If is the solution of then the value of is Trigonometric equations

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Solution

Q.
If $α, - \frac{π}{2} < α < \frac{π}{2}$ is the solution of $4 \cos θ + 5 \sin θ = 1,$ then the value of $\tan α$ is [2024]

1 $\frac{\sqrt{10} - 10}{12}$

2 $\frac{10 - \sqrt{10}}{6}$

3 $\frac{\sqrt{10} - 10}{6}$

4 $\frac{10 - \sqrt{10}}{12}$

Ans.
(1)

We have, $4 \cos θ + 5 \sin θ = 1$

$\Rightarrow 4 (\frac{1 - \tan^{2} \frac{θ}{2}}{1 + \tan^{2} \frac{θ}{2}}) + 5 (\frac{2 \tan \frac{θ}{2}}{1 + \tan^{2} \frac{θ}{2}}) = 1$

$\Rightarrow 5 \tan^{2} (θ / 2) - 10 \tan (θ / 2) - 3 = 0$

$\Rightarrow \tan (θ / 2) = \frac{10 \pm \sqrt{100 - 4 (5) (- 3)}}{2 \times 5} = \frac{10 \pm \sqrt{160}}{10} = \frac{5 \pm \sqrt{40}}{5}$

$∵$ $α \in (\frac{- π}{2}, \frac{π}{2})$ So, $\frac{α}{2} \in (\frac{- π}{4}, \frac{π}{4})$

$\Rightarrow \tan (\frac{α}{2}) \in (- 1, 1)$ $∴$ $\tan \frac{α}{2} = \frac{5 - \sqrt{40}}{5}$

$Hence, \tan α = \frac{2 \tan \frac{α}{2}}{1 - \tan^{2} \frac{α}{2}} = \frac{2 (\frac{5 - \sqrt{40}}{5})}{1 - {(1 - \frac{\sqrt{40}}{5})}^{2}}$

$= \frac{2 (\frac{5 - \sqrt{40}}{5})}{1 - (1 + \frac{40}{25} - \frac{2 \sqrt{40}}{5})}$

$= \frac{2 (\frac{5 - \sqrt{40}}{5})}{\frac{2 \sqrt{40}}{5} - \frac{8}{5}} = \frac{5 - \sqrt{40}}{\sqrt{40} - 4} \times \frac{\sqrt{40} + 4}{\sqrt{40} + 4}$

$= \frac{5 \sqrt{40} + 20 - 40 - 4 \sqrt{40}}{40 - 16} = \frac{\sqrt{40} - 20}{24} = \frac{\sqrt{10} - 10}{12}$

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