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Question: Evaluate \(\dfrac{{\tan 225 - \cot 81\cot 69}}{{\cot 261 + \tan 21}}\). A) \(1\) B) \(\dfrac{1}{...

Evaluate tan⁡225−cot⁡81cot⁡69cot⁡261+tan⁡21\dfrac{{\tan 225 - \cot 81\cot 69}}{{\cot 261 + \tan 21}}.
A) 11
B) 12\dfrac{1}{{\sqrt 2 }}
C) 3\sqrt 3
D) 13\dfrac{1}{{\sqrt 3 }}

Explanation

Solution

The given expression contains terms in the trigonometric ratios tan⁡\tan and cot⁡\cot . We have relations connecting these two. Thus we get every term with angles 99 or 2121. Then we can apply the sum formula of tan⁡\tan . Simplifying and substituting the known values we get the answer.

Useful formula:
For every angle θ\theta we have the following trigonometric relations.
tan⁡(270−θ)=cot⁡θ\tan (270 - \theta ) = \cot \theta
cot⁡(90−θ)=tan⁡θ\cot (90 - \theta ) = \tan \theta
Also we have,
tan⁡A+B=tan⁡A+tan⁡B1−tan⁡Atan⁡B\tan A + B = \dfrac{{\tan A + \tan B}}{{1 - \tan A\tan B}}

Complete step by step solution:
The given expression is tan⁡225−cot⁡81cot⁡69cot⁡261+tan⁡21\dfrac{{\tan 225 - \cot 81\cot 69}}{{\cot 261 + \tan 21}}.
We can consider the terms of the expression individually.
We know that tan⁡(270−θ)=cot⁡θ\tan (270 - \theta ) = \cot \theta
This gives, tan⁡225=tan⁡(270−45)=cot⁡45\tan 225 = \tan (270 - 45) = \cot 45
Also we have, cot⁡45=1tan⁡45\cot 45 = \dfrac{1}{{\tan 45}} and tan⁡45\tan 45 equals one.
So we get, cot⁡45=11=1\cot 45 = \dfrac{1}{1} = 1
This gives, tan⁡225=1\tan 225 = 1 _____(i)
Now we have,
cot⁡81\cot 81 can be written as cot⁡(90−9)\cot (90 - 9).
Also we know,
cot⁡(90−θ)=tan⁡θ\cot (90 - \theta ) = \tan \theta
So we get,
cot⁡(90−9)=tan⁡9\cot (90 - 9) = \tan 9
This gives,
cot⁡81=tan⁡9\cot 81 = \tan 9 __(ii)
Now consider the next term.
cot⁡69\cot 69 can be written as cot⁡(90−21)\cot (90 - 21).
And cot⁡(90−21)=tan⁡21\cot (90 - 21) = \tan 21
This gives,
cot⁡69=tan⁡21\cot 69 = \tan 21(iii)
Now we have 261=270−9261 = 270 - 9
This gives, cot⁡261=cot⁡(270−9)\cot 261 = \cot (270 - 9)
So we get, cot⁡261=tan⁡9\cot 261 = \tan 9 _____(iv)
Thus we converted the terms into our convenience to solve them.

Combining the equations (i), (ii), (iii) and (iv) we get,
tan⁡225−cot⁡81cot⁡69cot⁡261+tan⁡21=1−tan⁡9tan⁡21tan⁡9+tan⁡21\dfrac{{\tan 225 - \cot 81\cot 69}}{{\cot 261 + \tan 21}} = \dfrac{{1 - \tan 9\tan 21}}{{\tan 9 + \tan 21}} _____()
We know that tan⁡A+B=tan⁡A+tan⁡B1−tan⁡Atan⁡B\tan A + B = \dfrac{{\tan A + \tan B}}{{1 - \tan A\tan B}}
Substituting A=9,B=21A = 9,B = 21 in the above result we have,
tan⁡(9+21)=tan⁡9+tan⁡211−tan⁡9tan⁡21\tan (9 + 21) = \dfrac{{\tan 9 + \tan 21}}{{1 - \tan 9\tan 21}}
Comparing it with equation () we have,
tan⁡225−cot⁡81cot⁡69cot⁡261+tan⁡21=1tan⁡(9+21)\dfrac{{\tan 225 - \cot 81\cot 69}}{{\cot 261 + \tan 21}} = \dfrac{1}{{\tan (9 + 21)}}
⇒tan⁡225−cot⁡81cot⁡69cot⁡261+tan⁡21=1tan⁡30\Rightarrow \dfrac{{\tan 225 - \cot 81\cot 69}}{{\cot 261 + \tan 21}} = \dfrac{1}{{\tan 30}}
We know that tan⁡30=13\tan 30 = \dfrac{1}{{\sqrt 3 }}
Substituting this in the above equation we get,
⇒tan⁡225−cot⁡81cot⁡69cot⁡261+tan⁡21=113\Rightarrow \dfrac{{\tan 225 - \cot 81\cot 69}}{{\cot 261 + \tan 21}} = \dfrac{1}{{\dfrac{1}{{\sqrt 3 }}}}
⇒tan⁡225−cot⁡81cot⁡69cot⁡261+tan⁡21=3\Rightarrow \dfrac{{\tan 225 - \cot 81\cot 69}}{{\cot 261 + \tan 21}} = \sqrt 3

Therefore the answer is option C.

Note:
We converted the ratios of tan⁡\tan and cot⁡\cot into each other. Since tan⁡\tan and cot⁡\cot are positive in the first and third quadrant we get positive results. Thus we could apply the sum formula. Identifying the given expressions and their conversion is important here.