Question

prove that 1+Tan² A=sec² A
​

Answer 1

in ∆ABC

tanA = BC/AB =>tan²A = BC²/AB² ... (i)

secA = AC/AB =>sec²A = AC²/AB² ... (ii)

using pythagoreas theorem in ∆ABC, we get

AB² + BC² = AC²

dividing throughout by AB², we get

1 + BC²/AB² = AC²/AB²

substituting values of i and ii

1 + tan²A = sec²A

... Hence Proved!!

OR (alternate method)

RHS

= sec²A ... (i)

LHS

= 1 + tan²A

=

$= 1 + \frac{ {sin}^{2}a }{ {cos}^{2}a } \: \: \: \: \: \: \: \: \: ...(tana = \frac{sina}{cosa}) \\ = \frac{ {cos}^{2} a + {sin}^{2}a }{ {cos}^{2}a } \: \: \\ = \frac{1}{ {cos}^{2}a } \:\: \: \: \: \: \: \: ...( {sin}^{2} x + {cos}^{2} x = 1)$

= sec²A ... (ii)

On comparing i and ii

LHS = RHS

hence,

1 + tan²A = sec²A

... Hence Proved!

Answer 2

Answer:

proved

Step-by-step explanation:

cos²A+sin²A = 1

Divide both sides by cos²A

cos²A/cos²A+sin²A/cos²A = 1/cos²A

1+Tan² A=sec² A