Given that #cos x =p# find an expression, in terms of #p#, for #tan^2 x#?

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Answer 1 · 2015-10-12T14:38:17

#tan^2x= (1-p^2)/p^2#

If cos x =p, #sinx= sqrt(1-p^2)#

tan x = #sqrt(1-p^2)/p #

#tan^2x= (1-p^2)/p^2#

Answer 2 · 2015-10-12T21:35:47

#tan^2 x = (1-p^2)/p^2 = 1/p^2 - 1#

If #cos x = p#, then

#cos^2 x = p^2#.

By Pythagoras, #sin^2 x + cos^2 x = 1#, so:

#sin^2 x = 1 - cos^2 x = 1 - p^2#

Now #tan x = sin x / cos x#, so

#tan^2 x = sin^2 x / cos^2 x = (1 - p^2)/p^2 = 1/p^2 - 1#