Very Short Answer on Trigonometric Equations Class 11 OP Malhotra Exe-6B ISC Maths Solutions Ch-6. In this article you would learn to solve all type questions on Trigonometric Equations. Step by step solutions of latest textbook has been given as latest syllabus. Visit official Website CISCE for detail information about ISC Board Class-11
Trigonometric Equations Class 11 OP Malhotra Very Short Answer ISC Maths Solutions Ch-6
| Board | ISC |
| Publications | S Chand |
| Subject | Maths |
| Class | 11th |
| Chapter-6 | Trigonometric Equations. |
| Writer | OP Malhotra |
| Exe-6(B) | Very Short Answer Type Questions. |
Very Short Answer on Trigonometric Equations
OP Malhotra ISC Class 11 Maths Solutions
Que-1: If sinθ = 1/√2, then its principal solutions are _____________________
Sol: sinθ = 1/√2 ⇒ θ = 45° , 135°
(In radians: θ = π/4 , 3π/4)
Que-2: If √3 cosecθ = 2, then its principal solutions are ___________________
Sol: √3 cosecθ = 2 ⇒ cosecθ = 2/√3 ⇒ sinθ = √3/2
⇒ θ = 60° , 120°
(In radians: θ = π/3 , 2π/3)
Que-3: The value of x in (0, π/2) satisfying the equation sinx cosx = 1/4 is __________________________
Sol: sinx cosx = 1/4
⇒ 2sinx cosx = 1/2
⇒ sin2x = 1/2
⇒ 2x = π/6
⇒ x = π/12
Que-4: The principal solution of the equations 2 sin²θ + √3 cosθ + 1 = 0 are __________________________
Sol: 2 sin²θ + √3 cosθ + 1 = 0
2(1 − cos²θ) + √3 cosθ + 1 = 0
2 − 2cos²θ + √3 cosθ + 1 = 0
−2cos²θ + √3 cosθ + 3 = 0
2cos²θ − √3 cosθ − 3 = 0
(2cosθ + √3)(cosθ − √3) = 0
cosθ = −√3/2 or cosθ = √3 (not possible)
∴ cosθ = −√3/2
Principal solutions: θ = 150°, 210°
Que-5: If sinθ + cosθ = 0 and 0 < θ < π, then θ is equal to _________________
Sol: sinθ + cosθ = 0
sinθ = −cosθ
tanθ = −1
In (0, π), θ = 3π/4
Que-6: If cosecθ − cotθ = 1/√3, when 0 < θ < π/2, then find the value of cosθ.
Sol: cosecθ − cotθ = (1 − cosθ)/sinθ = 1/√3
Multiply: √3(1 − cosθ) = sinθ
Try standard angle → θ = 60°
∴ cosθ = 1/2
Que-7: Find the principal solutions of the equation tan2θ = 1.
Sol: tan 2θ = 1
2θ = 45° + n180°
Principal solutions: 2θ = 45°, 225°
θ = 22.5°, 112.5°
Write the general solution of the following equations:
Que-8: cosx = 1/2
Sol: cos x = 1/2
x = 60°, 300°
General solution: x = 2nπ ± π/3 (or x = 60° + 360°n, 300° + 360°n)
Que-9: tan²θ = 1
Sol: tan²θ = 1 ⇒ tanθ = ±1
⇒ θ = π/4 + nπ/2, n ∈ Z
Que-10: cotθ = −√3
Sol: cotθ = −√3 ⇒ tanθ = −1/√3
⇒ θ = −π/6 + nπ
Que-11: cosec x = −2
Sol: Cosec x = −2 ⇒ sinx = −1/2
⇒ x = 7π/6 + 2nπ or 11π/6 + 2nπ
Que-12: tan 3x = 1
Sol: tan3x = 1 ⇒ 3x = π/4 + nπ
⇒ x = π/12 + nπ/3
Que-13: tanθ = cotα
Sol: cotα = tan(π/2 − α)
⇒ tanθ = tan(π/2 − α)
⇒ θ = π/2 − α + nπ
Que-14: If cotθ + tanθ = 2 cosecθ, then find value of θ between 0 and 2π.
Sol: cotθ + tanθ = (cosθ/sinθ + sinθ/cosθ)
= (sin²θ + cos²θ)/(sinθcosθ) = 1/(sinθcosθ)
Given 2cosecθ = 2/sinθ
⇒ 1/(sinθcosθ) = 2/sinθ
⇒ 1/cosθ = 2 ⇒ cosθ = 1/2
⇒ θ = π/3 or 5π/3 (0 ≤ θ ≤ 2π)
Que-15: Find the general solution of the equation tan θ + tan 2θ + √3 tan θ tan 2θ = √3
Sol: Use identity:
tanA + tanB + √3 tanA tanB = √3
⇒ A + B = π/3
Here A = θ, B = 2θ
⇒ 3θ = π/3
⇒ θ = π/9 + nπ/3
Que-16: Answer true or false, if cosec x = 1 + cot x, then x = 2nπ, 2nπ + π/2.
Sol: cosec x = 1 + cot x
1/sin x = 1 + cos x/sin x
1/sin x = (sin x + cos x)/sin x
⇒ 1 = sin x + cos x
Now, sin x + cos x = 1
⇒ sin x = 1 and cos x = 0
⇒ x = 2nπ or x = 2nπ + π/2
So, It is True.
Que-17: Show that one of the principal values of √3 sec x = −2 is equal to 5π/6.
Sol: √3 sec x = −2 ⇒ sec x = −2/√3 ⇒ cos x = −√3/2
Principal value in [0, 2π] is x = 5π/6.
Que-18: Find the principal solutions of the equation sin 2x + cos 2x = 0, π < x < 2π.
Sol: sin2x + cos2x = 0 ⇒ tan2x = −1
⇒ 2x = −π/4 + nπ
⇒ x = −π/8 + nπ/2
In (π, 2π): x = 7π/8, 11π/8, 15π/8.
Que-19: Find the most general solution of tanθ = −1 and cosθ = 1/√2.
Sol: (i) tanθ = −1 ⇒ θ = −π/4 + nπ
(ii) cosθ = 1/√2 ⇒ θ = 2nπ ± π/4
Que-20: Least value of 3 sin²θ + 4 cos²θ is ______.
Sol: 3sin²θ + 4cos²θ = 3(1−cos²θ) + 4cos²θ = 3 + cos²θ
Minimum when cos²θ = 0
⇒ Least value = 3
Que-21: Show that the equation cosecθ secθ = 1 has no solution.
Sol: cosecθ secθ = 1 ⇒ 1/(sinθ cosθ) = 1
⇒ sinθ cosθ = 1
But max value of sinθ cosθ = 1/2
So equation has No solution.
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Return to :- OP Malhotra ISC Class-11 S Chand Publication Maths Solutions
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