First slide

Trigonometric Identities

Question

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

Moderate

A

$\frac{3}{4}$
B

$\frac{4}{3}$
C

$\frac{2 a}{(a^{2} + 1)}$
D

$\frac{2 a}{(a^{2} - 1)}$

Solution

Divide by cos a and square both sides and let $\tan α = t$ so that $\sec^{2} α = 1 + t^{2}$
$\begin{array}{ll} \Rightarrow [(a + 2) t + (2 a - 1)]^{2} = [(2 a + 1)^{2} (1 + t^{2})] \\ or t^{2} [(a + 2)^{2} - (2 a + 1)^{2}] + 2 (a + 2) (2 a - 1) t + [(2 a - 1)^{2} - (2 a + 1)^{2}] = 0 \\ or 3 (1 - a^{2}) t^{2} + 2 (2 a^{2} + 3 a - 2) t - 4 \times 2 a = 0 \\ or 3 (1 - a^{2}) t^{2} - 4 (1 - a^{2}) t + 6 at - 8 a = 0 \\ or t (1 - a^{2}) (3 t - 4) + 2 a (3 t - 4) = 0 \\ or (3 t - 4) [(1 - a^{2}) t + 2 a] = 0 \\ or t = \tan α = \frac{4}{3} or \frac{2 a}{a^{2} - 1} \end{array}$

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