QQuestion
If sin (A + B) = 1 and cos (A − B) = 1, then find the value of cot 2A, where 0° ≤ (A + B) ≤ 90° and A ≥ B
✓Answer
Given: sin (A + B) = 1
⇒ A + B = 90° —— (i)
Also, cos (A − B) = 1
⇒ A − B = 0°
⇒ A = B.
Now, A + B = 90° and A = B
⇒ 2A = 90°
⇒ A = 45°.
So, cot 2A = cot 90° = 0
Related Questions
More WBBSE Easy level questions
If sinθ = √3 cosθ , Find the value of tanθ + cotθ
If sin²θ + 2x cos²θ = 1, then value of x is ______.
The lengths of three sides of a triangle are sec θ, 1, and tan θ (θ ≠ 90°). The value of the greatest angle of the triangle is (a) 30° (b) 45° (c) 60° (d) 90°
For any value of θ the maximum value of 5 + 4 sin θ is (a) 9 (b) 1 (c) 0 (d) 5
If 5 sin² θ + 4 cos² θ = , then from this relation find the value of tan θ.
If sin (θ - 30°) = , then cos θ = ____
The minimum value of tanθ + cotθ is (a) 0 (c) -2 (b) 2 (d) 1
If cos⁴θ – sin⁴θ = , find the value of 1 – 2 sin²θ.
If tan θ = , find the value of
If sin x = m sin y and tan x = n tan y, then show that cos² x =
Show that =
If cos θ = , then prove that x sin θ = y cos θ