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 $$\mathrm{\angle }C=\theta$$.
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 $$\sin \theta ={\displaystyle \frac{AB}{AC}}$$………(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 $$\cos \theta ={\displaystyle \frac{BC}{AC}}$$………..(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 $$\tan \theta ={\displaystyle \frac{\sin \theta }{\cos \theta }}$$……….(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 $$\cot \theta ={\displaystyle \frac{\cos \theta }{\sin \theta }}$$………….(4)
Now, we will divide equation (1) by equation (2).
$$\Rightarrow {\displaystyle \frac{\sin \theta }{\cos \theta }}={\displaystyle \frac{\left({\displaystyle \frac{AB}{AC}}\right)}{\left({\displaystyle \frac{BC}{AC}}\right)}}$$
And we know that the same terms from numerator and denominator cancel out. Therefore, we get
$$\Rightarrow {\displaystyle \frac{\sin \theta }{\cos \theta }}={\displaystyle \frac{AB}{BC}}$$
From equation (3), we can write
$$\tan \theta ={\displaystyle \frac{AB}{BC}}$$……….(5)
Now, we will take the reciprocal of equation (5).
$$\Rightarrow {\displaystyle \frac{1}{{\displaystyle \frac{\sin \theta }{\cos \theta }}}}={\displaystyle \frac{1}{{\displaystyle \frac{AB}{BC}}}}$$
$$\Rightarrow {\displaystyle \frac{\cos \theta }{\sin \theta }}={\displaystyle \frac{BC}{AB}}$$
But from equation (4), we can say
$$\cot \theta ={\displaystyle \frac{BC}{AB}}$$……………(6)
Now, we will multiply equation (5) by (6).
$$\Rightarrow \tan \theta \cdot \cot \theta ={\displaystyle \frac{AB}{BC}}\cdot {\displaystyle \frac{BC}{AB}}$$
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.