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Q: What is whole under root of sec theta - 1 over sec theta 1?

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The derivative of (sin (theta))^.5 is (cos(theta))/(2sin(theta))

The only real solution is theta = 0For theta < 0 square root of 3 theta is not defined.For theta > 0, sin theta increases slower than 3*theta and so the sum is always negative.The only real solution is theta = 0For theta < 0 square root of 3 theta is not defined.For theta > 0, sin theta increases slower than 3*theta and so the sum is always negative.The only real solution is theta = 0For theta < 0 square root of 3 theta is not defined.For theta > 0, sin theta increases slower than 3*theta and so the sum is always negative.The only real solution is theta = 0For theta < 0 square root of 3 theta is not defined.For theta > 0, sin theta increases slower than 3*theta and so the sum is always negative.

sine[theta]=opposite/hypotenuse=square root of (1-[cos[theta]]^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.

No, it's not. The cube root of 16 is just under 2.52

The four roots are cos(theta)+i*sin(theta) where theta = pi/4, 3*pi/4, 5*pi/4 and 7*pi/4.

Nope - it's a fraction over 7.

under root 2, under root 3, under root 5,under root 6.

The square root of 16 is 4 which is a whole number.

The square root of 26 is not a whole number

formula of under root

The square root of 15 is not a whole number.

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