Question

Prove that tan²θ - sec²θ = 1.

Answer 1

Answer:

1 + tan²θ =

1 + (sin²θ/cos²θ) =

(cos²θ/cos²θ) + (sin²θ/cos²θ) =

(cos²θ + sin²θ) / cos²θ =

1/cos²θ =

sec²θ

sec²θ-tan²θ = 1

Hope it helps you

Answer 2

let triangle. abc,where ab is perpendicular,bc is base,acis hypotenuse.

Step-by-step explanation:

by Pythagoras theoram,

ab^2+bc^2=ac^2

dividing both sides by ac^2;

ab^2/ac^2+bc^2/ac^2=ac^2/ac^2

sin^2+cos^2=1

dividing both sides by cos^2;

sin^2/cos^2+cos^2/cos^2=1/cos^2

tan^2+1=sec^2

tan^2-sec^2=1;

hence proved,,,,