How do you verify #cos^2x + tan^2x - 1 / cos^2x + sin^2x = 0#?

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Answer 1 · 2018-05-28T04:56:51

Please see below.

We know that,

#color(blue)((1)1/costheta=sectheta#

#color(red)((2)sin^2theta+cos^2theta=1#

#color(red)((3)sec^2theta-tan^2theta=1#

We have to verify :

#cos^2x+tan^2x-color(blue)(1/cos^2x)+sin^2x=0#

We take left hand side :

#LHS=cos^2x+tan^2x-color(blue)(1/cos^2x)+sin^2x...tocolor(blue)(Apply(1)#

#LHS=cos^2x+tan^2x-color(blue)(sec^2x)+sin^2x#

#LHS=cos^2x+sin^2x-sec^2x+tan^2x#

#LHS={color(red)(cos^2x+sin^2x)}-{color(red)(sec^2x-tan^2x)}#

#LHS=1-1...tocolor(red)(Apply(2) and (3)#

#LHS=0#

#LHS=RHS#

Answer 2 · 2018-05-28T04:57:37

#color(green)(=> 0 = R H S#

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

#cos ^2 x + sin ^2 x + tan ^2 x - (1/cos^2 x)#, rearranging terms

#cos^2 x + sin ^2 = 1#, identity

# sec x -= 1/cos x#, identity

#:. cancel(cos^2 x + sin^2 x)^color(red)(1) + tan^2 x - sec^2 x#

#1 + tan^2 x = sec ^2x#, identity

#=> cancel(1 + tan^2 x )- cancel(sec ^2 x )= 0#

#color(green)(=> R H S#