Answer
Verified
410.1k+ views
Hint: For solving this question we will use the formula $2{{\tan }^{-1}}x={{\tan }^{-1}}\left( \dfrac{2x}{1-{{x}^{2}}} \right)$ , ${{\tan }^{2}}\theta \text{=se}{{\text{c}}^{2}}\theta \text{-1}$ and ${{\tan }^{-1}}x+{{\tan }^{-1}}y={{\tan }^{-1}}\left( \dfrac{x+y}{1-xy} \right)$ . After that, we will try to solve $2{{\tan }^{-1}}\left( \dfrac{1}{5} \right)+{{\sec }^{-1}}\left( \dfrac{5\sqrt{2}}{7} \right)+2{{\tan }^{-1}}\left( \dfrac{1}{8} \right)$ and prove that, it is equal to ${{\tan }^{-1}}\left( 1 \right)$ . Then, we will prove the desired result easily.
Complete step-by-step solution -
Given:
We have to prove the following equation:
$2{{\tan }^{-1}}\left( \dfrac{1}{5} \right)+{{\sec }^{-1}}\left( \dfrac{5\sqrt{2}}{7} \right)+2{{\tan }^{-1}}\left( \dfrac{1}{8} \right)=\dfrac{\pi }{4}$
Now, let $\alpha =2{{\tan }^{-1}}\left( \dfrac{1}{5} \right)$ , $\beta ={{\sec }^{-1}}\left( \dfrac{5\sqrt{2}}{7} \right)$ and $\gamma =2{{\tan }^{-1}}\left( \dfrac{1}{8} \right)$ . So, we will solve for $\alpha $ , $\beta $ and $\gamma $ separately. Then, we will find the value of $\alpha +\beta +\gamma $ .
Now, before we proceed further we should know the following formulas:
$ 2{{\tan }^{-1}}x={{\tan }^{-1}}\left( \dfrac{2x}{1-{{x}^{2}}} \right)\text{ }\left( \text{if }-1 < x < 1 \right)...............\left( 1 \right) $
$ {{\tan }^{2}}\theta \text{=se}{{\text{c}}^{2}}\theta \text{-1 }....................\left( 2 \right) $
$ {{\tan }^{-1}}x+{{\tan }^{-1}}y={{\tan }^{-1}}\left( \dfrac{x+y}{1-xy} \right)\text{ }\left( \text{if }xy<1 \right).............................\left( 3 \right) $
$ {{\tan }^{-1}}1=\dfrac{\pi }{4}..................................\left( 4 \right) $
Now, we will simplify $\alpha $ , $\beta $ and $\gamma $ separately with the help of the above four formulas.
Simplification of $\alpha =2{{\tan }^{-1}}\left( \dfrac{1}{5} \right)$ :
Now, we will use the formula from the equation (1) as $-1 < \dfrac{1}{5} < 1$ to write $ 2{{\tan }^{-1}}\left( \dfrac{1}{5} \right)={{\tan }^{-1}}\left( \dfrac{5}{12} \right)$ . Then,
$\begin{align}
& \alpha =2{{\tan }^{-1}}\left( \dfrac{1}{5} \right) \\
& \Rightarrow \alpha ={{\tan }^{-1}}\left( \dfrac{2\times \dfrac{1}{5}}{1-{{\left( \dfrac{1}{5} \right)}^{2}}} \right) \\
& \Rightarrow \alpha ={{\tan }^{-1}}\left( \dfrac{2\times 5}{{{5}^{2}}-1} \right) \\
& \Rightarrow \alpha ={{\tan }^{-1}}\left( \dfrac{10}{25-1} \right) \\
& \Rightarrow \alpha ={{\tan }^{-1}}\left( \dfrac{10}{24} \right) \\
& \Rightarrow \alpha ={{\tan }^{-1}}\left( \dfrac{5}{12} \right)........................\left( 5 \right) \\
\end{align}$
Simplification of $\gamma =2{{\tan }^{-1}}\left( \dfrac{1}{8} \right)$ :
Now, we will use the formula from the equation (1) as $-1<\dfrac{1}{8}<1$ to write $2{{\tan }^{-1}}\left( \dfrac{1}{8} \right)={{\tan }^{-1}}\left( \dfrac{16}{63} \right)$ . Then,
$\begin{align}
& \gamma =2{{\tan }^{-1}}\left( \dfrac{1}{8} \right) \\
& \Rightarrow \gamma ={{\tan }^{-1}}\left( \dfrac{2\times \dfrac{1}{8}}{1-{{\left( \dfrac{1}{8} \right)}^{2}}} \right) \\
& \Rightarrow \gamma ={{\tan }^{-1}}\left( \dfrac{2\times 8}{{{8}^{2}}-1} \right) \\
& \Rightarrow \gamma ={{\tan }^{-1}}\left( \dfrac{16}{64-1} \right) \\
& \Rightarrow \gamma ={{\tan }^{-1}}\left( \dfrac{16}{63} \right)........................\left( 6 \right) \\
\end{align}$
Simplification of $\beta ={{\sec }^{-1}}\left( \dfrac{5\sqrt{2}}{7} \right)$ :
Now, we write $\sec \beta =\dfrac{5\sqrt{2}}{7}$ and apply the formula from the equation (2) to get the value of $\tan \beta $ . Then,
$\begin{align}
& {{\tan }^{2}}\beta ={{\sec }^{2}}\beta -1 \\
& \Rightarrow {{\tan }^{2}}\beta ={{\left( \dfrac{5\sqrt{2}}{7} \right)}^{2}}-1 \\
& \Rightarrow {{\tan }^{2}}\beta =\dfrac{50}{49}-1 \\
& \Rightarrow {{\tan }^{2}}\beta =\dfrac{1}{49} \\
& \Rightarrow \tan \beta =\sqrt{\dfrac{1}{49}} \\
& \Rightarrow \tan \beta =\dfrac{1}{7} \\
\end{align}$
Now, as $\dfrac{5\sqrt{2}}{7}>1$ so, we can say that, $0<{{\sec }^{-1}}\left( \dfrac{5\sqrt{2}}{7} \right)<\dfrac{\pi }{2}$ and as $\tan \beta =\dfrac{1}{7}$ so, we can also write \[\beta ={{\tan }^{-1}}\left( \dfrac{1}{7} \right)\] . Then,
\[\beta ={{\tan }^{-1}}\left( \dfrac{1}{7} \right)..........................\left( 7 \right)\]
Now, we will calculate the value of $2{{\tan }^{-1}}\left( \dfrac{1}{5} \right)+{{\sec }^{-1}}\left( \dfrac{5\sqrt{2}}{7} \right)+2{{\tan }^{-1}}\left( \dfrac{1}{8} \right)$ . And as per our assumption $\alpha =2{{\tan }^{-1}}\left( \dfrac{1}{5} \right)$ , $\beta ={{\sec }^{-1}}\left( \dfrac{5\sqrt{2}}{7} \right)$ and $\gamma =2{{\tan }^{-1}}\left( \dfrac{1}{8} \right)$ . Then,
$\begin{align}
& 2{{\tan }^{-1}}\left( \dfrac{1}{5} \right)+{{\sec }^{-1}}\left( \dfrac{5\sqrt{2}}{7} \right)+2{{\tan }^{-1}}\left( \dfrac{1}{8} \right) \\
& \Rightarrow \alpha +\beta +\gamma \\
\end{align}$
Now, we put $\alpha ={{\tan }^{-1}}\left( \dfrac{5}{12} \right)$ from equation (5), $\gamma ={{\tan }^{-1}}\left( \dfrac{16}{63} \right)$ from equation (6) and \[\beta ={{\tan }^{-1}}\left( \dfrac{1}{7} \right)\] from equation (7) in the above expression. Then,
$\begin{align}
& \alpha +\beta +\gamma \\
& \Rightarrow {{\tan }^{-1}}\left( \dfrac{5}{12} \right)+{{\tan }^{-1}}\left( \dfrac{1}{7} \right)+{{\tan }^{-1}}\left( \dfrac{16}{63} \right) \\
\end{align}$
Now, we will use the formula from equation (3) as $\dfrac{5}{12}\times \dfrac{1}{7}=\dfrac{5}{84}<1$ to write ${{\tan }^{-1}}\left( \dfrac{5}{12} \right)+{{\tan }^{-1}}\left( \dfrac{1}{7} \right)={{\tan }^{-1}}\left( \dfrac{47}{79} \right)$ in the above line. Then,
$\begin{align}
& {{\tan }^{-1}}\left( \dfrac{5}{12} \right)+{{\tan }^{-1}}\left( \dfrac{1}{7} \right)+{{\tan }^{-1}}\left( \dfrac{16}{63} \right) \\
& \Rightarrow {{\tan }^{-1}}\left( \dfrac{\dfrac{5}{12}+\dfrac{1}{7}}{1-\dfrac{5}{12}\times \dfrac{1}{7}} \right)+{{\tan }^{-1}}\left( \dfrac{16}{63} \right) \\
& \Rightarrow {{\tan }^{-1}}\left( \dfrac{5\times 7+12}{7\times 12-5} \right)+{{\tan }^{-1}}\left( \dfrac{16}{63} \right) \\
& \Rightarrow {{\tan }^{-1}}\left( \dfrac{35+12}{84-5} \right)+{{\tan }^{-1}}\left( \dfrac{16}{63} \right) \\
& \Rightarrow {{\tan }^{-1}}\left( \dfrac{47}{79} \right)+{{\tan }^{-1}}\left( \dfrac{16}{63} \right) \\
\end{align}$
Now, we will use the formula from equation (3) as $\dfrac{47}{79}\times \dfrac{16}{63}=\dfrac{752}{4788}<1$ to write ${{\tan }^{-1}}\left( \dfrac{47}{79} \right)+{{\tan }^{-1}}\left( \dfrac{16}{63} \right)={{\tan }^{-1}}\left( 1 \right)$ in the above line. Then,
$\begin{align}
& {{\tan }^{-1}}\left( \dfrac{47}{79} \right)+{{\tan }^{-1}}\left( \dfrac{16}{63} \right) \\
& \Rightarrow {{\tan }^{-1}}\left( \dfrac{\dfrac{47}{79}+\dfrac{16}{63}}{1-\dfrac{47}{79}\times \dfrac{16}{63}} \right) \\
& \Rightarrow {{\tan }^{-1}}\left( \dfrac{47\times 63+16\times 79}{79\times 63-47\times 16} \right) \\
& \Rightarrow {{\tan }^{-1}}\left( \dfrac{2961+1264}{4977-752} \right) \\
& \Rightarrow {{\tan }^{-1}}\left( \dfrac{4225}{4225} \right) \\
& \Rightarrow {{\tan }^{-1}}\left( 1 \right) \\
\end{align}$
Now, from the formula from the equation (4), we can write ${{\tan }^{-1}}\left( 1 \right)=\dfrac{\pi }{4}$ . Then,
$\begin{align}
& {{\tan }^{-1}}\left( 1 \right) \\
& \Rightarrow \dfrac{\pi }{4} \\
\end{align}$
Now, from the above result, we conclude that, $2{{\tan }^{-1}}\left( \dfrac{1}{5} \right)+{{\sec }^{-1}}\left( \dfrac{5\sqrt{2}}{7} \right)+2{{\tan }^{-1}}\left( \dfrac{1}{8} \right)=\dfrac{\pi }{4}$ .
Thus, $2{{\tan }^{-1}}\left( \dfrac{1}{5} \right)+{{\sec }^{-1}}\left( \dfrac{5\sqrt{2}}{7} \right)+2{{\tan }^{-1}}\left( \dfrac{1}{8} \right)=\dfrac{\pi }{4}$ .
Hence, proved.
Note: Here, the student should understand what is asked in the question and then proceed in the right direction to prove the desired result quickly. Moreover, we should use formulas ${{\tan }^{-1}}x+{{\tan }^{-1}}y={{\tan }^{-1}}\left( \dfrac{x+y}{1-xy} \right)$ and ${{\tan }^{2}}\theta \text{=se}{{\text{c}}^{2}}\theta \text{-1}$ to avoid tough calculation. And avoid calculation mistakes while solving.
Complete step-by-step solution -
Given:
We have to prove the following equation:
$2{{\tan }^{-1}}\left( \dfrac{1}{5} \right)+{{\sec }^{-1}}\left( \dfrac{5\sqrt{2}}{7} \right)+2{{\tan }^{-1}}\left( \dfrac{1}{8} \right)=\dfrac{\pi }{4}$
Now, let $\alpha =2{{\tan }^{-1}}\left( \dfrac{1}{5} \right)$ , $\beta ={{\sec }^{-1}}\left( \dfrac{5\sqrt{2}}{7} \right)$ and $\gamma =2{{\tan }^{-1}}\left( \dfrac{1}{8} \right)$ . So, we will solve for $\alpha $ , $\beta $ and $\gamma $ separately. Then, we will find the value of $\alpha +\beta +\gamma $ .
Now, before we proceed further we should know the following formulas:
$ 2{{\tan }^{-1}}x={{\tan }^{-1}}\left( \dfrac{2x}{1-{{x}^{2}}} \right)\text{ }\left( \text{if }-1 < x < 1 \right)...............\left( 1 \right) $
$ {{\tan }^{2}}\theta \text{=se}{{\text{c}}^{2}}\theta \text{-1 }....................\left( 2 \right) $
$ {{\tan }^{-1}}x+{{\tan }^{-1}}y={{\tan }^{-1}}\left( \dfrac{x+y}{1-xy} \right)\text{ }\left( \text{if }xy<1 \right).............................\left( 3 \right) $
$ {{\tan }^{-1}}1=\dfrac{\pi }{4}..................................\left( 4 \right) $
Now, we will simplify $\alpha $ , $\beta $ and $\gamma $ separately with the help of the above four formulas.
Simplification of $\alpha =2{{\tan }^{-1}}\left( \dfrac{1}{5} \right)$ :
Now, we will use the formula from the equation (1) as $-1 < \dfrac{1}{5} < 1$ to write $ 2{{\tan }^{-1}}\left( \dfrac{1}{5} \right)={{\tan }^{-1}}\left( \dfrac{5}{12} \right)$ . Then,
$\begin{align}
& \alpha =2{{\tan }^{-1}}\left( \dfrac{1}{5} \right) \\
& \Rightarrow \alpha ={{\tan }^{-1}}\left( \dfrac{2\times \dfrac{1}{5}}{1-{{\left( \dfrac{1}{5} \right)}^{2}}} \right) \\
& \Rightarrow \alpha ={{\tan }^{-1}}\left( \dfrac{2\times 5}{{{5}^{2}}-1} \right) \\
& \Rightarrow \alpha ={{\tan }^{-1}}\left( \dfrac{10}{25-1} \right) \\
& \Rightarrow \alpha ={{\tan }^{-1}}\left( \dfrac{10}{24} \right) \\
& \Rightarrow \alpha ={{\tan }^{-1}}\left( \dfrac{5}{12} \right)........................\left( 5 \right) \\
\end{align}$
Simplification of $\gamma =2{{\tan }^{-1}}\left( \dfrac{1}{8} \right)$ :
Now, we will use the formula from the equation (1) as $-1<\dfrac{1}{8}<1$ to write $2{{\tan }^{-1}}\left( \dfrac{1}{8} \right)={{\tan }^{-1}}\left( \dfrac{16}{63} \right)$ . Then,
$\begin{align}
& \gamma =2{{\tan }^{-1}}\left( \dfrac{1}{8} \right) \\
& \Rightarrow \gamma ={{\tan }^{-1}}\left( \dfrac{2\times \dfrac{1}{8}}{1-{{\left( \dfrac{1}{8} \right)}^{2}}} \right) \\
& \Rightarrow \gamma ={{\tan }^{-1}}\left( \dfrac{2\times 8}{{{8}^{2}}-1} \right) \\
& \Rightarrow \gamma ={{\tan }^{-1}}\left( \dfrac{16}{64-1} \right) \\
& \Rightarrow \gamma ={{\tan }^{-1}}\left( \dfrac{16}{63} \right)........................\left( 6 \right) \\
\end{align}$
Simplification of $\beta ={{\sec }^{-1}}\left( \dfrac{5\sqrt{2}}{7} \right)$ :
Now, we write $\sec \beta =\dfrac{5\sqrt{2}}{7}$ and apply the formula from the equation (2) to get the value of $\tan \beta $ . Then,
$\begin{align}
& {{\tan }^{2}}\beta ={{\sec }^{2}}\beta -1 \\
& \Rightarrow {{\tan }^{2}}\beta ={{\left( \dfrac{5\sqrt{2}}{7} \right)}^{2}}-1 \\
& \Rightarrow {{\tan }^{2}}\beta =\dfrac{50}{49}-1 \\
& \Rightarrow {{\tan }^{2}}\beta =\dfrac{1}{49} \\
& \Rightarrow \tan \beta =\sqrt{\dfrac{1}{49}} \\
& \Rightarrow \tan \beta =\dfrac{1}{7} \\
\end{align}$
Now, as $\dfrac{5\sqrt{2}}{7}>1$ so, we can say that, $0<{{\sec }^{-1}}\left( \dfrac{5\sqrt{2}}{7} \right)<\dfrac{\pi }{2}$ and as $\tan \beta =\dfrac{1}{7}$ so, we can also write \[\beta ={{\tan }^{-1}}\left( \dfrac{1}{7} \right)\] . Then,
\[\beta ={{\tan }^{-1}}\left( \dfrac{1}{7} \right)..........................\left( 7 \right)\]
Now, we will calculate the value of $2{{\tan }^{-1}}\left( \dfrac{1}{5} \right)+{{\sec }^{-1}}\left( \dfrac{5\sqrt{2}}{7} \right)+2{{\tan }^{-1}}\left( \dfrac{1}{8} \right)$ . And as per our assumption $\alpha =2{{\tan }^{-1}}\left( \dfrac{1}{5} \right)$ , $\beta ={{\sec }^{-1}}\left( \dfrac{5\sqrt{2}}{7} \right)$ and $\gamma =2{{\tan }^{-1}}\left( \dfrac{1}{8} \right)$ . Then,
$\begin{align}
& 2{{\tan }^{-1}}\left( \dfrac{1}{5} \right)+{{\sec }^{-1}}\left( \dfrac{5\sqrt{2}}{7} \right)+2{{\tan }^{-1}}\left( \dfrac{1}{8} \right) \\
& \Rightarrow \alpha +\beta +\gamma \\
\end{align}$
Now, we put $\alpha ={{\tan }^{-1}}\left( \dfrac{5}{12} \right)$ from equation (5), $\gamma ={{\tan }^{-1}}\left( \dfrac{16}{63} \right)$ from equation (6) and \[\beta ={{\tan }^{-1}}\left( \dfrac{1}{7} \right)\] from equation (7) in the above expression. Then,
$\begin{align}
& \alpha +\beta +\gamma \\
& \Rightarrow {{\tan }^{-1}}\left( \dfrac{5}{12} \right)+{{\tan }^{-1}}\left( \dfrac{1}{7} \right)+{{\tan }^{-1}}\left( \dfrac{16}{63} \right) \\
\end{align}$
Now, we will use the formula from equation (3) as $\dfrac{5}{12}\times \dfrac{1}{7}=\dfrac{5}{84}<1$ to write ${{\tan }^{-1}}\left( \dfrac{5}{12} \right)+{{\tan }^{-1}}\left( \dfrac{1}{7} \right)={{\tan }^{-1}}\left( \dfrac{47}{79} \right)$ in the above line. Then,
$\begin{align}
& {{\tan }^{-1}}\left( \dfrac{5}{12} \right)+{{\tan }^{-1}}\left( \dfrac{1}{7} \right)+{{\tan }^{-1}}\left( \dfrac{16}{63} \right) \\
& \Rightarrow {{\tan }^{-1}}\left( \dfrac{\dfrac{5}{12}+\dfrac{1}{7}}{1-\dfrac{5}{12}\times \dfrac{1}{7}} \right)+{{\tan }^{-1}}\left( \dfrac{16}{63} \right) \\
& \Rightarrow {{\tan }^{-1}}\left( \dfrac{5\times 7+12}{7\times 12-5} \right)+{{\tan }^{-1}}\left( \dfrac{16}{63} \right) \\
& \Rightarrow {{\tan }^{-1}}\left( \dfrac{35+12}{84-5} \right)+{{\tan }^{-1}}\left( \dfrac{16}{63} \right) \\
& \Rightarrow {{\tan }^{-1}}\left( \dfrac{47}{79} \right)+{{\tan }^{-1}}\left( \dfrac{16}{63} \right) \\
\end{align}$
Now, we will use the formula from equation (3) as $\dfrac{47}{79}\times \dfrac{16}{63}=\dfrac{752}{4788}<1$ to write ${{\tan }^{-1}}\left( \dfrac{47}{79} \right)+{{\tan }^{-1}}\left( \dfrac{16}{63} \right)={{\tan }^{-1}}\left( 1 \right)$ in the above line. Then,
$\begin{align}
& {{\tan }^{-1}}\left( \dfrac{47}{79} \right)+{{\tan }^{-1}}\left( \dfrac{16}{63} \right) \\
& \Rightarrow {{\tan }^{-1}}\left( \dfrac{\dfrac{47}{79}+\dfrac{16}{63}}{1-\dfrac{47}{79}\times \dfrac{16}{63}} \right) \\
& \Rightarrow {{\tan }^{-1}}\left( \dfrac{47\times 63+16\times 79}{79\times 63-47\times 16} \right) \\
& \Rightarrow {{\tan }^{-1}}\left( \dfrac{2961+1264}{4977-752} \right) \\
& \Rightarrow {{\tan }^{-1}}\left( \dfrac{4225}{4225} \right) \\
& \Rightarrow {{\tan }^{-1}}\left( 1 \right) \\
\end{align}$
Now, from the formula from the equation (4), we can write ${{\tan }^{-1}}\left( 1 \right)=\dfrac{\pi }{4}$ . Then,
$\begin{align}
& {{\tan }^{-1}}\left( 1 \right) \\
& \Rightarrow \dfrac{\pi }{4} \\
\end{align}$
Now, from the above result, we conclude that, $2{{\tan }^{-1}}\left( \dfrac{1}{5} \right)+{{\sec }^{-1}}\left( \dfrac{5\sqrt{2}}{7} \right)+2{{\tan }^{-1}}\left( \dfrac{1}{8} \right)=\dfrac{\pi }{4}$ .
Thus, $2{{\tan }^{-1}}\left( \dfrac{1}{5} \right)+{{\sec }^{-1}}\left( \dfrac{5\sqrt{2}}{7} \right)+2{{\tan }^{-1}}\left( \dfrac{1}{8} \right)=\dfrac{\pi }{4}$ .
Hence, proved.
Note: Here, the student should understand what is asked in the question and then proceed in the right direction to prove the desired result quickly. Moreover, we should use formulas ${{\tan }^{-1}}x+{{\tan }^{-1}}y={{\tan }^{-1}}\left( \dfrac{x+y}{1-xy} \right)$ and ${{\tan }^{2}}\theta \text{=se}{{\text{c}}^{2}}\theta \text{-1}$ to avoid tough calculation. And avoid calculation mistakes while solving.
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
Difference between Prokaryotic cell and Eukaryotic class 11 biology CBSE
Difference Between Plant Cell and Animal Cell
Select the word that is correctly spelled a Twelveth class 10 english CBSE
Fill the blanks with the suitable prepositions 1 The class 9 english CBSE
What is the z value for a 90 95 and 99 percent confidence class 11 maths CBSE
Give 10 examples for herbs , shrubs , climbers , creepers
What organs are located on the left side of your body class 11 biology CBSE
What is BLO What is the full form of BLO class 8 social science CBSE
Change the following sentences into negative and interrogative class 10 english CBSE