. Similarly

cos theta= adjacent side/ hypotenuse

So cos theta = BC/AC

Now sin^2 theta + cos^2 theeta

= (AB/AC)^2 + (BC/AC)^2

= AB^2/AC^2 + BC^2/AC^2

= (AB^2+BC^2)/AC^2

= AC^2/AC^2 = 1 ( by using (1) )

Step-by-step explanation:

Answered by Anonymous
53

\huge\underbrace\mathfrak{Answer}

let PQR a right angle triangle

and right angled at R.

and angle θ at P.

By Pythagoras theorem

 {qp}^{2}  =  {pr}^{2}  +  {qr}^{2}

So ,

sin \: θ =  \frac{ perpendicular}{hypotenuse}

 =  > sin \: θ =  \frac{qr}{qp}

Now, Squaring Both Sides

 {sin \:}^{2} θ=   { (\frac{qr}{qp} )}^{2}  =  \frac{ {qr}^{2} }{ {qp}^{2} }

Similarly,

cos \: θ =  \frac{base}{hypotenuse}

 =  > cos \: θ =  \frac{pr}{qp}

Squaring Both Sides

  =  > {cos}^{2} θ =    { (\frac{pr}{qp} )}^{2}   = \frac{ {pr}^{2} }{ {qp}^{2} }

Now Adding cos² θ and sin² θ

So,

  =  > {sin}^{2} θ + {cos}^{2} θ  =  \frac{ {qr}^{2} }{ {qr}^{2} }  +  \frac{ {pr}^{2} }{ {qr}^{2} }

  =  > {sin}^{2} θ + {cos}^{2} θ  =   \frac{ {pr}^{2} +  {qr}^{2}  }{ {qp}^{2} } ...(1)

we know that

 {qp}^{2}  =  {pr}^{2}  +  {qr}^{2}

By Pythagoras theorem

So, we Substitute the value of pr²+qr² In eq(1)

Now,

 =  > {sin}^{2} θ + {cos}^{2} θ  =  \frac{ {qp}^{2} }{ {qp}^{2} }

 =  > {sin}^{2} θ + {cos}^{2} θ  = 1

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