Question

How do you prove \[\tan \theta \cot \theta =1\]?

Answer 1

Hint: To prove this \[\tan \theta \cot \theta =1\], we will use a right-angled triangle with one angle \[\theta \]. In the right angle triangle, we will find the values of \[\tan \theta \] and \[\cot \theta \] with the help of \[\sin \theta \] and \[\cos \theta \]. And then we will simplify it to get the required statement.

Complete step-by-step solution:
In this question, we have been asked to prove that \[\tan \theta \cot \theta =1\].
For that, let us take a right angle triangle ABC.

In the above right angle triangle ABC, we have considered the right angle at B and \[\angle C=\theta \].
So we can say, AB is the perpendicular, BC is the base and AC is the hypotenuse of the right-angled triangle ABC.
Now, we will try to find or calculate \[\sin \theta \].
In the right angle triangle ABC,
As we know that \[\sin \theta \] is the ratio of the perpendicular to the hypotenuse.
We can mathematically represent it as \[\sin \theta =\dfrac{AB}{AC}\]………(1)
Now, we will try to find or calculate \[\cos \theta \].
In the right angle triangle ABC,
As we know that \[\cos \theta \] is the ratio of the base to the hypotenuse.
We can mathematically represent it as \[\cos \theta =\dfrac{BC}{AC}\]………..(2)
Now, we will try to find or calculate \[\tan \theta \].
In the right angle triangle ABC,
As we know that \[\tan \theta \] is the ratio of \[\sin \theta \] to \[\cos \theta \].
We can mathematically represent it as \[\tan \theta =\dfrac{\sin \theta }{\cos \theta }\]……….(3)
Now, we will try to find or calculate \[\cot \theta \].
In the right angle triangle ABC,
As we know that \[\cot \theta \] is the ratio of \[\cos \theta \] to \[\sin \theta \].
We can mathematically represent it as \[\cot \theta =\dfrac{\cos \theta }{\sin \theta }\]………….(4)
Now, we will divide equation (1) by equation (2).
\[\Rightarrow \dfrac{\sin \theta }{\cos \theta }=\dfrac{\left( \dfrac{AB}{AC} \right)}{\left( \dfrac{BC}{AC} \right)}\]
And we know that the same terms from numerator and denominator cancel out. Therefore, we get
\[\Rightarrow \dfrac{\sin \theta }{\cos \theta }=\dfrac{AB}{BC}\]
From equation (3), we can write
\[\tan \theta =\dfrac{AB}{BC}\]……….(5)
Now, we will take the reciprocal of equation (5).
\[\Rightarrow \dfrac{1}{\dfrac{\sin \theta }{\cos \theta }}=\dfrac{1}{\dfrac{AB}{BC}}\]
\[\Rightarrow \dfrac{\cos \theta }{\sin \theta }=\dfrac{BC}{AB}\]
But from equation (4), we can say
\[\cot \theta =\dfrac{BC}{AB}\]……………(6)
Now, we will multiply equation (5) by (6).
\[\Rightarrow \tan \theta \cdot \cot \theta =\dfrac{AB}{BC}\cdot \dfrac{BC}{AB}\]
And we can further simplify it as
\[\Rightarrow \tan \theta \cdot \cot \theta =1\]
Hence, we have proved that \[\tan \theta \cdot \cot \theta =1\]

Note: Whenever we get this type of problem, we will try to use the right angle triangle and then use trigonometry. We can also do this problem in one step because \[\tan \theta \] is the reciprocal of \[\cot \theta \]. Also, we need to keep this property in our mind because we might require this property for other questions.