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Question: How do you write the expression \(\dfrac{\tan 140^{\circ}-\tan 60^{\circ}}{1+\tan 140^{\circ}\tan 60...

How do you write the expression tan⁡140∘−tan⁡60∘1+tan⁡140∘tan⁡60∘\dfrac{\tan 140^{\circ}-\tan 60^{\circ}}{1+\tan 140^{\circ}\tan 60^{\circ}} as the sine, cosine or the tangent of an angle?

Explanation

Solution

In order to do this question, you need to know the tan(A + B) formula. By using this you can simplify the above equation. tan⁡(a−b)=tan⁡a−tan⁡b1+tan⁡atan⁡b \tan \left( a-b \right)=\dfrac{\tan a-\tan b}{1+\tan a\tan b}. Here a is 140 degrees and b is 60 degrees. Therefore, you can simplify it. Next you need to know the formula of tan in terms of sine and cos that is tan⁡x=sin⁡xcos⁡x\tan x=\dfrac{\sin x}{\cos x}. Here x will be a - b . Therefore, you can write tan in terms of sine and cosine.

Complete step-by-step solution:
The first step in order to solve this question is to first simplify the tan⁡140∘−tan⁡60∘1+tan⁡140∘tan⁡60∘\dfrac{\tan 140^{\circ}-\tan 60^{\circ}}{1+\tan 140^{\circ}\tan 60^{\circ}}. Here we use the formula of tan (a - b) which is given by tan⁡(a−b)=tan⁡a−tan⁡b1+tan⁡atan⁡b \tan \left( a-b \right)=\dfrac{\tan a-\tan b}{1+\tan a\tan b}. Now, we have to take a is 140 and b is 60. Therefore, we can simplify the expression as
⇒tan⁡(140∘−60∘)=tan⁡140∘−tan⁡60∘1+tan⁡140∘tan⁡60∘\Rightarrow \tan \left( 140^{\circ}-60^{\circ} \right)= \dfrac{\tan 140^{\circ}-\tan 60^{\circ}}{1+\tan 140^{\circ}\tan 60^{\circ}}
⇒tan⁡(80∘)=tan⁡140∘−tan⁡60∘1+tan⁡140∘tan⁡60∘\Rightarrow \tan \left( 80^{\circ}\right)= \dfrac{\tan 140^{\circ}-\tan 60^{\circ}}{1+\tan 140^{\circ}\tan 60^{\circ}}

Therefore, in terms of tan, we can write the above expression as tan(3x).
Next step we have to take is that we have to write tan in terms of sin and cos. For this, we use the formula tan⁡x=sin⁡xcos⁡x\tan x=\dfrac{\sin x}{\cos x}. Therefore we get the answer as
⇒tan⁡(80∘)=sin⁡80∘cos⁡80∘\Rightarrow \tan \left(80^{\circ} \right)=\dfrac{\sin 80^{\circ}}{\cos 80^{\circ}}.
Therefore, we get the final answer of the question how do you write the expression tan⁡140∘−tan⁡60∘1+tan⁡140∘tan⁡60∘\dfrac{\tan 140^{\circ}-\tan 60^{\circ}}{1+\tan 140^{\circ}\tan 60^{\circ}} as the sine, cosine or the tangent of an angle as tan⁡(80∘)=sin⁡80∘cos⁡80∘\tan \left(80^{\circ} \right)=\dfrac{\sin 80^{\circ}}{\cos 80^{\circ}}.

Note: For solving these kinds of questions, you need to know the trigonometric identities and the sum of angles for tan and cosine and sine. You will not be able to solve this problem else. Also, be careful while using the substitutions or while using formulas.