56th Problem 2016

Geometry Level 2

tan ⁡ θ sin ⁡ θ + cos ⁡ θ = ? \large\tan { \theta \sin { \theta } } + \cos{ \theta =? }

Check out the set: 2016 Problems

sec ⁡ θ \sec\theta csc ⁡ θ \csc\theta sin ⁡ θ \sin\theta cos ⁡ θ \cos\theta

Angela Fajardo by Angela Fajardo , 19, Philippines

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2 solutions

Abhiram Rao
Apr 24, 2016

I really like this set. It's completely based on what we did in our 8th grade. By the way , can others contribute problems to this set?

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Hmmm, I haven't thought of that. Let me think about it :)

The set is based on what I've learned/know. Glad to help :D

Angela Fajardo - 5 years, 1 month ago

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Ya 'cause then there is a chance that there will be 2016 problems in this set.

Abhiram Rao - 5 years, 1 month ago

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Well, it is already 5% complete (LOL). I think you're on Slack; are you?

Angela Fajardo - 5 years, 1 month ago

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@Angela Fajardo – Yeah , I am on Slack.

Abhiram Rao - 5 years, 1 month ago
Munem Shahriar
Mar 28, 2018

tan ⁡ θ ⋅ sin ⁡ θ + cos ⁡ θ = sin ⁡ θ cos ⁡ θ ⋅ sin ⁡ θ + cos ⁡ θ = sin ⁡ 2 θ cos ⁡ θ + cos ⁡ θ = sin ⁡ 2 θ + cos ⁡ 2 θ cos ⁡ θ = 1 cos ⁡ θ [ sin ⁡ 2 θ + cos ⁡ 2 θ = 1 ] = sec ⁡ θ \begin{aligned} \tan \theta \cdot \sin \theta + \cos \theta & = \dfrac{\sin \theta}{\cos \theta} \cdot \sin \theta + \cos \theta \\ & = \dfrac{\sin^2 \theta}{\cos \theta} + \cos \theta \\ & = \dfrac{\sin^2 \theta + \cos^2 \theta}{\cos \theta} \\ & = \dfrac 1{\cos \theta} ~~~~~~~~~~~~~~~~~~~~[\sin^2 \theta + \cos^2 \theta = 1] \\ & = \sec \theta \\ \end{aligned}

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