# Prove the following identity \[{{\sin }^{2}}\theta +{{\cos }^{2}}\theta =1\], geometrically.

Answer

Verified

327.3k+ views

Hint: To solve this problem use Pythagoras theorem i.e. the square of the hypotenuse of the right angled triangle is equal to the area to the sum of the square on the other two sides.

Let us consider a triangle ABC, where it is right angled at \[{{90}^{\circ }}\] and angle C is taken as \[\theta \].

According to angle C \[\theta \], AB is the opposite side, BC is the adjacent side and AC is the hypotenuse.

We know the Pythagoras theorem.

i.e. the square of the hypotenuse of the right angled triangle is equal to the area to the sum of the square on the other two sides.

i.e. \[A{{C}^{2}}=A{{B}^{2}}+B{{C}^{2}}-(1)\]

Now, divide both RHS and LHS with \[A{{C}^{2}}\]

\[\begin{align}

& \Rightarrow \dfrac{A{{B}^{2}}+B{{C}^{2}}}{A{{C}^{2}}}=\dfrac{A{{C}^{2}}}{A{{C}^{2}}} \\

& \therefore \dfrac{A{{B}^{2}}}{A{{C}^{2}}}+\dfrac{B{{C}^{2}}}{A{{C}^{2}}}=1\Rightarrow {{\left( \dfrac{AB}{AC} \right)}^{2}}+{{\left( \dfrac{BC}{AC} \right)}^{2}}=1-(2) \\

\end{align}\]

where, \[\dfrac{AB}{AC}=\dfrac{opposite\ side}{hypotenuse}=\sin \theta \]

\[\begin{align}

& \dfrac{BC}{AC}=\dfrac{adjacent\ side}{hypotenuse}=\cos \theta \\

& \therefore {{\left( \dfrac{AB}{AC} \right)}^{2}}+{{\left( \dfrac{BC}{AC} \right)}^{2}}=1\Rightarrow {{\left( \sin \theta \right)}^{2}}+{{\left( \cos \theta \right)}^{2}}=1 \\

& \Rightarrow {{\sin }^{2}}\theta +{{\cos }^{2}}\theta =1 \\

\end{align}\]

Note: This is another method to prove \[{{\sin }^{2}}\theta +{{\cos }^{2}}\theta =1\]

From a right angle triangle, \[\cos \theta =\dfrac{adjacent\ side}{hypotenuse}=\dfrac{h}{r}\]

\[\sin \theta =\dfrac{opposite\ side}{hypotenuse}=\dfrac{v}{r}\]

From Pythagoras theorem, \[{{r}^{2}}={{v}^{2}}+{{h}^{2}}\]

Here, \[\cos \theta =\dfrac{h}{r}\Rightarrow h=r\cos \theta \]

\[\begin{align}

& \sin \theta =\dfrac{v}{r}\Rightarrow v=r\sin \theta \\

& \therefore {{r}^{2}}={{\left( r\cos \theta \right)}^{2}}+{{\left( r\sin \theta \right)}^{2}}\Rightarrow {{r}^{2}}={{r}^{2}}{{\cos }^{2}}\theta +{{r}^{2}}{{\sin }^{2}}\theta \\

& \therefore {{r}^{2}}{{\cos }^{2}}\theta +{{r}^{2}}{{\sin }^{2}}\theta ={{r}^{2}} \\

& {{r}^{2}}\left( {{\cos }^{2}}\theta +{{\sin }^{2}}\theta \right)={{r}^{2}} \\

& \Rightarrow {{\cos }^{2}}\theta +{{\sin }^{2}}\theta =1 \\

\end{align}\]

Let us consider a triangle ABC, where it is right angled at \[{{90}^{\circ }}\] and angle C is taken as \[\theta \].

According to angle C \[\theta \], AB is the opposite side, BC is the adjacent side and AC is the hypotenuse.

We know the Pythagoras theorem.

i.e. the square of the hypotenuse of the right angled triangle is equal to the area to the sum of the square on the other two sides.

i.e. \[A{{C}^{2}}=A{{B}^{2}}+B{{C}^{2}}-(1)\]

Now, divide both RHS and LHS with \[A{{C}^{2}}\]

\[\begin{align}

& \Rightarrow \dfrac{A{{B}^{2}}+B{{C}^{2}}}{A{{C}^{2}}}=\dfrac{A{{C}^{2}}}{A{{C}^{2}}} \\

& \therefore \dfrac{A{{B}^{2}}}{A{{C}^{2}}}+\dfrac{B{{C}^{2}}}{A{{C}^{2}}}=1\Rightarrow {{\left( \dfrac{AB}{AC} \right)}^{2}}+{{\left( \dfrac{BC}{AC} \right)}^{2}}=1-(2) \\

\end{align}\]

where, \[\dfrac{AB}{AC}=\dfrac{opposite\ side}{hypotenuse}=\sin \theta \]

\[\begin{align}

& \dfrac{BC}{AC}=\dfrac{adjacent\ side}{hypotenuse}=\cos \theta \\

& \therefore {{\left( \dfrac{AB}{AC} \right)}^{2}}+{{\left( \dfrac{BC}{AC} \right)}^{2}}=1\Rightarrow {{\left( \sin \theta \right)}^{2}}+{{\left( \cos \theta \right)}^{2}}=1 \\

& \Rightarrow {{\sin }^{2}}\theta +{{\cos }^{2}}\theta =1 \\

\end{align}\]

Note: This is another method to prove \[{{\sin }^{2}}\theta +{{\cos }^{2}}\theta =1\]

From a right angle triangle, \[\cos \theta =\dfrac{adjacent\ side}{hypotenuse}=\dfrac{h}{r}\]

\[\sin \theta =\dfrac{opposite\ side}{hypotenuse}=\dfrac{v}{r}\]

From Pythagoras theorem, \[{{r}^{2}}={{v}^{2}}+{{h}^{2}}\]

Here, \[\cos \theta =\dfrac{h}{r}\Rightarrow h=r\cos \theta \]

\[\begin{align}

& \sin \theta =\dfrac{v}{r}\Rightarrow v=r\sin \theta \\

& \therefore {{r}^{2}}={{\left( r\cos \theta \right)}^{2}}+{{\left( r\sin \theta \right)}^{2}}\Rightarrow {{r}^{2}}={{r}^{2}}{{\cos }^{2}}\theta +{{r}^{2}}{{\sin }^{2}}\theta \\

& \therefore {{r}^{2}}{{\cos }^{2}}\theta +{{r}^{2}}{{\sin }^{2}}\theta ={{r}^{2}} \\

& {{r}^{2}}\left( {{\cos }^{2}}\theta +{{\sin }^{2}}\theta \right)={{r}^{2}} \\

& \Rightarrow {{\cos }^{2}}\theta +{{\sin }^{2}}\theta =1 \\

\end{align}\]

Last updated date: 29th May 2023

â€¢

Total views: 327.3k

â€¢

Views today: 7.85k

Recently Updated Pages

If a spring has a period T and is cut into the n equal class 11 physics CBSE

A planet moves around the sun in nearly circular orbit class 11 physics CBSE

In any triangle AB2 BC4 CA3 and D is the midpoint of class 11 maths JEE_Main

In a Delta ABC 2asin dfracAB+C2 is equal to IIT Screening class 11 maths JEE_Main

If in aDelta ABCangle A 45circ angle C 60circ then class 11 maths JEE_Main

If in a triangle rmABC side a sqrt 3 + 1rmcm and angle class 11 maths JEE_Main

Trending doubts

Difference Between Plant Cell and Animal Cell

Write an application to the principal requesting five class 10 english CBSE

Ray optics is valid when characteristic dimensions class 12 physics CBSE

Give 10 examples for herbs , shrubs , climbers , creepers

Write the 6 fundamental rights of India and explain in detail

Write a letter to the principal requesting him to grant class 10 english CBSE

List out three methods of soil conservation

Fill in the blanks A 1 lakh ten thousand B 1 million class 9 maths CBSE

Epipetalous and syngenesious stamens occur in aSolanaceae class 11 biology CBSE