Let 'theta' = A [as 'A' is easier to type]
sec A - 1/(sec A)
= 1/(cos A) - cos A
= (1 - cos^2 A)/(cos A)
= (sin^2 A)/(cos A)
= (tan A)*(sin A)
Then you can swap back the 'A' with theta
Ut is equual to tan(theta) / (sec(theta) + 1)
zero
For such simplifications, it is usually convenient to convert any trigonometric function that is not sine or cosine, into sine or cosine. In this case, you have: sin theta / sec theta = sin theta / (1/cos theta) = sin theta cos theta.
No.
By converting cosecants and secants to the equivalent sine and cosine functions. For example, csc theta is the same as 1 / sin thetha.
Tan^2
It also equals 13 12.
csc[]tan[] = sec[]. L: Change csc[] into one over sin[]. Change tan[] into sin[] over cos[]. R: Change sec[] into one over cos[]. 1/sin[] times sin[]/cos[] = 1/cos[]. L: To multiply 2 fractions, multiply the numerators, and multiply the denominators, and put the numerators' product over the denominators' product. R: Nothing more to do. sin[]/sin[]cos[] = 1/cos[]. L: You have a sin[] on both top and bottom. Cross them off to get a one on the top. 1/cos[] = 1/cos[]. Done. [] is theta. L is the left side of the equation. R is the right side.
If tan theta equals 2, then the sides of the triangle could be -2, -1, and square root of 5 (I used the Pythagorean Theorem to get this). From this, sec theta is negative square root of 5. It is negative because theta is in the third quadrant, where cosine, secant, sine, and cosecant are all negative.
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1 - sin2(q) = cos2(q)dividing through by cos2(q),sec2(q) - tan2(q) = 1
There is none because it is not true.