How would you solve this problum?

Question

sin(squard)x divided by cos(squard)=sec(squard)x-1. This is Trig Identities and i do not understand them at all. Can you please help me?

Answer 1

Go back to your trig function definitions. Suppose you have a right triangle with sides of length a and b, and hypotenuse of length c. Let x be the angle between side b and hypotenuse c. Then, using the symbol ^ to denote an exponent,

sin (x) = a/c
cos (x) = b/c
sec (x) = 1/cos (x) = c/b

The squares are:

sin^2 (x) = a^2 / c^2
cos^2 (x) = b^2 / c^2
sec^2 (x) = c^2 / b^2

Plugging these definitions into your problem,

(a^2 / c^2) / (b^2 / c^2) = (c^2 / b^2) - 1

Simplifying,

a^2 / b^2 = c^2 / b^2 - 1

Multiplying by b^2,

a^2 = c^2 - b^2

Rearranging terms,

a^2 + b^2 = c^2

which may be recognized as the Pythagorean Theorem.

Answer 2

one million. 3/4 of 40 4 Convert 3/4 to decimal. = 3/4 (Divide 3 via 4) = 0.seventy 5 Multiply this with 40 4 to get the ultimate answer. = 0.seventy 5 * 40 4 = 33 (answer) comparable is going for something of the numbers. 2. 2/5 of 20 answer: = 2/5 = 0.4 = 0.4 x 10 = 4 (answer) 3. 3/5 of 20 answer: = 3/5 = 0.6 = 0.6 x 20 = 12 (answer) desire this facilitates.

Answer 3

Ok.. so the way you wrote gave the idea that is sec²(x-1) anyway..

sin²x / cos²x = sec²x - 1

as sec x = 1/cos x and sin²x + cos²x = 1 (sin²x = 1 - cos²x)
substitute and you will have:

1/cos²x - cos²x/cos²x = 1/cos²x - 1


1/cos²x - 1 = 1/cos²x - 1 indeed