Question

Prove that \[\dfrac{{\tan A}}{{1 - \cot A}} + \dfrac{{\cot A}}{{1 - \tan A}} = 1 + \sec A\cos ecA\].

Answer 1

Hint:
Here we need to prove the given trigonometric expression. For that, we will consider the left hand side of the expression and apply the trigonometric identities to simplify the given expression. Then we will use some basic algebraic identities to simplify the expression further.

Complete step by step solution:
Let’s consider the left hand side of the given expression:
\[\dfrac{{\tan A}}{{1 - \cot A}} + \dfrac{{\cot A}}{{1 - \tan A}}\]
Now, we will break the terms using the reciprocal trigonometric identities.
Substituting \[\cot A = \dfrac{1}{{\tan A}}\] in the above expression, we get
\[ \Rightarrow \dfrac{{\tan A}}{{1 - \cot A}} + \dfrac{{\cot A}}{{1 - \tan A}} = \dfrac{{\tan A}}{{1 - \dfrac{1}{{\tan A}}}} + \dfrac{{\dfrac{1}{{\tan A}}}}{{1 - \tan A}}\]
On simplifying the terms, we get
\[ \Rightarrow \dfrac{{\tan A}}{{1 - \cot A}} + \dfrac{{\cot A}}{{1 - \tan A}} = \dfrac{{\tan A}}{{\dfrac{{\tan A - 1}}{{\tan A}}}} + \dfrac{1}{{\tan A\left( {1 - \tan A} \right)}}\]
\[ \Rightarrow \dfrac{{\tan A}}{{1 - \cot A}} + \dfrac{{\cot A}}{{1 - \tan A}} = \dfrac{{{{\tan }^2}A}}{{\tan A - 1}} + \dfrac{1}{{\tan A\left( {1 - \tan A} \right)}}\]
On further simplification, we get
 \[ \Rightarrow \dfrac{{\tan A}}{{1 - \cot A}} + \dfrac{{\cot A}}{{1 - \tan A}} = \dfrac{{1 - {{\tan }^3}A}}{{\tan A\left( {1 - \tan A} \right)}}\]
Using the algebraic identity \[{a^3} - {b^3} = \left( {a - b} \right)\left( {{a^2} + ab + {b^2}} \right)\] in the above equation, we get
\[ \Rightarrow \dfrac{{\tan A}}{{1 - \cot A}} + \dfrac{{\cot A}}{{1 - \tan A}} = \dfrac{{\left( {1 - \tan A} \right)\left( {{1^2} + 1 \cdot \tan A + {{\tan }^2}A} \right)}}{{\tan A\left( {1 - \tan A} \right)}}\]
Simplifying the terms, we get
\[ \Rightarrow \dfrac{{\tan A}}{{1 - \cot A}} + \dfrac{{\cot A}}{{1 - \tan A}} = \dfrac{{1 + \tan A + {{\tan }^2}A}}{{\tan A}}\] ……… \[\left( 1 \right)\]
Using the trigonometric identity \[1 + {\tan ^2}A = {\sec ^2}A\] in equation \[\left( 1 \right)\], we get
\[ \Rightarrow \dfrac{{\tan A}}{{1 - \cot A}} + \dfrac{{\cot A}}{{1 - \tan A}} = \dfrac{{\left( {{{\sec }^2}A + \tan A} \right)}}{{\tan A}}\]
On separating the terms, we get
\[ \Rightarrow \dfrac{{\tan A}}{{1 - \cot A}} + \dfrac{{\cot A}}{{1 - \tan A}} = \dfrac{{{{\sec }^2}A}}{{\tan A}} + 1\]
On using the trigonometric identities, we get
\[ \Rightarrow \dfrac{{\tan A}}{{1 - \cot A}} + \dfrac{{\cot A}}{{1 - \tan A}} = \dfrac{{{{\sec }^2}A}}{{\dfrac{{\sec A}}{{\cos ecA}}}} + 1\]
On further simplifying the terms, we get
\[ \Rightarrow \dfrac{{\tan A}}{{1 - \cot A}} + \dfrac{{\cot A}}{{1 - \tan A}} = \sec A \cdot \cos ecA + 1\]
We can also write this trigonometric expression as
\[ \Rightarrow \dfrac{{\tan A}}{{1 - \cot A}} + \dfrac{{\cot A}}{{1 - \tan A}} = 1 + \sec A\cos ecA\]
We have got left hand side expression equal to right hand side expression.
Hence, we have proved the given trigonometric expression.

Note:
We need to know the meaning of the trigonometric identities as we have used the trigonometric identities in this question. Trigonometric identities are defined as the equalities which involve the trigonometric identities and they are true for every value of the occurring variables for which both sides of the equality are defined. All the trigonometric identities are periodic in nature. They repeat their values after a certain interval.