Question

If 11x is an acute angle and $\tan 11x = \cot 7x$, then what is the value of x ?

$$A.{\text{ }}5^\circ \\\ B.{\text{ }}6^\circ \\\ C.{\text{ }}7^\circ \\\ D.{\text{ }}8^\circ \\\$$

Answer 1

Hint:- Here 11x is an acute angle. It means that $11x < 90^\circ$. In this question we can use the properties of $90^\circ - \theta$ to change the $\tan \theta$ into $\cot \theta$ or vice versa. Trigonometry properties such as $\tan A - \tan B$ will also help us in solving such problems.

Complete step-by-step answer:
As we all know that here $\tan 11x = \cot 7x$
And by property of $\cot \theta = \tan 90^\circ - \theta$ we can write the above equation as
$\Rightarrow$ $\tan 11x = \tan (90^\circ - 7x)$
\Rightarrow $$$$\tan 11x - \tan (90^\circ - 7x) = 0
Now applying the property of $\tan A - \tan B = \tan (A - B) \times (1 + \tan A\tan B)$in above equation
\Rightarrow $$$$\tan (11x - 90^\circ + 7x) \times (1 + \tan 11x \times \tan (90^\circ - 7x) = 0
\Rightarrow $$$$\tan (18x - 90^\circ ) \times (1 + \tan 11x \times \tan (90^\circ - 7x) = 0
Now on further equating we can observe that $\tan (18x - 90^\circ )$ can be equal to 0 but $(1 + \tan 11x \times \tan (90^\circ - 7x)$cannot be equal to 0. This is because we had given that
11x < ${\text{ 90}}^\circ$so 18x must also be smaller than ${\text{ 90}}^\circ$.
Now let us assume that $\tan 11x \times \tan (90^\circ - 7x)$= 0 but still also $(1 + \tan 11x \times \tan (90^\circ - 7x)$ =
$1 + 0 \ne 0$ so, from here we come to know that
$\Rightarrow$ $(1 + \tan 11x \times \tan (90^\circ - 7x) \ne 0$
So now let us solve $\tan (18x - 90^\circ ) = 0$
As we know that ( $\tan 0^\circ = \tan 180^\circ = 0$) but here the value of 0 cannot be $\tan 180^\circ$ because if 11x < $90^\circ$ than 18x must be smaller than $180^\circ$.
So, $\tan (18x - 90^\circ ) = \tan 0^\circ$
Now, $18x - 90 = 0 \Rightarrow 18x = 90 \Rightarrow x = 5$
Hence A is the correct option.

Note :- Whenever we come up with this type of problem we can also use the identity of $\tan A - \tan B$ by changing $\tan A$ into $\dfrac{{\sin A}}{{\cos A}}$ and similarly changing the other terms in $\dfrac{{\sin \theta }}{{\cos \theta }}$ and this will also give us the same value of x but the solution will expand and the explanation with this type will be a bit complicated . So this is the easiest and efficient method to solve problems like this.