Answer
Verified
484.5k+ 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, \[\cos \theta =\dfrac{h}{r}\Rightarrow h=r\cos \theta \]
\[\begin{align}
& \sin \theta =\dfrac{v}{r}\Rightarrow v=r\sin \theta \\
& \therefore {{r}^{2}}={{\left( r\cos \theta \right)}^{2}}+{{\left( r\sin \theta \right)}^{2}}\Rightarrow {{r}^{2}}={{r}^{2}}{{\cos }^{2}}\theta +{{r}^{2}}{{\sin }^{2}}\theta \\
& \therefore {{r}^{2}}{{\cos }^{2}}\theta +{{r}^{2}}{{\sin }^{2}}\theta ={{r}^{2}} \\
& {{r}^{2}}\left( {{\cos }^{2}}\theta +{{\sin }^{2}}\theta \right)={{r}^{2}} \\
& \Rightarrow {{\cos }^{2}}\theta +{{\sin }^{2}}\theta =1 \\
\end{align}\]
Recently Updated Pages
what is the correct chronological order of the following class 10 social science CBSE
Which of the following was not the actual cause for class 10 social science CBSE
Which of the following statements is not correct A class 10 social science CBSE
Which of the following leaders was not present in the class 10 social science CBSE
Garampani Sanctuary is located at A Diphu Assam B Gangtok class 10 social science CBSE
Which one of the following places is not covered by class 10 social science CBSE
Trending doubts
Which are the Top 10 Largest Countries of the World?
How do you graph the function fx 4x class 9 maths CBSE
The Equation xxx + 2 is Satisfied when x is Equal to Class 10 Maths
Fill the blanks with the suitable prepositions 1 The class 9 english CBSE
Difference Between Plant Cell and Animal Cell
Why is there a time difference of about 5 hours between class 10 social science CBSE
Difference between Prokaryotic cell and Eukaryotic class 11 biology CBSE
Give 10 examples for herbs , shrubs , climbers , creepers
Draw a labelled sketch of the human eye class 12 physics CBSE