Question

(a + 2) sin $\alpha$ + (2a - 1) cos $\alpha$ = (2a + 1) if tan $\alpha$ =

• A. $\frac{3}{4}$
• B. $\frac{4}{3}$
• C. $\frac{2a}{a^2+1}$
• D. $\frac{2a}{a^2-1}$

Answer 1

Answer: (B)

$\frac{4}{3}$

Answer 2

The given equation is $(a + 2) \sin \alpha + (2a - 1) \cos \alpha = (2a + 1)$.

We can solve this equation by using the tangent half-angle substitution: Let $t = \tan(\alpha/2)$. Then $\sin \alpha = \frac{2t}{1 + t^2}$ and $\cos \alpha = \frac{1 - t^2}{1 + t^2}$.

Substitute these into the given equation: $(a + 2) \left(\frac{2t}{1 + t^2}\right) + (2a - 1) \left(\frac{1 - t^2}{1 + t^2}\right) = (2a + 1)$

Multiply both sides by $(1 + t^2)$ to clear the denominators: $2t(a + 2) + (2a - 1)(1 - t^2) = (2a + 1)(1 + t^2)$

Expand the terms: $2at + 4t + 2a - 2at^2 - 1 + t^2 = 2a + 2at^2 + 1 + t^2$

Rearrange the terms to form a quadratic equation in $t$: Collect $t^2$ terms: $-2at^2 + t^2 - 2at^2 - t^2 = -4at^2$ Collect $t$ terms: $2at + 4t = (2a + 4)t$ Collect constant terms: $2a - 1 - 2a - 1 = -2$

So the equation becomes: $-4at^2 + (2a + 4)t - 2 = 0$ Divide by -2: $2at^2 - (a + 2)t + 1 = 0$

This is a quadratic equation in $t$. We can solve for $t$ using the quadratic formula $t = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}$: $t = \frac{-(-(a+2)) \pm \sqrt{(-(a+2))^2 - 4(2a)(1)}}{2(2a)}$ $t = \frac{a+2 \pm \sqrt{(a+2)^2 - 8a}}{4a}$ $t = \frac{a+2 \pm \sqrt{a^2 + 4a + 4 - 8a}}{4a}$ $t = \frac{a+2 \pm \sqrt{a^2 - 4a + 4}}{4a}$ $t = \frac{a+2 \pm \sqrt{(a-2)^2}}{4a}$ $t = \frac{a+2 \pm |a-2|}{4a}$

This gives two possible values for $t$:

1. $t_1 = \frac{a+2 + (a-2)}{4a}$ (assuming $a-2 \ge 0$, i.e., $a \ge 2$) $t_1 = \frac{2a}{4a} = \frac{1}{2}$
2. $t_2 = \frac{a+2 - (a-2)}{4a}$ (assuming $a-2 \ge 0$, i.e., $a \ge 2$) $t_2 = \frac{4}{4a} = \frac{1}{a}$

If $a-2 < 0$, i.e., $a < 2$, then $|a-2| = -(a-2) = 2-a$.

1. $t_1 = \frac{a+2 + (2-a)}{4a} = \frac{4}{4a} = \frac{1}{a}$
2. $t_2 = \frac{a+2 - (2-a)}{4a} = \frac{a+2 - 2 + a}{4a} = \frac{2a}{4a} = \frac{1}{2}$

So, the two possible values for $t = \tan(\alpha/2)$ are $\frac{1}{2}$ and $\frac{1}{a}$.

Now we find $\tan \alpha$ using the formula $\tan \alpha = \frac{2t}{1 - t^2}$:

Case 1: If $t = \frac{1}{2}$ $\tan \alpha = \frac{2(1/2)}{1 - (1/2)^2} = \frac{1}{1 - 1/4} = \frac{1}{3/4} = \frac{4}{3}$

Case 2: If $t = \frac{1}{a}$ $\tan \alpha = \frac{2(1/a)}{1 - (1/a)^2} = \frac{2/a}{1 - 1/a^2} = \frac{2/a}{\frac{a^2 - 1}{a^2}} = \frac{2}{a} \cdot \frac{a^2}{a^2 - 1} = \frac{2a}{a^2 - 1}$

We have two potential solutions for $\tan \alpha$: $\frac{4}{3}$ and $\frac{2a}{a^2 - 1}$. Let's check which of these solutions are valid by substituting them back into the original equation.

For $\tan \alpha = \frac{4}{3}$: If $\tan \alpha = \frac{4}{3}$, then we can have $\sin \alpha = \frac{4}{5}, \cos \alpha = \frac{3}{5}$ (for $\alpha$ in Q1) or $\sin \alpha = -\frac{4}{5}, \cos \alpha = -\frac{3}{5}$ (for $\alpha$ in Q3).

Substitute $\sin \alpha = \frac{4}{5}$ and $\cos \alpha = \frac{3}{5}$ into the original equation: $(a+2)\left(\frac{4}{5}\right) + (2a-1)\left(\frac{3}{5}\right) = 2a+1$ $\frac{4a+8}{5} + \frac{6a-3}{5} = 2a+1$ $\frac{10a+5}{5} = 2a+1$ $2a+1 = 2a+1$ This is true for all values of $a$. Thus, $\tan \alpha = \frac{4}{3}$ is always a valid solution.

For $\tan \alpha = \frac{2a}{a^2 - 1}$: This value is option (D). The value $\frac{4}{3}$ is option (B). Since this is a single choice question, and both solutions are mathematically derived and valid, there might be a specific condition on $a$ that makes them equivalent, or one is preferred. If the two solutions are equal: $\frac{4}{3} = \frac{2a}{a^2 - 1}$ $4(a^2 - 1) = 6a$ $4a^2 - 4 = 6a$ $4a^2 - 6a - 4 = 0$ $2a^2 - 3a - 2 = 0$ Factoring the quadratic: $(2a+1)(a-2) = 0$ This implies $a = 2$ or $a = -\frac{1}{2}$. If $a=2$, then $\tan \alpha = \frac{2(2)}{2^2-1} = \frac{4}{3}$. So both solutions are identical. If $a=-1/2$, then $\tan \alpha = \frac{2(-1/2)}{(-1/2)^2-1} = \frac{-1}{1/4-1} = \frac{-1}{-3/4} = \frac{4}{3}$. So both solutions are identical.

Since the question asks for "if tan $\alpha$ =", it implies a unique value or a value that is always true. As $\tan \alpha = \frac{4}{3}$ is always a solution regardless of $a$ (as long as $a$ is such that the original expression is defined, which it is for any real $a$), it is the most general answer. The option $\frac{2a}{a^2-1}$ depends on $a$. In multiple choice questions, if a constant solution is derived that holds true for all values of the parameter, it is usually the intended answer unless specified otherwise.