Tangent , sine and cosine!

Level 2

If sin ⁡ θ + cos ⁡ θ = 1 5 \sin \theta + \cos \theta = \dfrac{1}{5} and θ ≠ π 2 \theta \neq \dfrac{π}{2} , what is tan ⁡ θ \tan \theta ?

− 7 2 -\dfrac{7}{2} or − 2 7 -\dfrac{2}{7} − 13 12 -\dfrac{13}{12} or − 12 13 -\dfrac{12}{13} − 3 4 -\dfrac{3}{4} or − 4 3 -\dfrac{4}{3} − 4 5 -\dfrac{4}{5} or − 5 4 -\dfrac{5}{4}

Prem Chebrolu by Prem Chebrolu , 17, India

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1 solution

Chew-Seong Cheong
Oct 22, 2018

sin ⁡ θ + cos ⁡ θ = 1 5 Divide both sides by cos ⁡ θ . tan ⁡ θ + 1 = sec ⁡ θ 5 Multiply both sides by 5. 5 tan ⁡ θ + 5 = sec ⁡ θ Squaring both sides 25 tan ⁡ 2 θ + 50 tan ⁡ θ + 25 = sec ⁡ 2 θ 25 tan ⁡ 2 θ + 50 tan ⁡ θ + 25 = tan ⁡ 2 θ + 1 24 tan ⁡ 2 θ + 50 tan ⁡ θ + 24 = 0 12 tan ⁡ 2 θ + 25 tan ⁡ θ + 12 = 0 ( 3 tan ⁡ θ + 4 ) ( 4 tan ⁡ θ + 3 ) = 0 \begin{aligned} \sin \theta + \cos \theta & = \frac 15 & \small \color{#3D99F6} \text{Divide both sides by }\cos \theta. \\ \tan \theta + 1 & = \frac {\sec \theta}5 & \small \color{#3D99F6} \text{Multiply both sides by }5. \\ 5\tan \theta + 5 & = \sec \theta & \small \color{#3D99F6} \text{Squaring both sides} \\ 25\tan^2 \theta + 50\tan \theta + 25 & = \sec^2 \theta \\ 25\tan^2 \theta + 50\tan \theta + 25 & = \tan^2 \theta + 1 \\ 24\tan^2 \theta + 50\tan \theta + 24 & = 0 \\ 12\tan^2 \theta + 25\tan \theta + 12 & = 0 \\ (3\tan \theta + 4)(4\tan \theta +3) & = 0 \end{aligned}

⟹ θ = − 3 4 or − 4 3 \implies \theta = \boxed{- \dfrac 34 \text{ or } - \dfrac 43}

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