Question

Evaluate:$\frac{\tan {26}^0}{\cot {64}^0}$

Answer 1

To evaluate : $\frac{\tan {26}^0}{\cot {64}^0}$

We can write ${26}^0$ as ${26}^0=({90}^0-{64}^0)$

So $\tan {26}^0=\tan ({90}^0-{64}^0)$

⇒ $\tan {26}^0=\cot {64}^0$ [Formula used: $\tan ({90}^0-\mathrm{\theta })=\cot \theta$]

Now, replacing the value of $\tan {26}^0$from the above, we get

$\frac{\tan {26}^0}{\cot {64}^0}$$=\frac{\cot {72}^0}{\cot {72}^0}$

Thus,$\frac{\tan {26}^0}{\cot {64}^0}=1$