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Geometry Level 2

sin ⁡ ( 18 0 ∘ + θ ) cos ⁡ ( 27 0 ∘ − θ ) cot ⁡ ( θ − 36 0 ∘ ) cos ⁡ ( θ − 9 0 ∘ ) sin ⁡ ( 36 0 ∘ − θ ) tan ⁡ ( 27 0 ∘ + θ ) \frac { \sin\left( 180^\circ+\theta \right) \cos(270^\circ-\theta )\cot\left( \theta -360^\circ \right) }{\cos\left( \theta -90^\circ \right) \sin(360^\circ-\theta )\tan(270^\circ+\theta ) } Find the value of the expression above, where θ ≠ 18 0 ∘ k \theta \neq 180^\circ k for k ∈ Z k \in \mathbb{Z} .


The answer is 1.

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Swapnil Das by Swapnil Das , 20, India

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5 solutions

Chew-Seong Cheong
Aug 18, 2015

sin ⁡ ( 18 0 ∘ + θ ) cos ⁡ ( 27 0 ∘ − θ ) cot ⁡ ( θ − 36 0 ∘ ) cos ⁡ ( θ − 9 0 ∘ ) sin ⁡ ( 36 0 ∘ − θ ) tan ⁡ ( 27 0 ∘ + θ ) = ( − sin ⁡ θ ) ( − sin ⁡ θ ) cot ⁡ θ sin ⁡ θ ( − sin ⁡ θ ) ( − cot ⁡ θ ) = 1 \begin{aligned} \dfrac{\sin{(180^\circ+\theta)} \cos{(270^\circ - \theta)} \cot{(\theta - 360^\circ)}} {\cos{(\theta - 90^\circ)} \sin{(360^\circ - \theta)} \tan{(270^\circ + \theta)}} & = \dfrac{(-\sin{\theta} )(- \sin{\theta}) \cot{\theta}} {\sin{\theta} (- \sin{\theta})( -\cot{\theta})} \\ & = \boxed{1} \end{aligned}

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The denominator of the question contains cos(theta-90 degrees) but the first step of your solution contains cos(theta-180 degrees). However, you have arrived at the right answer.

Devendra Kumar Singh - 5 years, 8 months ago

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Thanks. It should be cos ⁡ ( θ − 9 0 ∘ ) = cos ⁡ ( 9 0 ∘ − θ ) = sin ⁡ θ \cos (\theta-90^\circ) = \cos(90^\circ - \theta) = \sin \theta

Chew-Seong Cheong - 5 years, 8 months ago

Doesn't θ = 9 0 ∘ \theta=90^{\circ} make cot ⁡ ( − 27 0 ∘ ) = 0 \cot (-270^{\circ})=0 and thereby making the whole expression 0 0 ?

Rishik Jain - 5 years, 3 months ago

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When θ = 9 0 ∘ \theta = 90^\circ , tan ⁡ ( 27 0 ∘ + θ ) \tan(270^\circ + \theta) in the denominator is also 0 0 cancelling it out. Note that cot ⁡ ( θ − 36 0 ∘ ) tan ⁡ ( 27 0 ∘ + θ ) = cot ⁡ θ − cot ⁡ θ = − 1 \dfrac{\cot (\theta - 360^\circ)}{\tan (270^\circ + \theta)} = \dfrac{\cot \theta}{-\cot \theta} = - 1

Chew-Seong Cheong - 5 years, 3 months ago

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Shouldn't cot ⁡ θ − cot ⁡ θ \dfrac{\cot \theta}{-\cot \theta} only cancel when θ ≠ 9 0 ∘ , 27 0 ∘ \theta \ne 90^{\circ},270^{\circ}

Rishik Jain - 5 years, 3 months ago

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@Rishik Jain – Nope, it is for all θ \theta . cot ⁡ θ = x ⇒ − cot ⁡ θ = − x \cot \theta = x\quad \Rightarrow -\cot \theta = -x . cot ⁡ θ − cot ⁡ θ = x − x = − 1 \dfrac{\cot \theta}{-\cot \theta} = \dfrac{x}{-x} = -1

Chew-Seong Cheong - 5 years, 3 months ago

Shouldn't there be restrictions for the value of θ \theta ?

Eric Jan Escober - 5 years, 9 months ago

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Yes, θ ≠ 180 k ∘ \theta \ne 180k^\circ for k ∈ Z k \in Z

Chew-Seong Cheong - 5 years, 9 months ago

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Generally θ ≠ 360 k ∘ \theta \neq 360k^\circ for k ∈ Z k \in Z

Eric Jan Escober - 5 years, 9 months ago

Please elaborate a little...

Ritesh G - 5 years, 9 months ago
Tushar Kaushik
Sep 9, 2015

In any kind of problem which appears long the answer is always 0 or1

Everyone read the problem and solved it I gave the correct answer even without readingthe question

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Terry Pitman
Aug 22, 2015

I am stuped

2 Helpful 0 Interesting 0 Brilliant 0 Confused

We all know that No need to mention

tushar kaushik - 5 years, 9 months ago
Gaurav Tripathi
Aug 20, 2015

Nice solution

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Pratyush Sahoo
Aug 27, 2015

take out all theta and then try .

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