Help please to simplify as far as possible? #sin^2t +cos^2t+tan^2t#

Question

603 views around the world

You can reuse this answer
Creative Commons License

Answer 1 · 2018-02-26T19:48:25

It simplifies to #sec^2t#

We know that #sin^2x+ cos^2x= 1#.

#= 1 + tan^2t#

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

#=(cos^2t + sin^2t)/cos^2t#

#= 1/cos^2t#

#= sec^2t#

Hopefully this helps!

Answer 2 · 2018-02-26T19:58:50

#sec^2t#

#color(red)(sin^2t+cos^2t)+tan^2t=color(red)(1)+tan^2t=sec^2t#
Hint:
1. #color(red)(sin^2t+cos^2t=1)#
2. #color(red)(1+tan^2t=sec^2t)#