Question

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

Answer 1

Hint:
The given expression contains terms in the trigonometric ratios $\tan $ and $\cot $. We have relations connecting these two. Thus we get every term with angles $9$ or $21$. Then we can apply the sum formula of $\tan $. Simplifying and substituting the known values we get the answer.

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

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

Combining the equations (i), (ii), (iii) and (iv) we get,
$\dfrac{{\tan 225 - \cot 81\cot 69}}{{\cot 261 + \tan 21}} = \dfrac{{1 - \tan 9\tan 21}}{{\tan 9 + \tan 21}}$ _____(*)
We know that $\tan A + B = \dfrac{{\tan A + \tan B}}{{1 - \tan A\tan B}}$
Substituting $A = 9,B = 21$ in the above result we have,
$\tan (9 + 21) = \dfrac{{\tan 9 + \tan 21}}{{1 - \tan 9\tan 21}}$
Comparing it with equation (*) we have,
$\dfrac{{\tan 225 - \cot 81\cot 69}}{{\cot 261 + \tan 21}} = \dfrac{1}{{\tan (9 + 21)}}$
$ \Rightarrow \dfrac{{\tan 225 - \cot 81\cot 69}}{{\cot 261 + \tan 21}} = \dfrac{1}{{\tan 30}}$
We know that $\tan 30 = \dfrac{1}{{\sqrt 3 }}$
Substituting this in the above equation we get,
$ \Rightarrow \dfrac{{\tan 225 - \cot 81\cot 69}}{{\cot 261 + \tan 21}} = \dfrac{1}{{\dfrac{1}{{\sqrt 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 $ and $\cot $ into each other. Since $\tan $ and $\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.