Question

How can you prove that $\cot \left( {a - 45^\circ } \right) = \dfrac{{1 + \tan \left( a \right)}}{{\tan \left( a \right) - 1}}$ ?

Answer 1

In the above question, we are given a trigonometric equation containing the tangent and cotangent function. We have to prove that $\cot \left( {a - 45^\circ } \right) = \dfrac{{1 + \tan \left( a \right)}}{{\tan \left( a \right) - 1}}$ . In order to approach the solution, we can use the trigonometric identity of the cotangent function which is given as,
$\Rightarrow \cot \left( {A - B} \right) = \dfrac{{\cot A\cot B + 1}}{{\cot B - \cot A}}$
Use the above identity in the LHS of the equation $\cot \left( {a - 45^\circ } \right) = \dfrac{{1 + \tan \left( a \right)}}{{\tan \left( a \right) - 1}}$ and then put the value for $\cot 45^\circ = 1$ in the RHS. Further after substituting $\cot a = \dfrac{1}{{\tan a}}$ and $1 = \dfrac{{\tan a}}{{\tan a}}$ and solving we can obtain the required RHS equal to the LHS.

Complete step by step answer:
Given trigonometric equation is $\cot \left( {a - 45^\circ } \right) = \dfrac{{1 + \tan \left( a \right)}}{{\tan \left( a \right) - 1}}$ .
We have to prove that the LHS is equal to the RHS.
Since, we that the trigonometric identity for a cotangent function which is given as,
$\Rightarrow \cot \left( {A - B} \right) = \dfrac{{\cot A\cot B + 1}}{{\cot B - \cot A}}$
Now putting $A = a$ and $B = 45^\circ$ in the above identity, we get
$\Rightarrow \cot \left( {a - 45^\circ } \right) = \dfrac{{\cot a\cot 45^\circ + 1}}{{\cot 45^\circ - \cot a}}$
Since we know that $\cot 45^\circ = 1$ , therefore now we have
$\Rightarrow \cot \left( {a - 45^\circ } \right) = \dfrac{{\cot a + 1}}{{1 - \cot a}}$
Putting $\cot a = \dfrac{1}{{\tan a}}$ and $1 = \dfrac{{\tan a}}{{\tan a}}$ in the RHS of the above equation, we can write
$\Rightarrow \cot \left( {a - 45^\circ } \right) = \dfrac{{\dfrac{1}{{\tan a}} + \dfrac{{\tan a}}{{\tan a}}}}{{\dfrac{{\tan a}}{{\tan a}} - \dfrac{1}{{\tan a}}}}$
Solving the numerator and the denominator in the RHS, we get
$\Rightarrow \cot \left( {a - 45^\circ } \right) = \dfrac{{\dfrac{{1 + \tan a}}{{\tan a}}}}{{\dfrac{{\tan a - 1}}{{\tan a}}}}$
Cancelling the $\tan a$ term from the numerator and the denominator, we can write the above equation as,
$\Rightarrow \cot \left( {a - 45^\circ } \right) = \dfrac{{1 + \tan a}}{{\tan a - 1}}$
Hence,
$LHS = RHS$
That is the required proof of the above given problem.
Therefore, we have solved $\cot \left( {a - 45^\circ } \right) = \dfrac{{1 + \tan a}}{{\tan a - 1}}$.

Note:
The cotangent function or the cot function is exactly the opposite of the tangent function or the tan function. Therefore it is written as
$\Rightarrow \cot x = \dfrac{1}{{\tan x}}$
If $\theta$ is an acute angle at the vertex A in a right angled triangle $\vartriangle ABC$ , right angled at B then the cotangent function is defined as the ratio of the base AB and the perpendicular height BC, i.e.
$\Rightarrow \cot \theta = \dfrac{{base}}{{height}}$
Or,
$\Rightarrow \cot \theta = \dfrac{{AB}}{{BC}}$