
How do you prove that \[\tan^{2}A\ + \cot^{2}A\ = 1\] is not an identity by showing a counter example?
Answer
489.3k+ views
Hint: In this question, we have to prove that \[\tan^{2}A\ + \cot^{2}A\ = 1\] is not an identity by showing a counter example. First we need to consider the left hand side of the given expression. Then we can consider \[A=45^{o}\] . By substituting and simplifying the expression, we get the value of the left hand side. Then we need to compare the value with the right hand side of the given expression. From that we can conclude that the given expression is not an identity .
Complete step by step solution:
Given, \[\tan^{2}A\ + \cot^{2}A\ = 1\]
Here we need to prove that the given identity is not an identity by showing a counter example.
Now we can consider the left hand side of the given expression,
\[\Rightarrow \tan^{2}A\ + \cot^{2}A\]
Let us consider \[A = 45^{o}\]
On substituting \[A = 45^{o}\] ,
We get,
\[\Rightarrow \tan^{2}\left( 45^{o} \right) + \cot^{2}\left( 45^{o} \right)\]
We know that the value of \[\tan(45^{o}) = 1\] and \[\cot(45^{o}) = 1\] ,
On substituting the values,
We get,
\[\Rightarrow \left( 1 \right)^{2} + \left( 1 \right)^{2}\]
On simplifying,
We get,
\[\Rightarrow \ 2\]
Therefore, the value of the left hand side of the given expression when \[A = 45^{o}\] is \[2\] .
But the right hand side of the given expression is \[1\] .
Thus we can conclude that
\[\tan^{2}\left( 45^{o} \right) + \cot^{2}\left( 45^{o} \right) \neq \ 1\]
So this converse to the given expression . And hence The given expression is not always true .
Hence we can conclude that it is not a trigonometric identity.
Thus we get \[\tan^{2}A\ + \cot^{2}A\ = 1\] is not an identity
\[\tan^{2}A\ + \cot^{2}A\ = 1\] is not an identity.
Note:
First, we need to know what the counterexample means. A counterexample is nothing but a method used to counter the given statement by taking a random value for the given quantity and disproving the left and right side of the equation. A trigonometric identity is an equation whose left hand side is always to the right hand side for any value of the given angle. While solving such trigonometric identities problems, we need to have a good knowledge about the trigonometric identities.
Complete step by step solution:
Given, \[\tan^{2}A\ + \cot^{2}A\ = 1\]
Here we need to prove that the given identity is not an identity by showing a counter example.
Now we can consider the left hand side of the given expression,
\[\Rightarrow \tan^{2}A\ + \cot^{2}A\]
Let us consider \[A = 45^{o}\]
On substituting \[A = 45^{o}\] ,
We get,
\[\Rightarrow \tan^{2}\left( 45^{o} \right) + \cot^{2}\left( 45^{o} \right)\]
We know that the value of \[\tan(45^{o}) = 1\] and \[\cot(45^{o}) = 1\] ,
On substituting the values,
We get,
\[\Rightarrow \left( 1 \right)^{2} + \left( 1 \right)^{2}\]
On simplifying,
We get,
\[\Rightarrow \ 2\]
Therefore, the value of the left hand side of the given expression when \[A = 45^{o}\] is \[2\] .
But the right hand side of the given expression is \[1\] .
Thus we can conclude that
\[\tan^{2}\left( 45^{o} \right) + \cot^{2}\left( 45^{o} \right) \neq \ 1\]
So this converse to the given expression . And hence The given expression is not always true .
Hence we can conclude that it is not a trigonometric identity.
Thus we get \[\tan^{2}A\ + \cot^{2}A\ = 1\] is not an identity
\[\tan^{2}A\ + \cot^{2}A\ = 1\] is not an identity.
Note:
First, we need to know what the counterexample means. A counterexample is nothing but a method used to counter the given statement by taking a random value for the given quantity and disproving the left and right side of the equation. A trigonometric identity is an equation whose left hand side is always to the right hand side for any value of the given angle. While solving such trigonometric identities problems, we need to have a good knowledge about the trigonometric identities.
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