sin (45° + θ) – cos (45° – θ) is equal to
(A) 2cosθ (B) 0 (C) 2sinθ (D) 1
If sinθ – cosθ = 0, then the value of (sin4θ + cos4θ) is
(A) 1 (B)3/4 (C)1/2 (D)1/4
Given that sinα = 1/2 and cosβ =1/2 , then the value of (α + β) is
(A) 0° (B) 30° (C) 60° (D) 90°
If sinA + sin2A = 1, then the value of the expression (cos2A + cos4A) is
(A) 1 (B) 1/2(C) 2 (D) 3 2
The value of (tan1° tan2° tan3° … tan89°) is
(A) 0 (B) 1 (C) 2 (D) 1/2
If cos (α + β) = 0, then sin (α – β) can be reduced to
(A) cos β (B) cos 2β (C) sin α (D) sin 2α
The value of the expression [cosec (75° + θ) – sec (15° – θ) – tan (55° + θ) + cot (35° – θ)] is
(A) – 1 (B) 0 (C) 1 (D) 3/2
If cos A = 4/5, , then the value of tan A is
(A) 3/5 (B)/34 (C)4/3 (D)5/3
The value of (sin30° + cos30°) – (sin60° + cos60°) is
(A) – 1 (B) 0 (C) 1 (D) 2
Ayush starts walking from his house to office. Instead of going to the office directly, he goes to a bank first, from there to his daughter’s school and then reaches the office. What is the extra distance travelled by Ayush in reaching his office? (Assume that all distances covered are in straight lines).
If the house is situated at (2, 4), bank at (5, 8), school at (13, 14) and office at (13, 26) and coordinates are in km.
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