We know that sin θ = cos (90° - θ) and cos θ = sin (90° - θ) Therefore, sin 67° + cos75° = cos (90°- 67°) + sin (90° - 75°) = cos 23° + sin 15° Explanation video🙌😀
We know that sin θ = cos (90° – θ) and cos θ = sin (90° – θ)
Therefore, sin 67° + cos75° = cos (90°- 67°) + sin (90° – 75°)
= cos 23° + sin 15°
To verify that a given number is a zero(root) of a cubic polynomial, we can substitute the number into the polynomial and see if the result is zero. However, also the number is a zero of the polynomial, If the result is zero. See here the Solution 👀
To verify that a given number is a zero(root) of a cubic polynomial, we can substitute the number into the polynomial and see if the result is zero. However, also the number is a zero of the polynomial, If the result is zero.
Cubic polynomials can have one, two, three, or no real zeros (roots), depending on the values of the constants a, b, c, and d. For example, the cubic polynomial x³ + 2x² - 4x + 4 has one real zero (2), while the cubic polynomial x³ - 2x² + x - 1 has no real zeros. See here for Solution👇
Cubic polynomials can have one, two, three, or no real zeros (roots), depending on the values of the constants a, b, c, and d. For example, the cubic polynomial x³ + 2x² – 4x + 4 has one real zero (2), while the cubic polynomial x³ – 2x² + x – 1 has no real zeros.
Express sin 67° + cos 75° in terms of trigonometric ratios of angles between 0° and 45°.
We know that sin θ = cos (90° - θ) and cos θ = sin (90° - θ) Therefore, sin 67° + cos75° = cos (90°- 67°) + sin (90° - 75°) = cos 23° + sin 15° Explanation video🙌😀
We know that sin θ = cos (90° – θ) and cos θ = sin (90° – θ)
Therefore, sin 67° + cos75° = cos (90°- 67°) + sin (90° – 75°)
= cos 23° + sin 15°
Explanation video🙌😀
See lessIf tan A = cot B, prove that A + B = 90°.
Given that: tan A = cot B ⇒ cot (90° - A) = cot B [∵ cot (90° - θ) = tan θ] ⇒ 90° - A = B ⇒ 90° = A + B Hence, A + B = 90° Video explanation 👇
Given that: tan A = cot B
⇒ cot (90° – A) = cot B [∵ cot (90° – θ) = tan θ]
⇒ 90° – A = B ⇒ 90° = A + B
Hence, A + B = 90°
Video explanation 👇
See lessIf sec 4A = cosec (A – 20°), where 4A is an acute angle, find the value of A.
Given that: sec 4A = cosec(A - 20°) ⇒ cosec(90° - 4A) = cosec(A - 20°) [∵ cosec(90°- θ) = sec θ] ⇒ 90°-4A = A - 20° ⇒ 90°+20° = 5A ⇒ 5A = 110° A = 22° Hence, A = 22° See this for better understanding😮👇
Given that: sec 4A = cosec(A – 20°)
⇒ cosec(90° – 4A) = cosec(A – 20°) [∵ cosec(90°- θ) = sec θ]
⇒ 90°-4A = A – 20°
⇒ 90°+20° = 5A ⇒ 5A = 110° A = 22°
Hence, A = 22°
See this for better understanding😮👇
See lessVerify that the numbers given alongside of the cubic polynomials below are their zeroes. Also verify the relationship between the zeroes and the coefficients in each case: 2x³ + x² – 5x + 2; ½,1, -2
To verify that a given number is a zero(root) of a cubic polynomial, we can substitute the number into the polynomial and see if the result is zero. However, also the number is a zero of the polynomial, If the result is zero. See here the Solution 👀
To verify that a given number is a zero(root) of a cubic polynomial, we can substitute the number into the polynomial and see if the result is zero. However, also the number is a zero of the polynomial, If the result is zero.
See here the Solution 👀
See lessVerify that the numbers given alongside of the cubic polynomial below are their zeroes. Also verify the relationship between the zeroes and the coefficients in each case: x³ – 4x² + 5x – 2; 2, 1, 1
Cubic polynomials can have one, two, three, or no real zeros (roots), depending on the values of the constants a, b, c, and d. For example, the cubic polynomial x³ + 2x² - 4x + 4 has one real zero (2), while the cubic polynomial x³ - 2x² + x - 1 has no real zeros. See here for Solution👇
Cubic polynomials can have one, two, three, or no real zeros (roots), depending on the values of the constants a, b, c, and d. For example, the cubic polynomial x³ + 2x² – 4x + 4 has one real zero (2), while the cubic polynomial x³ – 2x² + x – 1 has no real zeros.
See here for Solution👇
See less