How do you prove # sec^2(csc^2) = sec^2 + csc^2#?

2 Answers
Bdub
May 4, 2016

see below

Explanation:

Right Side: #=sec^2x+csc^2x#

#=1/cos^2x + 1/sin^2 x#

#=(sin^2x+cos^2x)/(cos^2xsin^2x)#

#=1/(cos^2xsin^2x)#

#=1/cos^2x * 1/sin^2x#

#=sec^2xcsc^2x#

#=# Left Side

Truong-Son N.
May 4, 2016

Note the following identity:

#csc^2theta = 1 + cot^2theta#

So, let's see how that works out.

#\mathbf(sec^2theta(csc^2theta) = sec^2theta + csc^2theta)#

#sec^2theta(1 + cot^2theta) = sec^2theta + csc^2theta#

#sec^2theta + sec^2thetacot^2theta = sec^2theta + csc^2theta#

Lastly, use the identities

  • #cot^2theta = cos^2theta/sin^2theta,#
  • #1/sin^2theta = csc^2theta,#
  • #1/cos^2theta = sec^2theta,#

to get:

#sec^2theta + sec^2theta(cos^2theta/sin^2theta) = sec^2theta + csc^2theta#

#sec^2theta + 1/cancel(cos^2theta)(cancel(cos^2theta)/sin^2theta) = sec^2theta + csc^2theta#

#sec^2theta + 1/sin^2theta = sec^2theta + csc^2theta#

#color(blue)(sec^2theta + csc^2theta = sec^2theta + csc^2theta)#

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