If Secθ=5 tanθ=2√6 How would I go about finding cot(90°-θ)?
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Originally Posted by PhillyViDat If Secθ=5 tanθ=2√6 How would I go about finding cot(90°-θ)? Hint: $\displaystyle \cot x = \frac 1{\tan x}$ and $\displaystyle \tan (A - B) = \frac {\tan A - \tan B}{1 + \tan A \tan B}$
A known complement identity is that $\displaystyle \tan{\theta} = \cot{\left(\frac{\pi}{2}-\theta\right)}$
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