Answer

Verified

342.3k+ views

**Hint:**Here in this question, first write the $\sec \left( -{{240}^{\circ }} \right)$ using reciprocal identities. Then convert the degrees to radians. After conversion, we have to use the trigonometric identities to solve the function. We will apply cos(x + y) = cosx.cosy – sinx.siny for further calculations. After that we will use trigonometric ratios to find the values for angles.

**Complete step by step answer:**

Now, let’s solve the question.

As we all know the basic functions and derived functions. Derived functions are obtained from basic functions itself. So, first we will write all of them.

$\Rightarrow $ Sine (sin)

$\Rightarrow $Cosine (cos)

$\Rightarrow $Tangent (tan)

$\Rightarrow $cosec$\theta $ = $\dfrac{1}{\sin \theta }$

$\Rightarrow $sec$\theta $ = $\dfrac{1}{\cos \theta }$

$\Rightarrow $tan$\theta $ = $\dfrac{\sin \theta }{\cos \theta }$ = $\dfrac{1}{\cot \theta }$

$\Rightarrow $cot$\theta $ = $\dfrac{1}{\tan \theta }$ = $\dfrac{\cos \theta }{\sin \theta }$

Where $\theta $ means angle in degrees or radians.

This is the right-angled triangle with an angle $\theta $ at C.

We can find different functions using this triangle. Let’s say we want to find cos$\theta $, then the value of cos$\theta $ will be:

$\Rightarrow $cos$\theta $ = $\dfrac{base(B)}{hypotenuse(H)}$

But if we want for sec$\theta $, we will reciprocate cos$\theta $, so:

$\Rightarrow $sec$\theta $ = $\dfrac{hypotenuse(H)}{base(B)}$

Let’s study some of the trigonometric ratios from the below table:

Trigonometric ratios(angle $\theta $ in radians) | 0 | $\dfrac{\pi }{6}$ | $\dfrac{\pi }{4}$ | $\dfrac{\pi }{3}$ | $\dfrac{\pi }{2}$ |

sin$\theta $ | 0 | $\dfrac{1}{2}$ | $\dfrac{1}{\sqrt{2}}$ | $\dfrac{\sqrt{3}}{2}$ | 1 |

cos$\theta $ | 1 | $\dfrac{\sqrt{3}}{2}$ | $\dfrac{1}{\sqrt{2}}$ | $\dfrac{1}{2}$ | 0 |

tan$\theta $ | 0 | $\dfrac{1}{\sqrt{3}}$ | 1 | $\sqrt{3}$ | $\infty $ |

cosec$\theta $ | $\infty $ | 2 | $\sqrt{2}$ | $\dfrac{2}{\sqrt{3}}$ | 1 |

sec$\theta $ | 1 | $\dfrac{2}{\sqrt{3}}$ | $\sqrt{2}$ | 2 | $\infty $ |

cot$\theta $ | $\infty $ | $\sqrt{3}$ | 1 | $\dfrac{1}{\sqrt{3}}$ | 0 |

We can also obtain some values by reciprocating the functions:

$\Rightarrow $sinx = $\dfrac{1}{\cos ecx}$ or cosecx = $\dfrac{1}{\sin x}$

$\Rightarrow $cosx = $\dfrac{1}{\sec x}$ or secx = $\dfrac{1}{\cos x}$

$\Rightarrow $tanx = $\dfrac{1}{\cot x}$ or cotx = $\dfrac{1}{\tan x}$

Now, let’s see some even and odd functions.

$\Rightarrow $sin(-x) = -sinx

$\Rightarrow $ cos(-x) = cosx

$\Rightarrow $ tan(-x) = -tanx

$\Rightarrow $ cot(-x) = -cotx

$\Rightarrow $ cosec(-x) = -cosecx

$\Rightarrow $sec(-x) = secx

Write the function given in the question.

$\Rightarrow \sec \left( -{{240}^{\circ }} \right)$

As sec(-x) = secx, so:

$\Rightarrow \sec \left( -{{240}^{\circ }} \right)$ = $\sec \left( {{240}^{\circ }} \right)$

We have to apply reciprocal identity to convert in terms of cos.

$\Rightarrow \dfrac{1}{\cos \left( {{240}^{\circ }} \right)}$

Now we have to convert the given degrees to radians.

As we know that one revolution of a circle is of ${{360}^{\circ }}$ which is equal to $2\pi $ radians.

$\Rightarrow {{360}^{\circ }}=2\pi $radians

Now if ${{360}^{\circ }}=2\pi $ radians and if we will divide both sides by 2, we get:

$\Rightarrow \dfrac{{{360}^{\circ }}}{2}=\dfrac{2\pi }{2}$

After reducing the terms, we will get:

$\Rightarrow {{180}^{\circ }}=\pi $ Radians

So, this is the conversion from degrees to radians. If we wish to convert any degree into radian, we have to multiply the degree with a factor of $\dfrac{\pi }{{{180}^{\circ }}}$.

For ${{240}^{\circ }}$, we will multiply it by $\dfrac{\pi }{{{180}^{\circ }}}$ to obtain in radians.

$\Rightarrow {{240}^{\circ }}\times \dfrac{\pi }{{{180}^{\circ }}}$

On reducing the terms, we will get the value $\dfrac{4\pi }{3}$ radians.

So now it will be:

$\Rightarrow \dfrac{1}{\cos \left( \dfrac{4\pi }{3} \right)}$

Can we write $\dfrac{4\pi }{3}$ as $\left( \pi +\dfrac{\pi }{3} \right)$. Place this in above expression:

$\Rightarrow \dfrac{1}{\cos \left( \pi +\dfrac{\pi }{3} \right)}$

After this, we will use sum and difference identities so solve further. The sum identity is:

$\Rightarrow $cos(x + y) = cosx.cosy – sinx.siny

Here, x = $\pi $ and y = $\dfrac{\pi }{3}$. Place the angles in the formula, we will get:

$\Rightarrow \dfrac{1}{\left( \cos \pi .\cos \dfrac{\pi }{3}-\sin \pi .\sin \dfrac{\pi }{3} \right)}$

cos$\dfrac{\pi }{3}$ and sin$\dfrac{\pi }{3}$ are given in trigonometric ratios table, and cos$\pi $ = -1 and sin$\pi $ = 0. Place all the values:

$\Rightarrow \dfrac{1}{\left( (-1).\dfrac{1}{2}-0.\sin \dfrac{\sqrt{3}}{2} \right)}$

After calculation we will get:

$\Rightarrow \dfrac{1}{\left( \dfrac{-1}{2} \right)}$

After reciprocating:

$\Rightarrow \sec \left( -{{240}^{\circ }} \right)$ = -2

This is the final answer.

**Note:**

There is no direct trigonometric ratio for $\sec \left( -{{240}^{\circ }} \right)$, that’s why we first reciprocated and then split the angles so that it becomes easier to find the value. All the trigonometric identities should be on the tips for solving any questions related to angles, equations or simplification of the expression.

Recently Updated Pages

The Equation xxx + 2 is Satisfied when x is Equal to Class 10 Maths

How do you arrange NH4 + BF3 H2O C2H2 in increasing class 11 chemistry CBSE

Is H mCT and q mCT the same thing If so which is more class 11 chemistry CBSE

What are the possible quantum number for the last outermost class 11 chemistry CBSE

Is C2 paramagnetic or diamagnetic class 11 chemistry CBSE

What happens when entropy reaches maximum class 11 chemistry JEE_Main

Trending doubts

How do you solve x2 11x + 28 0 using the quadratic class 10 maths CBSE

Why is the adrenaline hormone called fight or flight class 11 biology CBSE

Fill the blanks with the suitable prepositions 1 The class 9 english CBSE

Difference between Prokaryotic cell and Eukaryotic class 11 biology CBSE

Differentiate between lanthanoids and actinoids class 12 chemistry CBSE

What is the color of ferrous sulphate crystals? How does this color change after heating? Name the products formed on strongly heating ferrous sulphate crystals. What type of chemical reaction occurs in this type of change.

Give 10 examples of unisexual and bisexual flowers

Open circulatory system is present in I Arthropods class 12 biology CBSE

Name the highest peak of the Indian Himalayas class 8 social science CBSE