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Question: Solve the following : \[1+\sec 20{}^\circ -\sqrt{3}\cot 40{}^\circ \]...

Solve the following : 1+sec⁡20∘−3cot⁡40∘1+\sec 20{}^\circ -\sqrt{3}\cot 40{}^\circ

Explanation

Solution

Now to solve this question we need to simplify all the trigonometric functions using their formulas. We convert them into sine and cos functions using the formulas that sec⁡x=1cos⁡x\sec x=\dfrac{1}{\cos x} and cot⁡x=cos⁡xsin⁡x\cot x=\dfrac{\cos x}{\sin x} . Now students will also need to use the formula that sin⁡(a−b)=sin⁡acos⁡b−cos⁡asin⁡b\sin (a-b)=\sin a\cos b-\cos a\sin b to further simplify and solve this equation and find the necessary answer we need in this question.

Complete step-by-step answer:
Now here in this question we need to find the value of the expression 1+sec⁡20∘−3cot⁡40∘1+\sec 20{}^\circ -\sqrt{3}\cot 40{}^\circ . Now to find its value we can name the expression to be
E=1+sec⁡20∘−3cot⁡40∘E=1+\sec 20{}^\circ -\sqrt{3}\cot 40{}^\circ
Now to solve this further we will start by making these equations in the form of sine and cos functions to make it simpler to solve. To convert the equation we use the formula of secant and cot that sec⁡x=1cos⁡x\sec x=\dfrac{1}{\cos x} and cot⁡x=cos⁡xsin⁡x\cot x=\dfrac{\cos x}{\sin x} . By substituting these values in the above given expression we get
E=1+1cos⁡20∘−3cos⁡40∘sin⁡40∘E=1+\dfrac{1}{\cos 20{}^\circ }-\sqrt{3}\dfrac{\cos 40{}^\circ }{\sin 40{}^\circ }

Now we take the first and third term of this equation and taking its LCM we get
E=sin⁡40∘−3cos⁡40∘sin⁡40∘+1cos⁡20∘E=\dfrac{\sin 40{}^\circ -\sqrt{3}\cos 40{}^\circ }{\sin 40{}^\circ }+\dfrac{1}{\cos 20{}^\circ }
Now in first term to simplify it further we multiply and divide both the numerator and denominator by 2 giving us
E=2(12sin⁡40∘−32cos⁡40∘)sin⁡40∘+1cos⁡20∘E=2\dfrac{\left( \dfrac{1}{2}\sin 40{}^\circ -\dfrac{\sqrt{3}}{2}\cos 40{}^\circ \right)}{\sin 40{}^\circ }+\dfrac{1}{\cos 20{}^\circ }
Now we know that we can write cos⁡60∘=12\cos 60{}^\circ =\dfrac{1}{2} and we can also write sin⁡60∘=32\sin 60{}^\circ =\dfrac{\sqrt{3}}{2}
E=2(cos⁡60∘sin⁡40∘−sin⁡60∘cos⁡40∘)sin⁡40∘+1cos⁡20∘E=2\dfrac{\left( \cos 60{}^\circ \sin 40{}^\circ -\sin 60{}^\circ \cos 40{}^\circ \right)}{\sin 40{}^\circ }+\dfrac{1}{\cos 20{}^\circ }
Now we can see that the numerator of first term is in the form of sin⁡(a−b)=sin⁡acos⁡b−cos⁡asin⁡b\sin (a-b)=\sin a\cos b-\cos a\sin b therefore we can see that a is 40∘40{}^\circ and b is 60∘60{}^\circ . So we can write this equation is
E=2sin⁡(40∘−60∘)sin⁡40∘+1cos⁡20∘E=2\dfrac{\sin (40{}^\circ -60{}^\circ )}{\sin 40{}^\circ }+\dfrac{1}{\cos 20{}^\circ }
We can also write sin⁡40∘\sin 40{}^\circ in the form of half angle of sine which gives us sin⁡40∘=2sin⁡20∘cos⁡20∘\sin 40{}^\circ =2\sin 20{}^\circ \cos 20{}^\circ and we also know that negative angle of sine is always equal to negative of sine that is sin⁡(−x)=−sin⁡x\sin (-x)=-\sin x; Therefore putting these values we get;
E=−2sin⁡20∘2sin⁡20∘cos⁡20∘+1cos⁡20∘E=\dfrac{-2\sin 20{}^\circ }{2\sin 20{}^\circ \cos 20{}^\circ }+\dfrac{1}{\cos 20{}^\circ }
Now cancelling the common terms from both numerator and denominator
E=−1cos⁡20∘+1cos⁡20∘E=-\dfrac{1}{\cos 20{}^\circ }+\dfrac{1}{\cos 20{}^\circ }
Therefore we get
E=0E=0
So we can say that ; 1+sec⁡20∘−3cot⁡40∘=01+\sec 20{}^\circ -\sqrt{3}\cot 40{}^\circ =0

Note: To solve questions like this students should always try to remember the half angle and double angle formulas of functions. Converting them into sine and cos functions gives us a way to easily simplify these expressions. Students must also know the product and addition formulas of trigonometric functions. Another necessity required is to know the trigonometric values of all basic angles.