Question

If $3\cos A = 4\sin A$ find the value of $4{\cos ^2}A - 3{\sin ^2}A + 2$.
A) $3$
B) $3\dfrac{{12}}{{25}}$
C) $2$
D) $\dfrac{{12}}{{25}}$

Answer 1

Recall all the basic trigonometric identities and basic definitions of each ratio. Rearrange the given equation in a way that we can use it to our advantage. Transform everything to its basic form and then use given identity to simplify the problem. Make the appropriate substitutions to get to the final answer.

Complete step-by-step answer:
It is given that $3\cos A = 4\sin A$ .
Therefore, rearranging the terms we can transform the given ratio as follow:
$\dfrac{3}{4} = \dfrac{{\sin A}}{{\cos A}}$
We know that $\dfrac{{\sin A}}{{\cos A}} = \tan A$ .
$\tan A = \dfrac{3}{4}$
Squaring both sides, we get,
${\tan ^2}A = \dfrac{9}{{16}}$
We know that $1 + {\tan ^2}A = {\sec ^2}A$ .
Therefore, using the obtained value of ${\tan ^2}A$ we can write:
$1 + \dfrac{9}{{16}} = {\sec ^2}A$
Therefore,
${\sec ^2}A = \dfrac{{25}}{{16}}$
Now we know that ${\cos ^2}A = \dfrac{1}{{{{\sec }^2}A}}$.
Using the above obtained value of ${\sec ^2}A$ we can write:
${\cos ^2}A = \dfrac{{16}}{{25}}$ … (1)
The basic trigonometric identity states that ${\sin ^2}A + {\cos ^2}A = 1$ .
Therefore, from equation (1) we write:
${\sin ^2}A + \dfrac{{16}}{{25}} = 1$
Rearranging the terms, we write,
${\sin ^2}A = \dfrac{9}{{25}}$ … (2)
We need to find the value of $4{\cos ^2}A - 3{\sin ^2}A + 2$.
From equations (1) and (2) we get,
$4\left( {\dfrac{{16}}{{25}}} \right) - 3\left( {\dfrac{9}{{25}}} \right) + 2 = \dfrac{{64}}{{25}} - \dfrac{{27}}{{25}} + 2$
Simplify the right-hand side and write the following:
$4{\cos ^2}A - 3{\sin ^2}A + 2 = \dfrac{{87}}{{25}}$
The right-hand side can also be written as $3\dfrac{{12}}{{25}}$ .

So, the correct answer is “Option B”.

Note: The problem involves many basic trigonometric transformations so be careful about the signs and basic identities. It is always simple to solve any problem if we transform it to the basic trigonometric identities. Also, there is no need to find the value of the exact ratio as the problem involves all the square terms only.