Sujeet Kumar

Question 1.Solve the following linear equations by using Cramer’s Rule.x+y + z = 6, x – y + z = 2,.x + 2y – z = 2x + y – 2z = -10,2x +y – 3z = -19, 4x + 6y + z = 2 = 6(1 – 2) – 1(-2 – 2) + 1(4 … Read more

Question 1.Without expanding, evaluate the following determinants. Solution: Question 2. Solution: Question 3.Using properties of determinant, show that= -2[0(ab – 0) – a(bc – 0) + b(0 – ac)]= -2(0 – abc – abc)= -2(-2abc)= 4abc = R.H.S. ii. Question 4.Solve the following equations. Question 5. Solution:∴ x = 0 or 12 – x = … Read more

Question 1.Find the values of the determinants. Question 2. Question 3. Solution:Comparing with x + iy, we get x = 11, y = 52 Question 4.Find the minors and cofactors of elements of the determinant D= Question 5.Solution:= 2(0 – 20) + 3(- 42 – 4) + 5(30 – 0) = 2(-20) + 3(- 46) … Read more

I. Select the correct option from the given alternatives. Question 1.The value of sin(n + 1) A sin(n + 2) A + cos(n + 1) A cos(n + 2) A is equal to(a) sin A(b) cos A(c) -cos A(d) sin 2AAnswer:(b) cos AHint:L.H.S. = sin [(n + 1)A] . sin [(n + 2)A] + cos … Read more

Question 1.In Δ ABC, A + B + C = π, show thatcos2A + cos2B + cos2C = – 1 – 4 cosA cosB cosCSolution:L.H.S. = cos 2A + cos 2B + cos 2C = 2.cos(A + B).cos (A – B) + 2cos2C – 1In ΔABC, A + B + C = π∴ A + … Read more

Question 1.Express the following as a sum or difference of two trigonometric functions.i. 2sin 4x cos 2x iii. 2cos 4θ cos 2θiv. 2cos 35° cos 75°Solution:i. 2sin 4x cos 2x = sin(4x + 2x ) + sin (4x – 2x)= sin 6x + sin 2x ii. iii. 2cos 4θ cos 2θ = cos(4θ + 2θ)+cos … Read more

Question 1.Find the values of: Question 2. Question 3. v. tan x + cot x = 2 cosec 2xSolution:L.H.S. = tan x + cot x Solution: Solution:= 2 cos x= R.H.S.[Note : The question has been modified.] viii. 16 sin θ cos θ cos 2θ cos 4θ cos 8θ = sin 16θSolution:L.H.S. = 16 sin … Read more

Question 1.Find the values of:i. sin 690°ii. sin 495°iii. cos 315°iv. cos 600°v. tan 225°vi. tan (- 690°)vii. sec 240°viii. sec (- 855°)ix. cosec 780°x. cot (-1110°)Solution:i. sin 690° = sin (720° -30°)Solution:i. sin 690° = sin (720° -30°)= sin (2 x 360° – 30°)= – sin 30°= -1/2 ii. sin 495° = sin (360° + … Read more

Question 1.Find the values of:i. sin 150°ü. cos 75°iii. tan 105°iv. cot 225°Solution:i. sin 15° = sin (45° – 30°)= sin 45° cos 30° – cos 45° sin 30° ii. cos 75° = cos (45° + 30°)= cos 45° cos 30° – sin 45° sin 30° iii. tan 105° = tan (60° +45°) iv. cot … Read more

I. Select the correct option from the given alternatives. Question 1.The value of the expressioncos1°. cos2°. cos3° … cos 179° =(A) -1(B) 0 (D) 1Answer:(B) 0 Explanation:cos 1° cos 2° cos 3° … cos 179°= cos 1° cos 2° cos 3° … cos 90°… cos 179°= 0 …[∵ cos 90° = 0] Question 2.(A) 2cosec … Read more