Question

Prove that \[{\tan ^{ - 1}}\dfrac{3}{4} + {\tan ^{ - 1}}\dfrac{2}{3} = {\tan ^{ - 1}}\dfrac{{17}}{6}\]

Answer 1

Hint: Here in this question, given a problem based on the topic of inverse trigonometry. Here we have to prove the left hand side is equal to the right hand side \[\left( {i.e.,LHS = RHS} \right)\] . To solve this apply a formula \[{\tan ^{ - 1}}a + {\tan ^{ - 1}}b = {\tan ^{ - 1}}\left( {\dfrac{{a + b}}{{1 - ab}}} \right)\] on LHS and by further simplification using a basic arithmetic operation to get the required solution.

Complete step by step solution:
 In mathematics, the inverse trigonometric functions are the inverse functions of the trigonometric functions. Specifically, they are the inverses of the sine, cosine, tangent, cotangent, secant, and cosecant functions, and are used to obtain an angle from any of the angle's trigonometric ratios.
Consider the given inverse trigonometric equation:
 \[ \Rightarrow {\tan ^{ - 1}}\dfrac{3}{4} + {\tan ^{ - 1}}\dfrac{2}{3} = {\tan ^{ - 1}}\dfrac{{17}}{6}\]
It has an inverse tan function it is the inverse function of the trigonometric function ‘tangent’.
Here we have to prove the left hand side is equal to right hand side i.e., \[LHS = RHS\] .
Consider left hand side i.e., LHS of the equation
 \[ \Rightarrow {\tan ^{ - 1}}\dfrac{3}{4} + {\tan ^{ - 1}}\dfrac{2}{3}\]
Here to solve this use a addition formula of inverse tangent i.e., \[{\tan ^{ - 1}}a + {\tan ^{ - 1}}b = {\tan ^{ - 1}}\left( {\dfrac{{a + b}}{{1 - ab}}} \right)\]
Here \[a = \dfrac{3}{4}\] and \[b = \dfrac{2}{3}\] on substituting in formula, we have
 \[ \Rightarrow {\tan ^{ - 1}}\left( {\dfrac{{\dfrac{3}{4} + \dfrac{2}{3}}}{{1 - \left( {\dfrac{3}{4}} \right)\left( {\dfrac{2}{3}} \right)}}} \right)\]
 \[ \Rightarrow {\tan ^{ - 1}}\left( {\dfrac{{\dfrac{3}{4} + \dfrac{2}{3}}}{{1 - \dfrac{6}{{12}}}}} \right)\]
 \[ \Rightarrow {\tan ^{ - 1}}\left( {\dfrac{{\dfrac{3}{4} + \dfrac{2}{3}}}{{1 - \dfrac{6}{{12}}}}} \right)\]
Take 12 a LCM in both numerator and denominator, we get
 \[ \Rightarrow {\tan ^{ - 1}}\left( {\dfrac{{\dfrac{{9 + 8}}{{12}}}}{{\dfrac{{12 - 6}}{{12}}}}} \right)\]
On cancelling like terms in both numerator and denominator, then
 \[ \Rightarrow {\tan ^{ - 1}}\left( {\dfrac{{9 + 8}}{{12 - 6}}} \right)\]
On simplification, we get
 \[ \Rightarrow {\tan ^{ - 1}}\left( {\dfrac{{17}}{6}} \right)\]
Therefore, LHS=RHS
Hence proved


Note: Since the question involves the inverse trigonometry concept we have to know the formula \[{\tan ^{ - 1}}a + {\tan ^{ - 1}}b = {\tan ^{ - 1}}\left( {\dfrac{{a + b}}{{1 - ab}}} \right)\] , The formulas we use it will based on the trigonometry ratios. If the question involves the sine trigonometry ratio then we use the other formula and we show it LHS is equal to the RHS. While simplifying we must know about the simple arithmetic operations.