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.