SOLUTION: plz help verify this equation. thx (tan^2x + 1)*(cos^2x + 1) = tan^2 x + 2

Question 864393: plz help verify this equation. thx
(tan^2x + 1)*(cos^2x + 1) = tan^2 x + 2

Found 2 solutions by hamsanash1981@gmail.com, jim_thompson5910:
Answer by hamsanash1981@gmail.com(151) (Show Source): You can put this solution on YOUR website!
LHS
=(tan^2x + 1)*(cos^2x + 1)
=(sin^2x /cos^2x +1)*(cos^2x +1)
=(sin^2x+cos^2x)/cos^2x*(cos^2+1)
=1/cos^2x*(cos^2x+1)
=cos^2x/cos^2x+1/cos^2x
=1 + (sin^2x+cos^2x)/cos^2x
=1 + sin^2x/cos^2x + cos^2x/cos^2x
=1+ tan^2x + 1
=tan^2x+ 2
= RHS

Answer by jim_thompson5910(35256) (Show Source): You can put this solution on YOUR website!
(tan^2(x) + 1)*(cos^2(x) + 1) = tan^2(x) + 2

tan^2(x)(cos^2(x) + 1) + 1(cos^2(x) + 1) = tan^2(x) + 2

tan^2(x)*cos^2(x) + tan^2(x) + cos^2(x) + 1 = tan^2(x) + 2

[sin^2(x)/cos^2(x)]*cos^2(x) + tan^2(x) + cos^2(x) + 1 = tan^2(x) + 2

sin^2(x) + tan^2(x) + cos^2(x) + 1 = tan^2(x) + 2

(sin^2(x) + cos^2(x)) + tan^2(x) + 1 = tan^2(x) + 2

1 + tan^2(x) + 1 = tan^2(x) + 2

tan^2(x) + 2 = tan^2(x) + 2

So the identity is verified.
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