Trigonometry problem #5

Geometry Level 1

True or false:

1 + cos ⁡ θ 1 − cos ⁡ θ = cot ⁡ θ + csc ⁡ θ \large \sqrt{\frac{1+\cos \theta }{1-\cos \theta }} = \cot \theta + \csc \theta

Note: 0 ∘ < θ < 180 ∘ 0^\circ < \theta < {180}^\circ

False True

Munem Shahriar by Munem Shahriar , 18, Bangladesh

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1 solution

Munem Shahriar
Mar 25, 2018

1 + cos ⁡ θ 1 − cos ⁡ θ = ( 1 + cos ⁡ θ ) ( 1 + cos ⁡ θ ) ( 1 − cos ⁡ θ ) ( 1 + cos ⁡ θ ) = ( 1 + cos ⁡ θ ) 2 1 − cos ⁡ 2 θ = 1 + cos ⁡ θ sin ⁡ 2 θ [ 1 − cos ⁡ 2 θ = sin ⁡ 2 θ ] = 1 + cos ⁡ θ sin ⁡ θ = 1 sin ⁡ θ + cos ⁡ θ sin ⁡ θ = csc ⁡ θ + cot ⁡ θ \large \begin{aligned} \sqrt{\frac{1+\cos \theta }{1-\cos \theta }} & = \sqrt{\frac{(1+\cos \theta )(1+\cos \theta )}{(1-\cos \theta )(1+\cos \theta)}} \\ & = \frac{\sqrt{(1+\cos \theta)^2}}{\sqrt{1 - \cos^2 \theta}} \\ &= \frac{1+ \cos \theta}{\sqrt{\sin^2 \theta}}~~~~~~~~[1 - \cos^2 \theta = \sin^2 \theta] \\ & = \frac{1 + \cos \theta}{\sin \theta} \\ & = \frac1{\sin \theta} + \frac{\cos \theta}{\sin \theta} \\ & = \csc \theta + \cot \theta \\ \end{aligned}

The answer is true.

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