Hi there! Can anyone help solve this? :) Given tan theta = p and that theta is an acute angle, find sec 2 theta and cot (90-2theta)

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Answer 1 · 2018-05-01T04:22:19

#sec theta = 1/cos theta = \sqrt{p^2 + 1 }#

# cot (90^circ - 2 theta) = {2p}/{1-p^2}#

#tan theta = p = p/1 #

That's a right triangle with opposite #p# and adjacent #1#, so hypotenuse #\sqrt{p^2+1}.#

So #cos theta = 1/{pm \sqrt{p^2+1} }#

We choose the positive square root because we're told #theta# is acute.

#sec theta = 1/cos theta = \sqrt{p^2 + 1 }#

# cot (90^circ - 2 theta) #

# = tan(2 theta ) #

# = {2 tan theta}/{1 - tan^2 theta } #

# = {2p}/{1-p^2}#