Vardaan Learning Institute
Answer Key: Introduction to Trigonometry PYQ
Class: 10 (CBSE)
Subject: Mathematics
Chapter: 08
BOARD EXAM 2025 (Set 30/6/3)
Q3.
If \(x = 2 \sin 60� \cos 60�\) and \(y = \sin^2 30� - \cos^2 30�\) and \(x^2 = ky^2\), the value of \(k\) is
(A) \(\sqrt{3}\)
(B) \(-\sqrt{3}\)
(C) 3
(D) -3
Q9.
In a right triangle ABC, right-angled at A, if \(\sin B = \frac{1}{4}\), then the value of \(\sec B\) is
(A) 4
(B) \(\frac{\sqrt{15}}{4}\)
(C) \(\sqrt{15}\)
(D) \(\frac{4}{\sqrt{15}}\)
Q21 (a).
If \(a \sec \theta + b \tan \theta = m\) and \(b \sec \theta + a \tan \theta = n\), prove that \(a^2 + n^2 = b^2 + m^2\).
Expanding both sides and using \(1 + \tan^2 \theta = \sec^2 \theta\), both simplify to \((a^2 + b^2)\sec^2 \theta + 2ab \sec \theta \tan \theta\). Hence proved.
OR
Q21 (b).
Use \(\sin^2 A + \cos^2 A = 1\) to prove \(\tan^2 A + 1 = \sec^2 A\). Find \(\tan A\) when \(\sec A = \frac{5}{3}\).
Q28 (a).
Prove that: \(\frac{\cos \theta - 2 \cos^3 \theta}{\sin \theta - 2 \sin^3 \theta} + \cot \theta = 0\)
Factor: \(\frac{\cos \theta(\sin^2 \theta - \cos^2 \theta)}{\sin \theta(\cos^2 \theta - \sin^2 \theta)} = -\cot \theta\). So \(-\cot \theta + \cot \theta = 0\). Proved.
OR
Q28 (b).
Given \(\sin \theta + \cos \theta = x\), prove that \(\sin^4 \theta + \cos^4 \theta = \frac{2 - (x^2 - 1)^2}{2}\).
\(\sin \theta \cos \theta = \frac{x^2 - 1}{2}\). \(\sin^4 + \cos^4 = 1 - 2(\sin \theta \cos \theta)^2 = \frac{2 - (x^2 - 1)^2}{2}\). Proved.