15 multiple-choice questions and 2 case-based questions on the current (rationalised) syllabus, in the style of the board paper's Sections A and E. Try each one first; the answer, a one-line reason and a worked solution open on a tap.
Choose one option. Section A of the board paper has 18 MCQs of 1 mark each, spread over all chapters.
In \(\triangle ABC\), right-angled at \(B\), \(AB = 5\) cm and \(BC = 12\) cm. Then \(\sin A\) is
Why: sin A = opposite/hypotenuse = BC/AC, with AC = 13.
\(AC = \sqrt{25 + 144} = 13\) cm. \(\sin A = \dfrac{BC}{AC} = \dfrac{12}{13}\).
If \(\tan A = \dfrac{8}{15}\), where \(A\) is acute, then \(\sec A\) equals
Why: Opposite 8, adjacent 15, hypotenuse 17; sec A = hypotenuse/adjacent.
Take opposite \(= 8k\), adjacent \(= 15k\); hypotenuse \(= 17k\). \(\sec A = \dfrac{17}{15}\) (or \(\sec^2 A = 1 + \tfrac{64}{225} = \tfrac{289}{225}\)).
The value of \(\sin 30^\circ \cos 60^\circ + \cos 30^\circ \sin 60^\circ\) is
Why: (1/2)(1/2) + (√3/2)(√3/2) = 1/4 + 3/4.
\(\dfrac12 \cdot \dfrac12 + \dfrac{\sqrt3}{2} \cdot \dfrac{\sqrt3}{2} = \dfrac14 + \dfrac34 = 1\).
If \(\tan A = \sqrt3\) and \(\tan B = \dfrac{1}{\sqrt3}\), where \(A\) and \(B\) are acute, then \(A - B\) is
Why: tan A = √3 gives A = 60°; tan B = 1/√3 gives B = 30°.
\(\tan 60^\circ = \sqrt3\) and \(\tan 30^\circ = \dfrac{1}{\sqrt3}\), so \(A = 60^\circ\), \(B = 30^\circ\) and \(A - B = 30^\circ\).
Without a calculator, which of the following has the greatest value?
Why: sec 30° = 2/√3, which is more than 1; the others are √3/2, 1/2 and 1.
\(\sin 60^\circ = \tfrac{\sqrt3}{2} \lt 1\), \(\cos 60^\circ = \tfrac12\), \(\tan 45^\circ = 1\), \(\sec 30^\circ = \tfrac{2}{\sqrt3}\). Since \(2 \gt \sqrt3\), \(\sec 30^\circ \gt 1\), so it is the greatest.
If \(\sin\theta = \cos\theta\) and \(\theta\) is acute, then \(\theta\) is
Why: sin θ = cos θ means tan θ = 1.
Dividing by \(\cos\theta\): \(\tan\theta = 1\), so \(\theta = 45^\circ\).
If \(\sqrt3 \tan\theta = 3\), where \(\theta\) is acute, then \(\sin\theta\) is
Why: tan θ = 3/√3 = √3, so θ = 60°.
\(\tan\theta = \dfrac{3}{\sqrt3} = \sqrt3 \Rightarrow \theta = 60^\circ\), so \(\sin\theta = \dfrac{\sqrt3}{2}\).
\((1 + \tan^2 A)\cos^2 A\) equals
Why: 1 + tan² A = sec² A and sec² A · cos² A = 1.
\((1 + \tan^2 A)\cos^2 A = \cos^2 A + \sin^2 A = 1\).
\((\operatorname{cosec} A - \cot A)(1 + \cos A)\) equals
Why: Write it as (1 − cos A)(1 + cos A)/sin A = sin² A/sin A.
\(\dfrac{1 - \cos A}{\sin A} \cdot (1 + \cos A) = \dfrac{1 - \cos^2 A}{\sin A} = \dfrac{\sin^2 A}{\sin A} = \sin A\).
If \(\sin A = \dfrac35\), where \(A\) is acute, then \(\cos A + \tan A\) equals
Why: cos A = 4/5 and tan A = 3/4.
\(\cos A = \sqrt{1 - \tfrac{9}{25}} = \dfrac45\), \(\tan A = \dfrac{3/5}{4/5} = \dfrac34\). Sum \(= \dfrac{16 + 15}{20} = \dfrac{31}{20}\).
The value of \(\sin 0^\circ + \cos 0^\circ + \tan 45^\circ\) is
Why: 0 + 1 + 1.
\(\sin 0^\circ = 0\), \(\cos 0^\circ = 1\), \(\tan 45^\circ = 1\). Sum \(= 2\).
If \(\operatorname{cosec}\theta - \cot\theta = \dfrac14\), then \(\operatorname{cosec}\theta\) equals
Why: cosec² θ − cot² θ = 1, so cosec θ + cot θ = 4; add the two equations.
\((\operatorname{cosec}\theta - \cot\theta)(\operatorname{cosec}\theta + \cot\theta) = 1 \Rightarrow \operatorname{cosec}\theta + \cot\theta = 4\). Adding: \(2\operatorname{cosec}\theta = \dfrac{17}{4}\), so \(\operatorname{cosec}\theta = \dfrac{17}{8}\).
For \(A = 30^\circ\), which of the following is true?
Why: cos 60° = 1/2 and 1 − 2 sin² 30° = 1 − 1/2 = 1/2.
\(\sin 60^\circ = \tfrac{\sqrt3}{2} \ne 2 \cdot \tfrac{\sqrt3}{2}\); \(\cos 60^\circ = \tfrac12 = 1 - 2 \cdot \tfrac14\) ✓; \(\tan 60^\circ = \sqrt3 \ne \tfrac{2}{\sqrt3}\); \(\cos 60^\circ = \tfrac12 \ne \sqrt3\).
\(\sin^4\theta - \cos^4\theta\) equals
Why: Difference of squares: (sin²θ − cos²θ)(sin²θ + cos²θ).
\(\sin^4\theta - \cos^4\theta = (\sin^2\theta - \cos^2\theta)(\sin^2\theta + \cos^2\theta) = \sin^2\theta - \cos^2\theta = 1 - 2\cos^2\theta\).
If \(5\sin\theta = 4\), where \(\theta\) is acute, then \(\sec\theta + \tan\theta\) equals
Why: sin θ = 4/5 gives cos θ = 3/5, so sec θ = 5/3 and tan θ = 4/3.
\(\sin\theta = \dfrac45\), \(\cos\theta = \sqrt{1 - \tfrac{16}{25}} = \dfrac35\). \(\sec\theta + \tan\theta = \dfrac53 + \dfrac43 = 3\).
Section E of the board paper has three case-based questions of 4 marks: a real-life passage, then parts of 1, 1 and 2 marks, with a choice (OR) on the 2-mark part.
A solar panel on a flat roof rests on a frame shaped like a right triangle \(ABC\), right-angled at \(B\). The horizontal bar \(AB\) is \(12\) dm long and the vertical post \(BC\) is \(5\) dm tall. The panel lies along \(AC\), and \(\theta = \angle BAC\) is its angle of tilt.
A geometry box has two set squares. One is a right triangle with angles \(30^\circ\), \(60^\circ\) and \(90^\circ\), whose longest side (hypotenuse) is \(20\) cm. The other is a right isosceles triangle with angles \(45^\circ\), \(45^\circ\), \(90^\circ\), whose two equal sides are \(12\) cm each.
This free set is separate from the chapter's question bank. On the Introduction to Trigonometry chapter page the revision notes and the first step of the Route to 95 are free for everyone. CBSE Essentials adds all 38 questions in the chapter bank (short and long answers, assertion–reason and more case studies) with their full step-marking schemes, the Route to 95 checkpoints with your progress saved, Skill Builders and the full common-mistakes library. There is no AI marking on CBSE Math Revision: you check your work against the marking scheme.
Original questions written by CBSE Math Revision for the CBSE 2026-27 syllabus; not taken from NCERT, NCERT Exemplar or CBSE papers. Every answer was re-checked by computer algebra and by an independent reviewer. CBSE Math Revision is independent and not affiliated with CBSE or NCERT. Spotted a slip? Tell us and it goes in the corrections log.