Addition

Geometry Level 3

If sin ⁡ θ + cos ⁡ θ = 1 2 \sin{\theta}+\cos{\theta}=\frac{1}{2} , then tan ⁡ θ + cot ⁡ θ = − a b \tan{\theta}+\cot{\theta}=-\frac{a}{b} , where a a and b b are coprime positive integers. Find a + b a+b


The answer is 11.

Akeel Howell by Akeel Howell , 22, USA

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1 solution

Akeel Howell
Jan 19, 2017

sin ⁡ θ + cos ⁡ θ = 1 2 ⟹ sin ⁡ 2 θ + c o s 2 θ + 2 sin ⁡ θ cos ⁡ θ = 1 4 \sin{\theta}+\cos{\theta}=\frac{1}{2} \implies \sin^2{\theta}+cos^2{\theta}+2\sin{\theta}\cos{\theta}=\frac{1}{4}

So sin ⁡ θ cos ⁡ θ = − 3 8 \sin{\theta}\cos{\theta}=-\frac{3}{8}

tan ⁡ θ + cot ⁡ θ = sin ⁡ θ cos ⁡ θ + cos ⁡ θ sin ⁡ θ = sin ⁡ 2 θ + cos ⁡ 2 θ sin ⁡ θ cos ⁡ θ = 1 sin ⁡ θ cos ⁡ θ = − 1 ( 3 8 ) = − 8 3 \tan{\theta}+\cot{\theta}=\frac{\sin{\theta}}{\cos{\theta}}+\frac{\cos{\theta}}{\sin{\theta}} \\ =\dfrac{\sin^2{\theta}+\cos^2{\theta}}{\sin{\theta}\cos{\theta}} = \frac{1}{\sin{\theta}\cos{\theta}} \\ =-\dfrac{1}{\left(\frac{3}{8}\right)}=-\frac{8}{3}

So a + b = 8 + 3 = 11 . a+b=8+3=\boxed{11}.

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