Question

evaluate tan 15. tan 25. tan 60. tan 65. tan 75 - tan 30​

Answer 1

Answer:

Step-by-step explanation:

tan15 * tan75 * tan65*tan25*tan60 - tan30

= tan15* cot(90-75)*tan25*cot(90-65)*tan60 - tan30

=tan15*cot15*tan25*cot25*tan60-tan30

=1*1*tan60-tan30

= root3 - 1/root3

(3-1)/root3

Answer 2

Answer:

The required answer is $1$.

Step-by-step explanation:

Given: $\tan 15^{\circ} \tan 25^{\circ} \tan 60^{\circ} \tan 65^{\circ} \tan 75^{\circ}$

To find: Evaluate values.

Tangent theta: The side opposite theta divided by the side next to theta is known as the tangent theta. This is theta, then. This length divided by this length, or $y$ over $x$ is the tangent theta. However, because it can only be defined in this way on right triangles, this definition only applies to acute angles, which are angles between $0$ and $90$ degrees.

We have

$\tan 15^{\circ} \tan 25^{\circ} \tan 60^{\circ} \tan 65^{\circ} \tan 75^{\circ}$

Now calculate value.

$=\cot \left(90^{\circ}-15^{\circ}\right) \tan \left(90^{\circ}-25^{\circ}\right) \tan 60^{\circ} \tan 65^{\circ} \tan 75^{\circ}\left[\because \tan \theta=\cot \left(90^{\circ}-\theta\right)\right]$

Here,

$=\cot 75^{\circ} \cot 65^{\circ} \tan 60^{\circ} \tan 65^{\circ} \tan 75^{\circ}$

${\left[\because \tan \theta=\frac{1}{\cot \theta}\right]}$

$=\frac{1}{\tan 75^{\circ}} \times \frac{1}{\tan 65^{\circ}} \times \tan 60^{\circ} \times \tan 65^{\circ} \times \tan 75^{\circ}$

$=\tan 60^{\circ}\left[\because \tan 60^{\circ}=\sqrt{3}\right]$

$=\sqrt{3}$

Now, calculate the value of  $\tan 30^{\circ=\frac{1}{\sqrt{3}}$​

Then, $\tan 15^{\circ} \tan 25^{\circ} \tan 60^{\circ} \tan 65^{\circ} \tan 75^{\circ}-\tan 30^{\circ=\sqrt{3}*\frac{1}{\sqrt{3}}=1$

Hence, the required answer is $1$.

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