Question

1+tan^2a=sec^2a proved that

Answer 1

Heya !!!

from LHS

1 + tan²A

but

tan²A = sin²A/cos²A

•°•

1 + sin²A / cos²A

or, cos²A + sin²A / cos²A

As we know that , sin²A + cos²A = 1

since, 1 / cos²A = sec²A

LHS = RHS prooved ♻

______________________

@Mr. Rajukumar111

Hope it helps you ;!! ☺
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Answer 2

Hi,
Here is the answer to your query:-
In a right angled triangle ABC, right angled at B,
We have:

AB^2 + BC^2 = AC^2--------(1) [BY PYTHAGORAS THEOREM]

Dividing each term of(1) by AB^2,we get

AB^2/AB^2 + BC^2/AB^2 = AC^2/AB^2

or (AB/AB)^2 + (BC/AB)^2 = (AC/AB)2

or 1 + (Perpendicular/Base)^2 = (Hypotenuse/Base)^2

or 1 + Tan^2a = Sec^2a

Hence Proved.

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