Given a2+b2+2abcosθ=1,c2+d2+2cdcosθ=1a^{2} + b^{2} + 2ab \cos \theta = 1 , c^{2} + d^{2} + 2cd \cos \theta = 1a2+b2+2abcosθ=1,c2+d2+2cdcosθ=1 and (ac+bd)+(ad+bc)cosθ=0(ac + bd) + ( ad + bc) \cos \theta = 0(ac+bd)+(ad+bc)cosθ=0
Then prove that a2+c2=cosec2θa^{2} + c^{2} = cosec^{2} \thetaa2+c2=cosec2θ
Im struggling to prove this for a while.... Can anyone please post their proofs in the comments section? Thanks!
Note by Ojasee Duble 3 years, 1 month ago
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Easy Math Editor
This discussion board is a place to discuss our Daily Challenges and the math and science related to those challenges. Explanations are more than just a solution — they should explain the steps and thinking strategies that you used to obtain the solution. Comments should further the discussion of math and science.
When posting on Brilliant:
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to ensure proper formatting.2 \times 3
2^{34}
a_{i-1}
\frac{2}{3}
\sqrt{2}
\sum_{i=1}^3
\sin \theta
\boxed{123}
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