Home › Practice › Class 11 › Maths
Class 11 Maths — questions with answers and worked explanations.
These questions are drawn from the Pariksha Sutra question bank for Trigonometric Functions, part of the Class 11 Maths syllabus. Each one shows the correct answer and, where a method helps, the working behind it.
Read the question, decide your answer before looking, then check the explanation — that is what turns practice into marks. If a question catches you out, the explanation is the part worth re-reading.
Convert degrees to radians using 180° = π rad.
Set up proportion: 30° × π /180° = π/6.
Thus the radian measure of 30° equals π/6.
Use the double‑angle identity: sin(2x)=2·sin x·cos x.
It is derived from sin(A+B)=sinA cosB+cosA sinB by setting A=B=x.
Thus sin(2x)=sin(x+x)=sin x cos x+cos x sin x=2 sin x cos x.
Hence the correct choice is 2 sinx cosx.
tan(150°)=tan(180°−30°)=−tan30° because tan(π−θ)=−tanθ.
tan30°=1/√3 from the 30‑60‑90 triangle.
Thus tan150°=−1/√3, which matches the given answer.
cos2x = (1−tan²x)/(1+tan²x) follows from writing cos2x = (cos²x−sin²x)/ (cos²x+sin²x) and dividing numerator and denominator by cos²x, giving (1−tan²x)/(1+tan²x). This matches the given option. (1 - tan^2x)/(1 + tan^2x)
sin θ is the y‑coordinate of the point where the terminal side of angle θ meets the unit circle.
For θ = π/2 radians the point on the unit circle is (0, 1).
Thus the y‑coordinate, i.e. sin(π/2), equals 1.
Answer: 1
secθ = 1/cosθ, so evaluate cos 60° = ½.
Take the reciprocal: sec 60° = 1 ÷ (½) = 2.
Hence the correct answer is 2.
Use the cosine difference identity: cos(A−B)=cosA·cosB+sinA·sinB.
Derive it from the sum formula cos(A−B)=cosAcos(−B)+sinA sin(−B) and note cos(−B)=cosB, sin(−B)=−sinB, which changes the sign to plus.
Thus the expression matches the third option, cosA cosB + sinA sinB.
210° = 180° + 30°, so it lies in the third quadrant where sine is negative.
sin(180°+θ) = –sin θ, thus sin 210° = –sin 30°.
sin 30° = 1/2, so sin 210° = –1/2.
tan(A+B)=sin(A+B)/cos(A+B) = (sinAcosB+cosAsinB)/(cosAcosB−sinAsinB)
Divide numerator and denominator by cosAcosB → (tanA+tanB)/(1−tanA·tanB)
Thus the formula for tan of a sum is (tanA+tanB)/(1−tanA tanB).
The correct answer is (tanA + tanB)/(1 - tanA tanB).
cos π corresponds to the angle 180° on the unit circle.
The x‑coordinate of the point at 180° is –1, which is the definition of cosine.
Therefore cos π = –1. The correct answer is -1.
cosec θ = 1/sin θ
For θ = 30°, sin 30° = 1/2 (from the standard triangle)
Thus cosec 30° = 1 ÷ (1/2) = 2
Hence the correct answer is 2.
tan θ = opposite/adjacent in a 30°‑60°‑90° right triangle.
For θ = 60°, the sides are 1 (adjacent) and √3 (opposite).
Thus tan 60° = √3/1 = √3, which matches the given answer.
Convert radians to degrees using 180° = π rad.
Degree = (5π/6) × (180°/π).
Cancel π: (5/6) × 180° = 5 × 30° = 150°.
Thus the angle equals 150 degrees.
sin x = 0 when the angle corresponds to the x‑axis on the unit circle.
The sine function is zero at multiples of 180°, i.e. at angles 0°, 180°, 360°, …
In radians these are 0, π, 2π, 3π,… which can be written as x = n·π where n is any integer.
Thus the general solution is x = nπ.
The basic sine function satisfies \(-1\le\sin x\le 1\) for all real \(x\).
Multiplying by 3 scales the entire range by the factor 3, giving \(-3\le 3\sin x\le 3\).
Thus the set of possible values of \(3\sin x\) is the interval \([-3,3]\).
Use the double‑angle identity for tangent: tan(2x)=sin(2x)/cos(2x).
Express sin2x and cos2x in terms of tan x: sin2x=2tan x/(1+tan²x), cos2x=(1−tan²x)/(1+tan²x).
Divide sin2x by cos2x: tan(2x)=[2tan x/(1+tan²x)] ÷ [(1−tan²x)/(1+tan²x)] = (2tan x)/(1−tan²x).
Thus the correct choice is (2 tanx)/(1 - tan^2x).
Use the double‑angle identity: sin 2θ = 2 sinθ cosθ.
Here θ = x/2, so 2 sin(x/2) cos(x/2) = sin 2·(x/2) = sin x.
Thus the expression equals sin x.
cot θ = adjacent/ opposite = 1/tan θ.
For θ = 45°, tan 45° = 1 (since opposite = adjacent in a 45°‑45°‑90° triangle).
Thus cot 45° = 1/(tan 45°) = 1/1 = 1.
Answer: 1
sin θ is negative below the x‑axis (θ between 180° and 360°) and cos θ is negative left of the y‑axis (θ between 90° and 270°).
The only region where both conditions hold simultaneously is when θ lies between 180° and 270°.
That interval corresponds to the third quadrant, so both sin and cos are negative there.
sin is an odd function, so sin(−x)=−sin x.
Using the unit‑circle definition, the y‑coordinate for angle −x is the negative of that for x.
Thus the value equals −sin x. -sin x
The full chapter test is timed, gives an All-India rank and a subject-wise breakdown of where you lost marks.
See the test series →