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

Hint: 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,
\[ \Rightarrow \cot \theta = \dfrac{{AB}}{{BC}}\]