Question

Find the value of ‘\[{\tan ^{ - 4}}A + {\cot ^{ - 4}}A\]’ given that ‘\[\tan A + \cot A = 4\]’.
A. \[110\]
B. \[191\]
C. \[80\]
D. \[194\]

Answer 1

Hint: First squaring both sides of the given equation \[\tan A + \cot A = 4\] and after that simplifying this and again squaring this and we get the required value of \[{\tan ^{ - 4}}A + {\cot ^{ - 4}}A\].

Formula Used:
Square formula of algebra \[{\left( {A + B} \right)^2} = {A^2} + {B^2} + 2AB\]
Multiplication of formula of trigonometry \[\tan x \times \cot x = 1\]
Properties of \[\tan x\] and \[\cot x\] , \[\tan x = \dfrac{1}{{\cot x}}\] and \[\cot x = \dfrac{1}{{\tan x}}\]

Complete step by step solution:
Given \[\tan A + \cot A = 4\]
On squaring both the sides, we get:
\[{\left( {\tan A + \cot A} \right)^2} = {\tan ^2}A + {\cot ^2}A + 2 \times \tan A \times \cot A\]
We know that \[\tan A \times \cot A = 1\], substituting this and we get
\[ \Rightarrow {\left( {\tan A + \cot A} \right)^2} = {\tan ^2}A + {\cot ^2}A + 2 \times 1\]
Since, \[\tan A + \cot A = 4\], we have:
\[ \Rightarrow {\left( 4 \right)^2} = {\tan ^2}A + {\cot ^2}A + 2\]
\[ \Rightarrow 16 = {\tan ^2}A + {\cot ^2}A + 2\]
\[ \Rightarrow 14 = {\tan ^2}A + {\cot ^2}A\];
Again, squaring both the sides of the above equation, we get:
\[ \Rightarrow {\left( {14} \right)^2} = {\tan ^4}A + {\cot ^4}A + 2 \times {\tan ^2}A \times {\cot ^2}A\]………..(1)
We know that \[\tan A \times \cot A = 1\],
Squaring both sides of \[\tan A \times \cot A = 1\] and we get
\[ \Rightarrow {\left( {\tan A \times \cot A} \right)^2} = {1^2}\]
\[ \Rightarrow {\tan ^2}A \times {\cot ^2}A = 1\]
Substituting this in equation (1) and we get
\[ \Rightarrow {\left( {14} \right)^2} = {\tan ^4}A + {\cot ^4}A + 2 \times 1\]
\[ \Rightarrow 194 = {\tan ^4}A + {\cot ^4}A\]
Using the properties of \[\tan A\] and \[\cot A\] , we get
\[ \Rightarrow \left( {\dfrac{1}{{{{\cot }^{ - 4}}A}}} \right) + \left( {\dfrac{1}{{{{\tan }^{ - 4}}A}}} \right) = 194\]
\[ \Rightarrow \left( {\dfrac{{{{\tan }^{ - 4}}A + {{\cot }^{ - 4}}A}}{{{{\tan }^{ - 4}}A \times {{\cot }^{ - 4}}A}}} \right) = 194\]
Since \[\tan A \times \cot A = 1\] , then \[{\tan ^{ - 4}}A \times {\cot ^{ - 4}}A = 1\]
\[ \Rightarrow \left( {\dfrac{{{{\tan }^{ - 4}}A + {{\cot }^{ - 4}}A}}{1}} \right) = 194\]
\[ \Rightarrow {\tan ^{ - 4}}A + {\cot ^{ - 4}}A = 194\]
\[ = 194\]
The correct answer is option D.
Note: Students make mistakes as they try to solve \[{\tan ^{ - 4}}A + {\cot ^{ - 4}}A\] directly. But if we try to solve this equation directly then it will be complex to solve. So this is not a good method to solve this question.