Trigonometry expression

Geometry Level 1

tan ⁡ θ \tan{\theta}

Which identity is equal to the above expression?

cos ⁡ 2 θ + sin ⁡ 2 θ \text{}{\cos}^{2}{\theta}+{\sin}^{2}{\theta} 1 − cos ⁡ 2 θ \text{}1-{\cos}^{2}{\theta} 1 − sin ⁡ 2 θ \text{}1-{\sin}^{2}{\theta} sin ⁡ θ cos ⁡ θ \frac{\sin{\theta}}{\cos{\theta}} cos ⁡ θ sin ⁡ θ \frac{\cos{\theta}}{\sin{\theta}}

Matin Naseri by Matin Naseri , 17, Afghanistan

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2 solutions

Munem Shahriar
Jul 29, 2018

  • tan ⁡ θ = Opposite side Adjacent side = A B B C \tan \theta = \dfrac{\text{Opposite side}}{\text{Adjacent side}} = \dfrac{AB}{BC}

  • sin ⁡ θ = Opposite side Hypotenuse = A B A C \sin \theta = \dfrac{\text{Opposite side}}{\text{Hypotenuse}} = \dfrac{AB}{AC}

  • cos ⁡ θ = Adjacent side Hypotenuse = B C A C \cos \theta = \dfrac{\text{Adjacent side}}{\text{Hypotenuse}} = \dfrac{BC}{AC}

Now,

tan ⁡ θ = A B B C = A B A C B C A C [ Dividing the numerator and denominator by AC ] = sin ⁡ θ cos ⁡ θ \large \begin{aligned} \tan \theta & = \dfrac{AB}{BC} \\ & = \dfrac{\frac{AB}{AC}}{\frac{BC}{AC}} ~~~~[\text{Dividing the numerator and denominator by AC}] \\ & = \dfrac{\sin \theta}{\cos \theta} \\ \end{aligned}

Proved.

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tan ⁡ θ = sin ⁡ θ cos ⁡ θ \tan \theta =\dfrac{\sin \theta}{\cos \theta} is an identity . . . . . .

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