Question

If $\tan A=2\tan B+\cot B$then $2\tan (A-B)$ is equal to
A. $\tan B$
B. $2\tan B$
C. $\cot B$
D. $2\cot B$

Answer 1

Hint: As we are asked the tangent of the difference of the angles, we must know all the trigonometric formulas for the sine, cosine and tangent of sum and difference of angles. To solve this we must know the formulas that $2\tan (A-B)=2\times \dfrac{\tan A-\tan B}{1+\tan A\tan B}$ . And we must also know cot is the inverse of tan and vice versa .

Complete step by step answer:
To get the answer for this we must know what the formula for $2\tan (A-B)$ which is
$2\tan (A-B)=2\times \dfrac{\tan A-\tan B}{1+\tan A\tan B}$
 Now after we know this formula we can realize and know that we need the value of $tanA\;tanB$ and $\tan A-\tan B$ to be able to get the answer .That is why to get the answer of $tanA\;tanB$ we will first start this by multiplying $tanB$ on both sides of the equation. Now therefore the given equation is
 $\tan A=2\tan B+\cot B$
When we multiple $tanB$ on both sides of this equation we get,
$\tan A\tan B=2{{\tan }^{2}}B+1$
Now the only thing we are left to find here is $\tan A-\tan B$ which we can find when we subtract $tanB$ from both the sides of the equation.
$\tan A-\tan B=2\tan B+\cot B-\tan B$
This gives us,
$\tan A-\tan B=\tan B+\cot B$
Now since $cotB$ is the inverse of $tanB$ we get that
$\tan A-\tan B=\tan B+\dfrac{1}{\tan B}$
Taking LCM on the right side we get
$\tan A-\tan B=\dfrac{{{\tan }^{2}}B+1}{\tan B}$
Now that we know the value of both the things we need to get the answer for $2\tan (A-B)$ we can put the values into the equation,
$2\tan (A-B)=2\times \dfrac{\tan B+\cot B}{1+2{{\tan }^{2}}B+1}$
Trying to simplifying it;
$2\tan (A-B)=2\times \dfrac{\tan B+\cot B}{2(1+{{\tan }^{2}}B)}$
$cotB$ is the inverse of $tanB$ that’s why;
$2\tan (A-B)=2\times \dfrac{\tan B+\dfrac{1}{\tan B}}{2(1+{{\tan }^{2}}B)}$
Taking the LCM we get
$2\tan (A-B)=2\times \dfrac{{{\tan }^{2}}+1}{2(1+{{\tan }^{2}}B)\tan B}$
Simplifying the numerator and denominator we get
$2\tan (A-B)=\dfrac{1}{\tan B}$
Now as we know the inverse of $cot\;$ is $tan\;$ in the same way the inverse of $tanB\;$ is $cotB\;$ . Therefore we get
$2\tan (A-B)=\cot B$

So, the correct answer is “Option C”.

Note: Trigonometry functions can be defined as real functions where they relate an angle to a right angle triangle and then they are defined by the ratio of two side lengths and the angle between them.