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It depends if 1 plus tan theta is divided or multiplied by 1 minus tan theta.

Q: How can you verify 1 plus tan theta divided by 1 minus tan theta equals cot theta plus 1 divided by cot theta minus 1?

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- cos theta

sin(theta) = 15/17, cosec(theta) = 17/15 cos(theta) = -8/17, sec(theta) = -17/8 cotan(theta) = -8/15 theta = 2.0608 radians.

Since sin(theta) = 1/cosec(theta) the first two terms simply camcel out and you are left with 1 divided by tan(theta), which is cot(theta).

Tan^2

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.

Related questions

Using x instead of theta, cos2x/cosec2x + cos4x = cos2x*sin2x + cos4x = cos2x*(sin2x + cos2x) = cos2x*1 = cos2x

It also equals 13 12.

It is 1.

I'm asuming you meant "ten theta"the square of 5 is 2525 divided by 10 is 2.5so theta equals 2.5there you go =)

Zero. Anything minus itself is zero.

- cos theta

cosine (90- theta) = sine (theta)

sin(theta) = 15/17, cosec(theta) = 17/15 cos(theta) = -8/17, sec(theta) = -17/8 cotan(theta) = -8/15 theta = 2.0608 radians.

-2(cot2theta)

Yes. (Theta in radians, and then approximately, not exactly.)

sin (theta) = [13* sin (32o)]/8 = 13*0.529919264/8 = 0.861118804 [theta] = sin-1 (0.861118804) [theta] = 59.44o

theta = arcsin(0.0138) is the principal value.