Prove whether the given expressions are equal or not ${{\tan }^{-1}}\left( \dfrac{1}{4} \right)+{{\tan }^{-1}}\left( \dfrac{2}{9} \right)=\dfrac{1}{2}{{\cos }^{-1}}\left( \dfrac{3}{5} \right)=\dfrac{1}{2}{{\sin }^{-1}}\left( \dfrac{4}{5} \right)$.
VerifiedHint:We will apply the formulas which are part of trigonometry. We will use ${{\tan }^{-1}}\left( x \right)+{{\tan }^{-1}}\left( y \right)={{\tan }^{-1}}\left( \dfrac{x+y}{1-xy} \right)$ here to solve the question and also we will apply the Pythagoras theorem and substitution in order to solve the question further.
Complete step-by-step answer:
We will here consider the expression ${{\tan }^{-1}}\left( \dfrac{1}{4} \right)+{{\tan }^{-1}}\left( \dfrac{2}{9} \right)=\dfrac{1}{2}{{\cos }^{-1}}\left( \dfrac{3}{5} \right)=\dfrac{1}{2}{{\sin }^{-1}}\left( \dfrac{4}{5} \right)...(i)$. We have three sides here. One is on the left hand side and the other two are on the middle and right sides. We will first consider the left hand side of expression (i) which is ${{\tan }^{-1}}\left( \dfrac{1}{4} \right)+{{\tan }^{-1}}\left( \dfrac{2}{9} \right)$. Now we will apply the formula which is given by ${{\tan }^{-1}}\left( x \right)+{{\tan }^{-1}}\left( y \right)={{\tan }^{-1}}\left( \dfrac{x+y}{1-xy} \right)$. Therefore, we get
${{\tan }^{-1}}\left( \dfrac{1}{4} \right)+{{\tan }^{-1}}\left( \dfrac{2}{9} \right)={{\tan }^{-1}}\left( \dfrac{\dfrac{1}{4}+\dfrac{2}{9}}{1-\left( \dfrac{1}{4} \right)\left( \dfrac{2}{9} \right)} \right)$
As the lcm of 4 and 9 is 36 thus, we have
\[ {{\tan }^{-1}}\left( \dfrac{1}{4} \right)+{{\tan }^{-1}}\left( \dfrac{2}{9} \right)={{\tan }^{-1}}\left( \dfrac{\dfrac{1}{4}+\dfrac{2}{9}}{1-\left( \dfrac{1}{4} \right)\left( \dfrac{2}{9} \right)} \right) \]
\[ \Rightarrow {{\tan }^{-1}}\left( \dfrac{1}{4} \right)+{{\tan }^{-1}}\left( \dfrac{2}{9} \right)={{\tan }^{-1}}\left( \dfrac{\dfrac{9+8}{36}}{\dfrac{36-2}{36}} \right) \]
After further solving we have
\[ {{\tan }^{-1}}\left( \dfrac{1}{4} \right)+{{\tan }^{-1}}\left( \dfrac{2}{9} \right)={{\tan }^{-1}}\left( \dfrac{\dfrac{17}{36}}{\dfrac{34}{36}} \right) \]
\[ \Rightarrow {{\tan }^{-1}}\left( \dfrac{1}{4} \right)+{{\tan }^{-1}}\left( \dfrac{2}{9} \right)={{\tan }^{-1}}\left( \dfrac{17}{34} \right) \]
\[ \Rightarrow {{\tan }^{-1}}\left( \dfrac{1}{4} \right)+{{\tan }^{-1}}\left( \dfrac{2}{9} \right)={{\tan }^{-1}}\left( \dfrac{1}{2} \right) \]
Now, we will put ${{\tan }^{-1}}\left( \dfrac{1}{2} \right)=x$ and we will place the inverse tangent term to the right side of the equation. Thus, we have ${{\tan }^{-1}}\left( \dfrac{1}{2} \right)=x$. The diagram for this is given as,
Now we will apply the Pythagoras theorem. Therefore, we get
\[ {{y}^{2}}={{\left( 1 \right)}^{2}}+{{\left( 2 \right)}^{2}} \]
\[ \Rightarrow {{y}^{2}}=1+4 \]
\[ \Rightarrow {{y}^{2}}=5 \]
\[ \Rightarrow y=\pm \sqrt{5} \]
As the side of the triangle cannot be negative therefore we have y = 5 which is the hypotenuse of the triangle. Now, we will use the formula $\cos x=\dfrac{\text{Base}}{\text{Hypotenuse}}$. This results in $\cos x=\dfrac{2}{\sqrt{5}}$. So, our left hand side is converted into $\cos x=\dfrac{2}{\sqrt{5}}...(ii)$.
Also, we will use the formula $\sin \left( x \right)=\dfrac{\text{Perpendicular}}{\text{Hypotenuse}}$ resulting into $\sin \left( x \right)=\dfrac{1}{\sqrt{5}}...(iii)$.
Now we will consider the middle term which is $\dfrac{1}{2}{{\cos }^{-1}}\left( \dfrac{3}{5} \right)$ and will substitute t equal to z. Therefore, we get $\dfrac{1}{2}{{\cos }^{-1}}\left( \dfrac{3}{5} \right)=z$. Now, we will multiply the whole equation by 2. Thus we now have ${{\cos }^{-1}}\left( \dfrac{3}{5} \right)=2z$ which after taking the inverse cosine term to the right side of the equation is converted into $\dfrac{3}{5}=\cos \left( 2z \right)$. As we know that the formula for $\cos \left( 2z \right)={{\cos }^{2}}\left( z \right)-1$. Therefore we now have
\[ {{\cos }^{2}}\left( z \right)-1=\dfrac{3}{5} \]
\[ \Rightarrow {{\cos }^{2}}\left( z \right)=\dfrac{3}{5}+1 \]
\[ \Rightarrow {{\cos }^{2}}\left( z \right)=\dfrac{3+5}{5} \]
\[ \Rightarrow {{\cos }^{2}}\left( z \right)=\dfrac{8}{5} \]
\[ \Rightarrow \cos \left( z \right)=\sqrt{\dfrac{8}{5}} \]
\[ \Rightarrow \cos \left( z \right)=\dfrac{2}{\sqrt{5}} \]
And this is equal to the equation (ii) since, x = z only. Therefore the left hand side is equal to the middle term.
Now we will consider the right hand side of the expression (i) which is $\dfrac{1}{2}{{\sin }^{-1}}\left( \dfrac{4}{5} \right)$. After substituting it equal to p we have $\dfrac{1}{2}{{\sin }^{-1}}\left( \dfrac{4}{5} \right)=p$. Now, we will multiply the whole equation by 2. Thus we now have ${{\sin }^{-1}}\left( \dfrac{4}{5} \right)=2p$ which after taking the inverse sine term to the right side of the equation is converted into $\left( \dfrac{4}{5} \right)=\sin \left( 2p \right)$. As we know that the formula for $\sin \left( 2p \right)=2\sin \left( p \right)\cos \left( p \right)$. Therefore we now have
\[ \sin \left( 2p \right)=2\sin \left( p \right)\cos \left( p \right) \]
\[ \Rightarrow \dfrac{4}{5}=2\sin \left( p \right)\cos \left( p \right) \]
As we are taking the same angle that means y = z = p. Thus we can substitute the value of $\cos \left( z \right)=\dfrac{2}{\sqrt{5}}$ as z = p. Thus, we get
\[ 2\sin \left( p \right)\cos \left( p \right)=\dfrac{4}{5} \]
\[ \Rightarrow 2\sin \left( p \right)\dfrac{2}{\sqrt{5}}=\dfrac{4}{5} \]
\[ \Rightarrow 2\sin \left( p \right)=\dfrac{4}{5}\times \dfrac{\sqrt{5}}{2} \]
\[ \Rightarrow 2\sin \left( p \right)=\dfrac{2}{\sqrt{5}} \]
\[ \Rightarrow \sin \left( p \right)=\dfrac{1}{\sqrt{5}} \]
Now, we will take the sine term to the right side therefore we get $p={{\sin }^{-1}}\left( \dfrac{1}{\sqrt{5}} \right)$ which is equal to the equation (iii) since p = x.
Hence, we have proved the expression ${{\tan }^{-1}}\left( \dfrac{1}{4} \right)+{{\tan }^{-1}}\left( \dfrac{2}{9} \right)=\dfrac{1}{2}{{\cos }^{-1}}\left( \dfrac{3}{5} \right)=\dfrac{1}{2}{{\sin }^{-1}}\left( \dfrac{4}{5} \right)$.
Note: While substituting the terms equal to variables which represent a certain angle implies the same angle if it is in the same question. Alternatively we could have started the question solving with the right hand side first and then assigned a triangle for it. After applying the Pythagoras theorem, find the base of the triangle and proceed to the middle and then the left hand side of the expression. Also, starting from the middle will also result in the right answer if used the same steps done in this question.
