Question

PROVE THAT:tan^2@-1/cos^2@+1=0.​

Answer 1

Answer:

L.H.S = tan2 θ – (1/cos2 θ) + 1

= tan2 θ - sec2 θ + 1 [since, 1/cos θ = sec θ]  

= tan2 θ – (1 + tan2 θ) +1 [since, sec2 θ = 1 + tan2 θ]  

= tan2 θ – 1 – tan2 θ + 1  

= 0 = R.H.S.

Answer 2

$\bold{L.H.S = tan^2 θ – ( \frac{1}{cos^2 θ}) + 1}$

$\small \bold{= tan {}^{2} θ - sec {}^{2} θ + 1 [since, \frac{1}{cos θ} = sec θ] }$

$\small\bold{= tan {}^{2} θ – (1 + tan {}^{2} θ) +1 [since, sec {}^{2} θ = 1 + tan {}^{2} θ] }$

$\bold{= tan {}^{2} θ – 1 – tan {}^{2} θ + 1 }$

$\bold{= 0 = R.H.S.}$