Question

Find the value of ${\displaystyle \frac{\tan {65}^0}{\cot {25}^0}}$

Answer 1

Hint: To solve this question, we will use some basic trigonometric ratios of complementary angles to evaluate the given expression. If $\theta$ is an acute angle, then $\tan \left({90}^0-\theta \right)=\cot \theta$

Complete step-by-step answer:
We have,
${\displaystyle \frac{\tan {65}^0}{\cot {25}^0}}$ ……….. (i)
We can write $\tan {65}^0$ as:
$$\Rightarrow \tan \left({90}^0-{65}^0\right)$$
And we know that,
If $\theta$ is an acute angle, then $\tan \left({90}^0-\theta \right)=\cot \theta$
So,
$$\Rightarrow \tan \left({90}^0-{25}^0\right)=\cot {25}^0$$
Putting this in equation (i), we will get
$\Rightarrow {\displaystyle \frac{\cot {25}^0}{\cot {25}^0}}=1$
Hence, we can say that ${\displaystyle \frac{\cot {25}^0}{\cot {25}^0}}=1$

Note: Whenever we are asked such types of questions, first we have to find out which trigonometric identity is mentioned in the question. Then we will use basic trigonometric ratios and by putting them into that identity, we will get the required answer.