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Question: Prove that:- \[\dfrac{\tan {{69}^{\circ }}+\tan {{66}^{\circ }}}{1-\tan {{69}^{\circ }}\tan {{66}^...

Prove that:-
tan⁡69∘+tan⁡66∘1−tan⁡69∘tan⁡66∘=−1\dfrac{\tan {{69}^{\circ }}+\tan {{66}^{\circ }}}{1-\tan {{69}^{\circ }}\tan {{66}^{\circ }}}=-1

Explanation

Solution

Hint: In such questions, we prove them by either making the left hand side that is L.H.S. or by making the right hand side that is R.H.S. equal to the other in order to prove the proof that has been asked.
tan⁡(x+y)=tan⁡x+tan⁡y1−tan⁡x⋅tan⁡y\tan (x+y)=\dfrac{\tan x+\tan y}{1-\tan x\cdot \tan y}

Complete step-by-step answer:
Now, in such questions, if one tries to simplify the left hand side that is L.H.S., then first thing that is to be done is to convert the tan function in terms of sin and cos functions and that is done by using the following relations
tan⁡x=sin⁡xcos⁡x\tan x=\dfrac{\sin x}{\cos x}
Another important formulae that might be used in solving this question is as follows
tan⁡(x+y)=tan⁡x+tan⁡y1−tan⁡x⋅tan⁡y\tan (x+y)=\dfrac{\tan x+\tan y}{1-\tan x\cdot \tan y}
And also
tan⁡(180∘−45∘)=−1\tan ({{180}^{\circ }}-{{45}^{\circ }})=-1
Now, these are the results that would be used to prove the proof mentioned in this question as using these identities, we would convert the left hand side that is L.H.S. or the right hand side that is R.H.S. to make either of them equal to the other.

As mentioned in the question, we have to prove the given expression.
Now, we will start with the left hand side that is L.H.S. as follows

& =\dfrac{\tan {{69}^{\circ }}+\tan {{66}^{\circ }}}{1-\tan {{69}^{\circ }}\tan {{66}^{\circ }}} \\\ & =\tan ({{69}^{\circ }}+{{66}^{\circ }}) \\\ & =\tan ({{135}^{\circ }}) \\\ \end{aligned}$$ (Using the identities that are mentioned in the hint) Now, we know that we can write the above expression as follows $$\begin{aligned} & =\tan ({{135}^{\circ }}) \\\ & =\tan ({{180}^{\circ }}-{{45}^{\circ }}) \\\ & =\tan ({{45}^{\circ }}) \\\ & =-1 \\\ \end{aligned}$$ (Using the relation mentioned in the hint) Now, as the right hand side that is R.H.S. is equal to the left hand side that is L.H.S., hence, the expression has been proved. Note:Another method of attempting this question is by converting the right hand side that is R.H.S. to the left hand side that is L.H.S. by using the relations that are given in the hint. Through this method also, we could get to the correct answer and hence, we would be able to prove the required proof.