Question

Find the value of \[\tan A + \tan (60 + A) - \tan (60 - A)\]
(1) 3 tan 3A
(2) tan 3A
(3) cot 3A
(4) sin 3A

Answer 1

Hint: To solve the given above question we must know the formulas of tan(A-B), tan(A+B), and also tan 3A. Here we need to apply LCM and also need to know the values of tan60 degree to substitute in the formula.

Complete step-by-step answer:
Here they have given four option for the given question we need to find the correct answer
 Now, Consider the given question \[\tan A + \tan (60 + A) - \tan (60 - A)\]
 \[ = \] tan A+\[\dfrac{{\tan 60 + \tan A}}{{1 - \tan 60.\tan A}}\]- \[\dfrac{{\tan 60 - \tan A}}{{1 + \tan 60.\tan A}}\]
[here we have applied the formula of tan(A+B) and tan(A-B)]
\[ = \]tan A+\[\dfrac{{\sqrt 3 + \tan A}}{{1 - \sqrt 3 \tan A}}\]-\[\dfrac{{\sqrt 3 - \tan A}}{{1 + \sqrt 3 \tan A}}\] [here tan 60 =\[\sqrt 3 \]]
Taking L C M we get,
\[ = \]tan A+\[\left[ {\dfrac{{(1 + \sqrt 3 \tan A)(\sqrt 3 + \tan A) - ((1 - \sqrt 3 \tan A)(\sqrt 3 - \tan A))}}{{(1 - \sqrt 3 \tan A)(1 + \sqrt 3 \tan A)}}} \right]\]
\[ = \]tan A+\[\left[ {\dfrac{{(1 + \sqrt 3 \tan A)(\sqrt 3 + \tan A) - ((1 - \sqrt 3 \tan A)(\sqrt 3 - \tan A))}}{{(1 - 3{{\tan }^2}A)}}} \right]\]
\[ = \]tan A+\[\left[ {\dfrac{{\sqrt 3 + \tan A + 3\tan A + \sqrt 3 {{\tan }^2}A - (\sqrt 3 - \tan A - 3\tan A + \sqrt 3 {{\tan }^2}A}}{{(1 - 3{{\tan }^2}A)}}} \right]\]
\[ = \]tan A+\[\left[ {\dfrac{{\sqrt 3 + \tan A + 3\tan A + \sqrt 3 {{\tan }^2}A - \sqrt 3 + \tan A + 3\tan A - \sqrt 3 {{\tan }^2}A}}{{(1 - 3{{\tan }^2}A)}}} \right]\]
\[ = \]tan A+\[\left[ {\dfrac{{8\tan A}}{{1 - 3{{\tan }^2}A}}} \right]\]
Again, Taking LCM we get
\[ = \]\[\left[ {\dfrac{{\tan A - 3{{\tan }^3}A + 8\tan A}}{{1 - 3{{\tan }^2}A}}} \right]\]
\[ = \]\[\left[ {\dfrac{{9\tan A - 3{{\tan }^3}A}}{{1 - 3{{\tan }^2}A}}} \right]\]
Taking 3 as common in numerator we get,
\[ = \]\[3\left[ {\dfrac{{3\tan A - {{\tan }^3}A}}{{1 - 3{{\tan }^2}A}}} \right]\]
\[ = \]\[3\tan 3A\]
Therefore, In the given four option (1) 3 tan3A is correct.
Hence we got the required solution


So, the correct answer is “Option (1)”.

Note: Formulas:
tan(A+B) = \[\dfrac{{\tan A + \tan B}}{{1 - \tan A.\tan B}}\]
tan(A-B) = \[\dfrac{{\tan A - \tan B}}{{1 + \tan A.\tan B}}\]
tan 3A = \[\left[ {\dfrac{{3\tan A - {{\tan }^3}A}}{{1 - 3{{\tan }^2}A}}} \right]\]
Trigonometric functions are real functions which relate an angle of a right-angled triangle to ratios of two side lengths.
The trigonometric functions most widely used in modern mathematics are the sine, the cosine, and the tangent.
The angle measure is expressed in one of two units such as degrees or radians.
A degree is \[\dfrac{1}{{{{360}^{th}}}}\]of a complete rotation around a circle. Whereas, radians are alternate units used to measure angles. Sometimes they are referred to as Natural angles.
tan0=0
tan30=\[\dfrac{1}{{\sqrt 3 }}\]
tan45= 1
tan60=\[\sqrt 3 \]
tan90= \[\infty \]

[We know that L C M is the least common multiple].