Question

(Sec 30*)(Cos 30*) - (Tan 60*)(Cot 60*)

Answer 1

first of all, lemme tell u the values of sec and cos at 30° and tan and cot at 60°

sec 30° = 2/√3

cos 30° = √3/2

tan 60° = √3

cot 60° = √3/3

$\sf \therefore \: (\sec30 \degree)( \cos30 \degree) - ( \tan60 \degree) \\ \sf( \cot60 \degree) : - \\ \\ = \frac{ \cancel2}{ \cancel{\sqrt{3}} } \times \frac{ \cancel{ \sqrt{3}} }{ \cancel2} - ( \cancel{\sqrt{3} } \times \frac{1}{ \cancel{\sqrt{3}} } ) \\ \\ = 1 - 1 \\ \\ = 0$

Answer 2

Answer:

The value of $&\sec \left(30^{\circ}\right) \cos \left(30^{\circ}\right)-\tan \left(60^{\circ}\right) \cot \left(60^{\circ}\right)$ is equal to 0.

Step-by-step explanation:

Given:

$&\left(\sec \left(30^{\circ}\right)\right)\left(\cos \left(30^{\circ}\right)\right)-\left(\tan \left(60^{\circ}\right)\right)\left(\cot \left(60^{\circ}\right)\right)$

$&=\sec \left(30^{\circ}\right) \cos \left(30^{\circ}\right)-\tan \left(60^{\circ}\right) \cot \left(60^{\circ}\right)$

We have to find the value of

$&\sec \left(30^{\circ}\right) \cos \left(30^{\circ}\right)-\tan \left(60^{\circ}\right) \cot \left(60^{\circ}\right)$

Step 1

Express with sin, cos

$&\sec \left(30^{\circ}\right)=\frac{1}{\cos \left(30^{\circ}\right)}$

$&=\frac{1}{\cos \left(30^{\circ}\right)} \cos \left(30^{\circ}\right)-\tan \left(60^{\circ}\right) \cot \left(60^{\circ}\right)$

$&\frac{1}{\cos \left(30^{\circ}\right)} \cos \left(30^{\circ}\right)=1$

$&=1-\tan \left(60^{\circ}\right) \cot \left(60^{\circ}\right)$

Step 2

Use the following trivial identity:

Since the zero solution exists as the "obvious" solution, hence it exists named a trivial solution.

Simplifying the values of the above equation, we get

$\tan \left(60^{\circ}\right)=\sqrt{3}$

$\cot \left(60^{\circ}\right)=\frac{\sqrt{3}}{3}$

$=1-\sqrt{3} \frac{\sqrt{3}}{3}$

$1-\sqrt{3} \frac{\sqrt{3}}{3}=0$

$=0$.

Therefore, the value of

$&\sec \left(30^{\circ}\right) \cos \left(30^{\circ}\right)-\tan \left(60^{\circ}\right) \cot \left(60^{\circ}\right)$ is equal to 0.

#SPJ2