InterviewSolution
Saved Bookmarks
InterviewSolution
This section includes InterviewSolutions, each offering curated multiple-choice questions to sharpen your knowledge and support exam preparation. Choose a topic below to get started.
| 1. |
Prove the following trigonometric identities:cotθ - tanθ =\(\frac{2cos^2θ-1}{sinθ.cosθ}\) |
|
Answer» L.H.Scotθ - tanθ =\(\frac{cosθ}{sinθ}-\frac{sinθ}{cosθ}\) =\(\frac{cos^2θ-sin^2θ}{sinθ.cosθ}\) =\(\frac{cos^2θ-(1-cos^2θ)}{sinθ.cosθ}\) =\(\frac{cos^2θ-1+cos^2θ)}{sinθ.cosθ}\) = \(\frac{2cos^2θ-1}{sinθ.cosθ}\) = R.H.S Hence Proved. |
|
| 2. |
If sin 3θ = cos(θ – 6°) here 3θ and (θ – 6°) are acute angles, then find the value of θ. |
|
Answer» Given : sin 3θ = cos(θ – 6°) or cos(90° – 3θ) = cos(θ – 6°) [∵ cos(90° – θ) = sin θ] or 90° – 3θ = θ – 6° or 3θ + θ = 90° + 6° or 4θ = 96° or θ = 96/4 = 24° |
|
| 3. |
Verify the followingwith the help of identities.(i) ((1 + cotθ + tanθ)(sinθ - cosθ))/(sec3θ - cosec3θ) = sin2θ cos2θ(ii) (sinθ + cosθ)/(sinθ - cosθ) + (sinθ - cosθ)/(sinθ + cosθ) = 2/(1 - 2cos2θ) = 2/(2cos2θ - 1) |
|
Answer» (i) LHS = ((1 + cotθ + tanθ)(sinθ - cosθ))/(sec3θ - cosec3θ) = sin2θ cos2θ = R.H.S. (ii) LHS = (sinθ + cosθ)/(sinθ - cosθ) + (sinθ - cosθ)/(sinθ + cosθ) = 2/(1 - 2cos2θ) = R.H.S. Again by equation (i) = 2/(2cos2θ - 1) = R.H.S. |
|
| 4. |
In the adjoining figure, ∠B = 90°, ∠BAC = θ°, BC = CD = 4 cm and AD=10 cm. Find (i) sinθ and (ii) cosθ. |
|
Answer» Draw a triangle using given instructions: From figure: Δ ABC and Δ ABD are right angled triangles where AD = 10 cm BC = CD = 4 cm BD = BC + CD = 8 cm By Pythagoras theorem: AD2= BD2+ AB2 (10)2= (8)2+ AB2 100 = 64 + AB2 AB2= 36 = (6)2 or AB = 6 cm Again, AC2= BC2+ AB2 AC2= (4)2+ (6)2 AC2= 16 + 36 = 52 or AC = √52 = 2√13 cm (i) Find sin θ sin θ = BC/AC = 4/2√13 = 2√13/13 (ii) Find cos θ cosθ = AB/AC = 6/2√13 = 3/√13 = 3√13/13 |
|