Answer
Verified
394.2k+ views
Hint: We are going to substitute ${{\sin }^{-1}}\dfrac{3}{5}$ with $\alpha $and ${{\cot }^{-1}}\dfrac{3}{2}$ with$\beta $. Then we will use definitions of trigonometric ratios to determine the values required to use in the formula,
$\cos (\alpha +\beta )=\cos \alpha .\cos \beta -\sin \alpha .\sin \beta $.
Complete step by step answer:
In a right angled-triangle $\Delta \text{ABC}$ we denote the perpendicular $\overline{\text{AB}}$ by $p$, the $ \overline{\text{BC}}$ base by $b$ and the hypotenuse $ \overline{\text{AC}}$ by $h$.\[\]
From Pythagoras theorem,
\[{{p}^{2}}+{{b}^{2}}={{h}^{2}}\]
Let us also denote$\angle \text{ACB}=\theta $,
Then following the conventional the trigonometric definitions,
\[\sin \theta =\dfrac{p}{h},\cos \theta =\dfrac{b}{h},\tan \theta =\dfrac{p}{b},\cot \theta =\dfrac{b}{p}\]
Now using Pythagoras theorem,
\[\begin{align}
& {{\sin }^{2}}\theta +{{\cos }^{2}}\theta =\dfrac{{{p}^{2}}}{{{h}^{2}}}+\dfrac{{{b}^{2}}}{{{h}^{2}}}=\dfrac{{{p}^{2}}+{{b}^{2}}}{{{h}^{2}}}=1 \\
& \Rightarrow \cos \theta =\sqrt{1-{{\sin }^{2}}\theta }=\sqrt{1-\dfrac{{{p}^{2}}}{{{h}^{2}}}}......(1) \\
\end{align}\]
Again,
\[\cot \theta =\dfrac{b}{p}=\dfrac{\dfrac{b}{h}}{\dfrac{p}{h}}=\dfrac{\dfrac{b}{\sqrt{{{b}^{2}}+{{p}^{2}}}}}{\dfrac{p}{\sqrt{{{b}^{2}}+{{p}^{2}}}}}=\dfrac{\cos \theta }{\sin \theta }\]
From above we obtain,
\[\left. \begin{align}
& \cos \theta =\dfrac{b}{\sqrt{{{b}^{2}}+{{p}^{2}}}} \\
& \sin \theta =\dfrac{p}{\sqrt{{{b}^{2}}+{{p}^{2}}}} \\
\end{align} \right\}........(2)\]
Let us assign ${{\sin }^{-1}}\dfrac{3}{5}=\alpha $ and ${{\cot }^{-1}}\dfrac{3}{2}=\beta $.
Now taking $\sin$ both side of ${{\sin }^{-1}}\dfrac{3}{5}=\alpha $ as
\[\begin{align}
& {{\sin }^{-1}}\dfrac{3}{5}=\alpha \\
& \Rightarrow \sin ({{\sin }^{-1}}\dfrac{3}{5})=\sin \alpha \\
& \Rightarrow \sin \alpha =\dfrac{3}{5} \\
& \Rightarrow \text{cos}\alpha =\sqrt{1-{{\sin }^{2}}\alpha }=\sqrt{1-{{\left( \dfrac{3}{5} \right)}^{2}}}=\dfrac{4}{5}\left( \text{from}\left( 1 \right) \right) \\
\end{align}\]
Similarly taking cot both side of ${{\cot }^{-1}}\dfrac{3}{2}=\beta $,
\[\begin{align}
& {{\cot }^{-1}}\dfrac{3}{2}=\beta \\
& \Rightarrow \cot \left( {{\cot }^{-1}}\dfrac{3}{2} \right)=\cot \beta \\
& \Rightarrow \cot \beta =\dfrac{3}{2} \\
& \Rightarrow \cos \beta =\dfrac{3}{\sqrt{{{3}^{2}}+{{2}^{2}}}}=\dfrac{3}{\sqrt{13}},\sin \beta =\dfrac{2}{\sqrt{{{3}^{2}}+{{2}^{2}}}}=\dfrac{2}{\sqrt{13}}\left( \text{from}\left( 2 \right) \right) \\
& \\
\end{align}\]
Now we have all determined all the values to use in the formula,$\cos (\alpha +\beta )=\cos \alpha .\cos \beta -\sin \alpha .\sin \beta $.\[\]
Now putting all the values in the left hand side of the proof,
\[\begin{align}
& \cos \left( {{\sin }^{-1}}\dfrac{3}{5}+{{\cot }^{-1}}\dfrac{3}{2} \right) \\
& =\cos \left( \alpha +\beta \right) \\
& =\cos \alpha \cdot \cos \beta -\sin \alpha \cdot \sin \beta \\
& =\dfrac{4}{5}\cdot \dfrac{3}{\sqrt{13}}-\dfrac{3}{5}\cdot \dfrac{2}{\sqrt{13}}=\dfrac{6}{5\sqrt{13}} \\
\end{align}\]
The value obtained is equal to the right hand side of the proof. Hence Proved.
Note: In this problem we need to take care of proper placement during the calculation values while using the formula. The misplacement of values will result in hindrance in proof. We also need to be careful when we are taking inverse of any mathematical function because it may also be possible that the inverse may not be defined. In this problem the value of ${{\sin }^{-1}}\dfrac{3}{5}$ and ${{\cot }^{-1}}\dfrac{3}{2}$ lies domains of $\sin $and $\cot $functions respectively. In other problems if you ever find values which do not lie in the domain know that then the functions are not invertible and you cannot proceed further.
$\cos (\alpha +\beta )=\cos \alpha .\cos \beta -\sin \alpha .\sin \beta $.
Complete step by step answer:
In a right angled-triangle $\Delta \text{ABC}$ we denote the perpendicular $\overline{\text{AB}}$ by $p$, the $ \overline{\text{BC}}$ base by $b$ and the hypotenuse $ \overline{\text{AC}}$ by $h$.\[\]
From Pythagoras theorem,
\[{{p}^{2}}+{{b}^{2}}={{h}^{2}}\]
Let us also denote$\angle \text{ACB}=\theta $,
Then following the conventional the trigonometric definitions,
\[\sin \theta =\dfrac{p}{h},\cos \theta =\dfrac{b}{h},\tan \theta =\dfrac{p}{b},\cot \theta =\dfrac{b}{p}\]
Now using Pythagoras theorem,
\[\begin{align}
& {{\sin }^{2}}\theta +{{\cos }^{2}}\theta =\dfrac{{{p}^{2}}}{{{h}^{2}}}+\dfrac{{{b}^{2}}}{{{h}^{2}}}=\dfrac{{{p}^{2}}+{{b}^{2}}}{{{h}^{2}}}=1 \\
& \Rightarrow \cos \theta =\sqrt{1-{{\sin }^{2}}\theta }=\sqrt{1-\dfrac{{{p}^{2}}}{{{h}^{2}}}}......(1) \\
\end{align}\]
Again,
\[\cot \theta =\dfrac{b}{p}=\dfrac{\dfrac{b}{h}}{\dfrac{p}{h}}=\dfrac{\dfrac{b}{\sqrt{{{b}^{2}}+{{p}^{2}}}}}{\dfrac{p}{\sqrt{{{b}^{2}}+{{p}^{2}}}}}=\dfrac{\cos \theta }{\sin \theta }\]
From above we obtain,
\[\left. \begin{align}
& \cos \theta =\dfrac{b}{\sqrt{{{b}^{2}}+{{p}^{2}}}} \\
& \sin \theta =\dfrac{p}{\sqrt{{{b}^{2}}+{{p}^{2}}}} \\
\end{align} \right\}........(2)\]
Let us assign ${{\sin }^{-1}}\dfrac{3}{5}=\alpha $ and ${{\cot }^{-1}}\dfrac{3}{2}=\beta $.
Now taking $\sin$ both side of ${{\sin }^{-1}}\dfrac{3}{5}=\alpha $ as
\[\begin{align}
& {{\sin }^{-1}}\dfrac{3}{5}=\alpha \\
& \Rightarrow \sin ({{\sin }^{-1}}\dfrac{3}{5})=\sin \alpha \\
& \Rightarrow \sin \alpha =\dfrac{3}{5} \\
& \Rightarrow \text{cos}\alpha =\sqrt{1-{{\sin }^{2}}\alpha }=\sqrt{1-{{\left( \dfrac{3}{5} \right)}^{2}}}=\dfrac{4}{5}\left( \text{from}\left( 1 \right) \right) \\
\end{align}\]
Similarly taking cot both side of ${{\cot }^{-1}}\dfrac{3}{2}=\beta $,
\[\begin{align}
& {{\cot }^{-1}}\dfrac{3}{2}=\beta \\
& \Rightarrow \cot \left( {{\cot }^{-1}}\dfrac{3}{2} \right)=\cot \beta \\
& \Rightarrow \cot \beta =\dfrac{3}{2} \\
& \Rightarrow \cos \beta =\dfrac{3}{\sqrt{{{3}^{2}}+{{2}^{2}}}}=\dfrac{3}{\sqrt{13}},\sin \beta =\dfrac{2}{\sqrt{{{3}^{2}}+{{2}^{2}}}}=\dfrac{2}{\sqrt{13}}\left( \text{from}\left( 2 \right) \right) \\
& \\
\end{align}\]
Now we have all determined all the values to use in the formula,$\cos (\alpha +\beta )=\cos \alpha .\cos \beta -\sin \alpha .\sin \beta $.\[\]
Now putting all the values in the left hand side of the proof,
\[\begin{align}
& \cos \left( {{\sin }^{-1}}\dfrac{3}{5}+{{\cot }^{-1}}\dfrac{3}{2} \right) \\
& =\cos \left( \alpha +\beta \right) \\
& =\cos \alpha \cdot \cos \beta -\sin \alpha \cdot \sin \beta \\
& =\dfrac{4}{5}\cdot \dfrac{3}{\sqrt{13}}-\dfrac{3}{5}\cdot \dfrac{2}{\sqrt{13}}=\dfrac{6}{5\sqrt{13}} \\
\end{align}\]
The value obtained is equal to the right hand side of the proof. Hence Proved.
Note: In this problem we need to take care of proper placement during the calculation values while using the formula. The misplacement of values will result in hindrance in proof. We also need to be careful when we are taking inverse of any mathematical function because it may also be possible that the inverse may not be defined. In this problem the value of ${{\sin }^{-1}}\dfrac{3}{5}$ and ${{\cot }^{-1}}\dfrac{3}{2}$ lies domains of $\sin $and $\cot $functions respectively. In other problems if you ever find values which do not lie in the domain know that then the functions are not invertible and you cannot proceed further.
Recently Updated Pages
Basicity of sulphurous acid and sulphuric acid are
Assertion The resistivity of a semiconductor increases class 13 physics CBSE
The Equation xxx + 2 is Satisfied when x is Equal to Class 10 Maths
What is the stopping potential when the metal with class 12 physics JEE_Main
The momentum of a photon is 2 times 10 16gm cmsec Its class 12 physics JEE_Main
Using the following information to help you answer class 12 chemistry CBSE
Trending doubts
State the differences between manure and fertilize class 8 biology CBSE
Why are xylem and phloem called complex tissues aBoth class 11 biology CBSE
Difference between Prokaryotic cell and Eukaryotic class 11 biology CBSE
Difference Between Plant Cell and Animal Cell
What would happen if plasma membrane ruptures or breaks class 11 biology CBSE
Give 10 examples for herbs , shrubs , climbers , creepers
What precautions do you take while observing the nucleus class 11 biology CBSE
What would happen to the life of a cell if there was class 11 biology CBSE
Change the following sentences into negative and interrogative class 10 english CBSE