Question

Prove that ${\text{ta}}{{\text{n}}^{ - 1}}(1) + {\tan ^{ - 1}}(2) + {\tan ^{ - 1}}(3) = \pi$

Answer 1

We have given an inverse trigonometric expression. Which we have to prove equal to $\pi$. Firstly we take the left hand side and the left hand side is an expression of ${\text{ta}}{{\text{n}}^{ - 1}}$ function. First we take last two terms these two terms are in addition we change theme in cot function there again we convert theme in ${\text{ta}}{{\text{n}}^{ - 1}}$ function them we apply identity on them. We solve and put value so the inverse functions and we get the result.

Complete step-by-step answer:
The given expression is we have to prove this firstly we take the left hand side of the expression.
Left hand side${\text{ = ta}}{{\text{n}}^{ - 1}}(1) + {\tan ^{ - 1}}(2) + {\tan ^{ - 1}}(3)$
Since the value of ${\text{ta}}{{\text{n}}^{ - 1}}(1)$is equal to $\pi$ so the left hand side becomes.
$\Rightarrow \dfrac{\pi }{2} + {\text{ta}}{{\text{n}}^{ - 1}}(2) + {\tan ^{ - 1}}(3)$
Converting ${\text{ta}}{{\text{n}}^{ - 1}}$ into ${\text{co}}{{\text{t}}^{ - 1}}$
We have the relation ${\text{ta}}{{\text{n}}^{ - 1}}x + {\cot ^{ - 1}}x = \dfrac{\pi }{2}$
So ${\text{ta}}{{\text{n}}^{ - 1}}x = \dfrac{\pi }{2} - {\cot ^{ - 1}}(x)$
Therefore left hand side become
$\dfrac{\pi }{2} + \left[ {\dfrac{\pi }{2} - {{\cot }^{ - 1}}(2)} \right] - \left[ {\dfrac{\pi }{2} - {{\cot }^{ - 1}}(3)} \right]$
$\Rightarrow \dfrac{\pi }{2} + \dfrac{\pi }{2} + \dfrac{\pi }{2} - \left[ {{{\cot }^{ - 1}}(2) + {{\cot }^{ - 1}}(3)} \right]$
$\Rightarrow \dfrac{{5\pi }}{2} - \left[ {{{\cot }^{ - 1}}(2) + {{\cot }^{ - 1}}(3)} \right]$
Now we know that ${\cot ^{ - 1}}(x) = {\tan ^{ - 1}}\left( {\dfrac{1}{x}} \right)$
So ${\cot ^{ - 1}}(2)$become ${\tan ^{ - 1}}\left( {\dfrac{1}{2}} \right)$
and ${\cot ^{ - 1}}(3)$become ${\tan ^{ - 1}}\left( {\dfrac{1}{3}} \right)$
Therefore left hand side become
$\Rightarrow \dfrac{{5m}}{4} - \left[ {{{\tan }^{ - 1}}\left( {\dfrac{1}{2}} \right) + {{\tan }^{ - 1}}\left( {\dfrac{1}{3}} \right)} \right]$
The term in the bracket is in the form
${\tan ^{ - 1}}(x) + {\tan ^{ - 1}}(y){\text{ when }}x = \dfrac{1}{2}$and
$y = \dfrac{1}{3}{\text{ also }}xy = \dfrac{1}{2} \times \dfrac{1}{3} = \dfrac{1}{6} < 1$
So we can apply the formula
${\tan ^{ - 1}}(x) + {\tan ^{ - 1}}(y){\text{ = }}{\tan ^{ - 1}}\left( {\dfrac{{x + y}}{{1 - xy}}} \right)$
So ${\tan ^{ - 1}}\left( {\dfrac{1}{2}} \right) + {\tan ^{ - 1}}\left( {\dfrac{1}{3}} \right) = {\tan ^{ - 1}}\left( {\dfrac{{\dfrac{1}{2} \times \dfrac{1}{3}}}{{1 - \dfrac{1}{2} \times \dfrac{1}{3}}}} \right)$
$\Rightarrow {\tan ^{ - 1}}\left( {\dfrac{{\dfrac{{2 + 3}}{6}}}{{\dfrac{{6 - 1}}{6}}}} \right) = {\tan ^{ - 1}}\left( {\dfrac{5}{5}} \right) = {\tan ^{ - 1}}(1)$
We also know that ${\tan ^{ - 1}}1 = \left( {\dfrac{\pi }{4}} \right)$
So ${\tan ^{ - 1}}\left( {\dfrac{1}{2}} \right) + {\tan ^{ - 1}}\left( {\dfrac{1}{3}} \right) = \dfrac{\pi }{4}$
So left hand side become
$\Rightarrow \dfrac{{5\pi }}{4} - \dfrac{\pi }{4}$
$\Rightarrow \dfrac{{5\pi - \pi }}{4} = \dfrac{4}{4}\pi$
$\Rightarrow \pi$
Which is equal to Right hand side hence proved.

Note: Trigonometric is the branch of mathematics that studies the relationship between side lengths and angles of the triangle. Trigonometry has six Trigonometric functions. Which are sin, cos, tan, cosec, second cot. Trigonometric functions are the real functions which relate an angle of right angle triangles to the ratio of two sides of a triangle.
Trigonometric functions are also called circular functions. With the help of their trigonometric functions we can drive lots of Trigonometric formulas.