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Trigonometric Functions - Previous Year Questions

⏱️ Est. Revision Time: 15 Mins
📅 Year 2024
Q1
CBSE Board Exam 20243 Marks
Prove that sin x − sin y / cos x + cos y = tan(x − y)/2.
▶ Show Detailed Solution
sin x − sin y = 2 cos[(x+y)/2] sin[(x−y)/2]
cos x + cos y = 2 cos[(x+y)/2] cos[(x−y)/2]
Dividing: [2 cos((x+y)/2) sin((x−y)/2)] / [2 cos((x+y)/2) cos((x−y)/2)] = sin((x−y)/2)/cos((x−y)/2) = tan((x−y)/2)
Q2
CBSE Board Exam 20242 Marks
Find the value of tan 75° using the compound angle formula.
▶ Show Detailed Solution
tan 75° = tan(45° + 30°) = (tan 45° + tan 30°)/(1 − tan 45° tan 30°)
= (1 + 1/√3)/(1 − 1/√3) = (√3 + 1)/(√3 − 1)
Rationalising: (√3+1)²/((√3)²−1²) = (3 + 2√3 + 1)/2 = (4 + 2√3)/2
tan 75° = 2 + √3
📅 Year 2023
Q1
CBSE Board Exam 20233 Marks
If tan x = 3/4, π < x < 3π/2, find the value of sin x/2, cos x/2 and tan x/2.
▶ Show Detailed Solution
x in QIII ⇒ sin x < 0, cos x < 0. tan x = 3/4 ⇒ sin x = −3/5, cos x = −4/5.
Since π < x < 3π/2, π/2 < x/2 < 3π/4 ⇒ x/2 in QII.
sin(x/2) = √[(1−cos x)/2] = √[(1+4/5)/2] = √(9/10) = 3/√10
cos(x/2) = −√[(1+cos x)/2] = −√[(1−4/5)/2] = −√(1/10) = −1/√10
tan(x/2) = sin(x/2)/cos(x/2) = −3
sin(x/2)=3/√10, cos(x/2)=−1/√10, tan(x/2)=−3