Question

How do you find $\tan 2A$, given $\sin A=3/5$ and $A$ is $QII$ ?

Answer 1

In this question we will use the trigonometric identity to simplify the value of $\tan 2A$ In terms of $\sin A$ and $\cos A$, and substitute the values in the identity and simplify to get the required solution and then by using the value of $\sin A=3/5$, we will calculate the value of $\cos A$and then solve for the value of $\tan 2A$.

Complete step by step answer:
We have to find the value of $\tan 2A$ given that the value of $\sin A$ is given.
We know the trigonometric identity that
$\Rightarrow \tan 2A=\dfrac{2\tan A}{1-{{\tan }^{2}}A}\to (1)$
Now we know that $\tan A=\dfrac{\sin A}{\cos A}$
But from the question we have been only given the value of $\sin A$ which is $\dfrac{3}{5}$.
We know the trigonometric identity ${{\sin }^{2}}A+{{\cos }^{2}}A=1$ therefore, ${{\cos }^{2}}A=1-{{\sin }^{2}}A$ which means $\cos A=\sqrt{1-{{\sin }^{2}}A}\to (2)$
On substituting $\sin A=\dfrac{3}{5}$ in equation $(2)$, we get:
$\Rightarrow \cos A=\sqrt{1-{{\left( \dfrac{3}{5} \right)}^{2}}}$
On taking the square of the term, we get:
$\Rightarrow \cos A=\sqrt{1-\dfrac{9}{25}}$
On taking the lowest common multiple, we get:
$\Rightarrow \cos A=\sqrt{\dfrac{25-9}{25}}$
On simplifying, we get:
$\Rightarrow \cos A=\sqrt{\dfrac{16}{25}}$
On taking the square root, we get:
$\Rightarrow \cos A=\pm \dfrac{4}{5}$
Now we know from the question that the angle $A$ lies in the second quadrant and since in the second quadrant $\cos$is negative, we will consider the value of $\cos$as:
$\Rightarrow \cos A=-\dfrac{4}{5}$
Now $\tan A=\dfrac{3/5}{-4/5}$
On rearranging the terms and simplifying the expression, we get:
$\Rightarrow \tan A=-\dfrac{3}{4}$
Now on substituting the value of $\tan A$ in equation $(1)$, we get:
$\Rightarrow \tan 2A=\dfrac{2\left( -\dfrac{3}{4} \right)}{1-{{\left( -\dfrac{3}{4} \right)}^{2}}}$
On simplifying the numerator and taking the square in the denominator, we get:
$\Rightarrow \tan 2A=\dfrac{-\dfrac{3}{2}}{1-\dfrac{9}{16}}$
On taking the lowest common multiple in the denominator, we get:
$\Rightarrow \tan 2A=\dfrac{-\dfrac{3}{2}}{\dfrac{16-9}{16}}$
On simplifying, we get:
$\Rightarrow \tan 2A=\dfrac{-\dfrac{3}{2}}{\dfrac{7}{16}}$
On rearranging the terms in the expression, we get:
$\Rightarrow \tan 2A=-\dfrac{3}{2}\times \dfrac{16}{7}$
On simplifying the term, we get:
$\Rightarrow \tan 2A=-3\times \dfrac{8}{7}$
On multiplying, we get:
$\Rightarrow \tan 2A=-\dfrac{24}{7}$, which is the required solution.

Note:
It is to be remembered that which trigonometric identity is positive and negative in which quadrants, along with that the various trigonometric double angle formulas should be remembered.
While doing any trigonometric question, the question should be converted to $\sin$ and $\cos$, and then simplified further.