Question

prove that 1/sec x-tan x -1/cos x =1/cos x - 1/sec x +tan x​

Answer 1

Step-by-step explanation:

Multiply both sides by sec(x)-tan(x) so that the equation is now sec^2(x)-tan^2(x)=1

We know that sec(x) is the reciprocal of cos(x). Cos(x), in turn is adjacent/hypotenuse.

Tan(x) is opp/adj.

So square both sec(x) and tan(x) individually so we get (h/a)^2-(o/a)^2=1

From Pythagoras’ theorem we know that c^2=a^2+b^2, and c is interchangeable with h (hypotenuse) and a and b, either the opposite or the adjacent.

This means h^2-o^2=a^2

For obvious reasons a cannot be zero, and any other real number over itself is 1.

Answer 2

see the attachment..........