Question

Prove that, ${{\tan }^{-1}}\left( \dfrac{1}{4} \right)+{{\tan }^{-1}}\left( \dfrac{2}{9} \right)=\dfrac{1}{2}{{\cos }^{-1}}\left( \dfrac{3}{5} \right)$

Answer 1

In this question we have been given with a trigonometric expression for which we have to prove that the left-hand side is equal to the right-hand side. We will solve this question by considering the left-hand side and using the formula ${{\tan }^{-1}}x+{{\tan }^{-1}}y={{\tan }^{-1}}\left( \dfrac{x+y}{1-xy} \right)$ and ${{\tan }^{-1}}x=\dfrac{1}{2}{{\cos }^{-1}}\left( \dfrac{1-{{x}^{2}}}{1+{{x}^{2}}} \right)$ to simplify the terms and get the required solution.

Complete step by step answer:
We have the expression given to us as:
$\Rightarrow {{\tan }^{-1}}\left( \dfrac{1}{4} \right)+{{\tan }^{-1}}\left( \dfrac{2}{9} \right)=\dfrac{1}{2}{{\cos }^{-1}}\left( \dfrac{3}{5} \right)$
Consider the left-hand side of the expression, we get:
$\Rightarrow {{\tan }^{-1}}\left( \dfrac{1}{4} \right)+{{\tan }^{-1}}\left( \dfrac{2}{9} \right)$
Now, we know the formula ${{\tan }^{-1}}x+{{\tan }^{-1}}y={{\tan }^{-1}}\left( \dfrac{x+y}{1-xy} \right)$ therefore, on substituting the values, we get:
$\Rightarrow {{\tan }^{-1}}\left( \dfrac{\dfrac{1}{4}+\dfrac{2}{9}}{1-\dfrac{1}{4}\times \dfrac{2}{9}} \right)$
On multiplying the terms, we get:
$\Rightarrow {{\tan }^{-1}}\left( \dfrac{\dfrac{1}{4}+\dfrac{2}{9}}{1-\dfrac{2}{36}} \right)$
On taking the lowest common multiple, we get:
$\Rightarrow {{\tan }^{-1}}\left( \dfrac{\dfrac{9+8}{36}}{\dfrac{36-2}{36}} \right)$
On adding the terms, we get:
$\Rightarrow {{\tan }^{-1}}\left( \dfrac{\dfrac{17}{36}}{\dfrac{34}{36}} \right)$
Since both the denominators are the same, we cancel an write it as:
$\Rightarrow {{\tan }^{-1}}\left( \dfrac{17}{34} \right)$
On simplifying the fraction, we get:
$\Rightarrow {{\tan }^{-1}}\left( \dfrac{1}{2} \right)$
Now we know the formula ${{\tan }^{-1}}x=\dfrac{1}{2}{{\cos }^{-1}}\left( \dfrac{1-{{x}^{2}}}{1+{{x}^{2}}} \right)$ therefore, on substituting the values, we get:
$\Rightarrow {{\tan }^{-1}}\left( \dfrac{1}{2} \right)=\dfrac{1}{2}{{\cos }^{-1}}\left( \dfrac{1-{{\left( \dfrac{1}{2} \right)}^{2}}}{1+{{\left( \dfrac{1}{2} \right)}^{2}}} \right)$
On squaring, we get:
$\Rightarrow \dfrac{1}{2}{{\cos }^{-1}}\left( \dfrac{1-\dfrac{1}{4}}{1+\dfrac{1}{4}} \right)$
On taking the lowest common multiple, we get:
$\Rightarrow \dfrac{1}{2}{{\cos }^{-1}}\left( \dfrac{\dfrac{4-1}{4}}{\dfrac{4+1}{4}} \right)$
On simplifying the terms, we get:
$\Rightarrow \dfrac{1}{2}{{\cos }^{-1}}\left( \dfrac{\dfrac{3}{4}}{\dfrac{5}{4}} \right)$
On cancelling the denominators, we get:
$\Rightarrow \dfrac{1}{2}{{\cos }^{-1}}\left( \dfrac{3}{5} \right)$, which is the right-hand side, hence proved.

Note: The various trigonometric identities and formulae should be remembered while doing these types of sums. The various Pythagorean identities should also be remembered while doing these types of questions. It is to be remembered that to add two or more fractions, the denominator of both them should be the same, if the denominator is not same, the lowest common multiple known as L.C.M should be taken. If there is nothing to simplify, then only you should use the double angle formulas to expand the given equation.