Question

how to prove 1 + tan squared theta equal to sec squared theta​

Answer 1

Answer:

1 + Tan²θ = Sec²θ

Step-by-step explanation:

TO PROVE: 1 + Tan²θ = Sec²θ

Tan θ = (Sin θ) / (Cos θ)

∴ Tan²θ = (Sin θ / Cos θ)²  ..... (1)

L.H.S = 1 + Tan²θ

(substituting from equation 1 we get)

L.H.S. = 1 + (sin θ/cos θ)²

∴ L.H.S = 1 + (Sin²θ / Cos²θ)

∴ L.H.S. = (Cos²θ + Sin²θ) / Cos²θ

But, we know that - Cos²θ + Sin²θ = 1

∴ L.H.S. = 1 / Cos²θ

   ∵ 1 / Cosθ = Secθ

∴ L.H.S. = Sec²θ = R.H.S.

as LHS = RHS... hence prooved

1 + Tan²θ = Sec²θ

Answer 2

Answer:

$1+\tan^2\theta=sec^2\theta$

Step-by-step explanation:

Formula used:

In a right angled triangle square on the hypotenuse is equal to sum of the squares on the other two sides.

Let ΔABC be right angled triangle at B.

Let \theta be the one of the acute of ΔABC.

By pythagoras theorem

$AC^2=AB^2+BC^2$

$tan\theta=\frac{opposite\:side}{adjacent\:side}\\\\tan\theta=\frac{AB}{BC}\\\\sec\theta=\frac{hypotenuse}{adjacent\:side}\\\\sec\theta=\frac{AC}{BC}$

$Now,\\\\1+tan^2\theta\\\\=1+(\frac{AB}{BC})^2\\\\=1+\frac{(AB)^2}{(BC)^2}\\\\=\frac{(BC)^2+(AB)^2}{(BC)^2}\\\\=\frac{(AC)^2}{(BC)^2}\\\\=(\frac{AC}{BC})^2\\\\=sec^2\theta$