CBSE & State Board Question Bank | 9th - 12th

Deepa Maths Academy • Global Examination Portal

CBSE Class 10 Maths Chapter 8 Trigonometry Model Questions - 2 Marks - Part 1

Secure University-Grade Repository for Model Assessments, Board Examinations, and Step-by-Step Solutions.

Portal ID: DMA-EXAM-2026 • Security: Encrypted • Status: Active Examination

SECTION A — Multiple Choice Questions [1 Mark Each]

• 1 Mark Q1. If $\tan \theta + \sin \theta = m$ and $\tan \theta - \sin \theta = n$, then the value of $m^2 - n^2$ is:
  (a) $4\sqrt{mn}$

  (b) $2\sqrt{mn}$

  (c) $mn$

  (d) $4mn$
  Solution:
  $m^2 - n^2 = (m + n)(m - n) = (\tan\theta + \sin\theta + \tan\theta - \sin\theta)(\tan\theta + \sin\theta - (\tan\theta - \sin\theta))$
  $= (2\tan\theta)(2\sin\theta) = 4\tan\theta\sin\theta$.
  Also, $4\sqrt{mn} = 4\sqrt{(\tan\theta+\sin\theta)(\tan\theta-\sin\theta)} = 4\sqrt{\tan^2\theta - \sin^2\theta} = 4\sqrt{\sin^2\theta(\sec^2\theta - 1)} = 4\sin\theta\tan\theta$.
  Answer: (a) $4\sqrt{mn}$
• 1 Mark Q2. If $\sec \theta + \tan \theta = x$, then the value of $\sin \theta$ is:
  (a) $\frac{x^2 - 1}{x^2 + 1}$

  (b) $\frac{x^2 + 1}{x^2 - 1}$

  (c) $\frac{x - 1}{x + 1}$

  (d) $\frac{2x}{x^2 + 1}$
  Solution:
  Since $\sec\theta - \tan\theta = \frac{1}{x}$, subtracting it from $\sec\theta + \tan\theta = x$ gives $2\tan\theta = x - \frac{1}{x} = \frac{x^2-1}{x} \implies \tan\theta = \frac{x^2-1}{2x}$.
  Adding them gives $2\sec\theta = x + \frac{1}{x} = \frac{x^2+1}{x} \implies \cos\theta = \frac{2x}{x^2+1}$.
  $\sin\theta = \tan\theta \times \cos\theta = \frac{x^2-1}{2x} \times \frac{2x}{x^2+1} = \frac{x^2-1}{x^2+1}$.
  Answer: (a) $\frac{x^2 - 1}{x^2 + 1}$
• 1 Mark Q3. If $x = a \cos^3 \theta$ and $y = b \sin^3 \theta$, then the relation between $x$ and $y$ independent of $\theta$ is:
  (a) $\left(\frac{x}{a}\right)^{\frac{2}{3}} + \left(\frac{y}{b}\right)^{\frac{2}{3}} = 1$

  (b) $\left(\frac{x}{a}\right)^{\frac{3}{2}} + \left(\frac{y}{b}\right)^{\frac{3}{2}} = 1$

  (c) $\frac{x^2}{a^2} + \frac{y^2}{b^2} = 1$

  (d) $\frac{x}{a} + \frac{y}{b} = 1$
  Solution:
  $\frac{x}{a} = \cos^3\theta \implies \left(\frac{x}{a}\right)^{\frac{1}{3}} = \cos\theta \implies \left(\frac{x}{a}\right)^{\frac{2}{3}} = \cos^2\theta$.
  $\frac{y}{b} = \sin^3\theta \implies \left(\frac{y}{b}\right)^{\frac{1}{3}} = \sin\theta \implies \left(\frac{y}{b}\right)^{\frac{2}{3}} = \sin^2\theta$.
  Adding both gives $\left(\frac{x}{a}\right)^{\frac{2}{3}} + \left(\frac{y}{b}\right)^{\frac{2}{3}} = \cos^2\theta + \sin^2\theta = 1$.
  Answer: (a) $\left(\frac{x}{a}\right)^{\frac{2}{3}} + \left(\frac{y}{b}\right)^{\frac{2}{3}} = 1$
• 1 Mark Q4. If $\sin \theta + \cos \theta = \sqrt{2} \cos \theta$, then the value of $\tan \theta$ is:
  (a) $\sqrt{2} - 1$

  (b) $\sqrt{2} + 1$

  (c) $\sqrt{2}$

  (d) $1$
  Solution:
  $\sin\theta = \sqrt{2}\cos\theta - \cos\theta = (\sqrt{2} - 1)\cos\theta$.
  $\tan\theta = \frac{\sin\theta}{\cos\theta} = \sqrt{2} - 1$.
  Answer: (a) $\sqrt{2} - 1$
• 1 Mark Q5. If $a \cos \theta - b \sin \theta = c$, then $(a \sin \theta + b \cos \theta)^2$ is equal to:
  (a) $a^2 + b^2 - c^2$

  (b) $a^2 + b^2 + c^2$

  (c) $a^2 - b^2 + c^2$

  (d) $c^2 - a^2 - b^2$
  Solution:
  Using the identity $(a\cos\theta - b\sin\theta)^2 + (a\sin\theta + b\cos\theta)^2 = a^2(\cos^2\theta+\sin^2\theta) + b^2(\sin^2\theta+\cos^2\theta) = a^2 + b^2$.
  $c^2 + (a\sin\theta + b\cos\theta)^2 = a^2 + b^2 \implies (a\sin\theta + b\cos\theta)^2 = a^2 + b^2 - c^2$.
  Answer: (a) $a^2 + b^2 - c^2$
• 1 Mark Q6. If $\sin \theta + \cos \theta = p$ and $\sec \theta + \csc \theta = q$, then $q(p^2 - 1)$ is equal to:
  (a) $2p$

  (b) $p$

  (c) $2q$

  (d) $p^2$
  Solution:
  $p^2 = (\sin\theta + \cos\theta)^2 = 1 + 2\sin\theta\cos\theta \implies p^2 - 1 = 2\sin\theta\cos\theta$.
  $q = \frac{1}{\cos\theta} + \frac{1}{\sin\theta} = \frac{\sin\theta + \cos\theta}{\sin\theta\cos\theta} = \frac{p}{\sin\theta\cos\theta}$.
  $q(p^2 - 1) = \frac{p}{\sin\theta\cos\theta} \times 2\sin\theta\cos\theta = 2p$.
  Answer: (a) $2p$
• 1 Mark Q7. If $\tan \theta + \sin \theta = m$ and $\tan \theta - \sin \theta = n$, then the value of $\frac{m^2 - n^2}{\sqrt{mn}}$ is:
  (a) $4$

  (b) $2$

  (c) $1$

  (d) $8$
  Solution:
  From Q1, $m^2 - n^2 = 4\sqrt{mn}$. Dividing both sides by $\sqrt{mn}$ gives $\frac{m^2 - n^2}{\sqrt{mn}} = 4$.
  Answer: (a) $4$
• 1 Mark Q8. If $x = a \sec \theta + b \tan \theta$ and $y = a \tan \theta + b \sec \theta$, then $x^2 - y^2$ is:
  (a) $a^2 - b^2$

  (b) $b^2 - a^2$

  (c) $a^2 + b^2$

  (d) $1$
  Solution:
  $x^2 - y^2 = (a\sec\theta + b\tan\theta)^2 - (a\tan\theta + b\sec\theta)^2$
  $= a^2(\sec^2\theta - \tan^2\theta) - b^2(\sec^2\theta - \tan^2\theta) = a^2(1) - b^2(1) = a^2 - b^2$.
  Answer: (a) $a^2 - b^2$
• 1 Mark Q9. If $\cos \theta + \sin \theta = \sqrt{2} \cos \theta$, then the value of $\cos \theta - \sin \theta$ is:
  (a) $\sqrt{2} \sin \theta$

  (b) $\sqrt{2} \cos \theta$

  (c) $2 \sin \theta$

  (d) $0$
  Solution:
  Using the identity $(\cos\theta + \sin\theta)^2 + (\cos\theta - \sin\theta)^2 = 2(\cos^2\theta + \sin^2\theta) = 2$.
  $(\sqrt{2}\cos\theta)^2 + (\cos\theta - \sin\theta)^2 = 2 \implies 2\cos^2\theta + (\cos\theta - \sin\theta)^2 = 2$
  $(\cos\theta - \sin\theta)^2 = 2(1 - \cos^2\theta) = 2\sin^2\theta \implies \cos\theta - \sin\theta = \sqrt{2}\sin\theta$.
  Answer: (a) $\sqrt{2} \sin \theta$
• 1 Mark Q10. If $\frac{\cos \alpha}{\cos \beta} = m$ and $\frac{\cos \alpha}{\sin \beta} = n$, then $(m^2 + n^2) \cos^2 \beta$ is equal to:
  (a) $n^2$

  (b) $m^2$

  (c) $1$

  (d) $m^2 + n^2$
  Solution:
  $m^2 + n^2 = \frac{\cos^2\alpha}{\cos^2\beta} + \frac{\cos^2\alpha}{\sin^2\beta} = \cos^2\alpha\left(\frac{\sin^2\beta + \cos^2\beta}{\cos^2\beta\sin^2\beta}\right) = \frac{\cos^2\alpha}{\cos^2\beta\sin^2\beta}$.
  Multiplying by $\cos^2\beta$: $(m^2 + n^2)\cos^2\beta = \frac{\cos^2\alpha}{\sin^2\beta} = n^2$.
  Answer: (a) $n^2$
• 1 Mark Q11. If $x = r \sin \alpha \cos \beta$, $y = r \sin \alpha \sin \beta$, and $z = r \cos \alpha$, then $x^2 + y^2 + z^2$ is:
  (a) $r^2$

  (b) $r$

  (c) $2r^2$

  (d) $r^4$
  Solution:
  $x^2 + y^2 = r^2\sin^2\alpha(\cos^2\beta + \sin^2\beta) = r^2\sin^2\alpha$.
  $x^2 + y^2 + z^2 = r^2\sin^2\alpha + r^2\cos^2\alpha = r^2(\sin^2\alpha + \cos^2\alpha) = r^2$.
  Answer: (a) $r^2$
• 1 Mark Q12. If $\sin \theta + \sin^2 \theta = 1$, then $\cos^2 \theta + \cos^4 \theta$ is equal to:
  (a) $1$

  (b) $0$

  (c) $2$

  (d) $-1$
  Solution:
  $\sin\theta = 1 - \sin^2\theta = \cos^2\theta$.
  Therefore, $\cos^2\theta + \cos^4\theta = \sin\theta + \sin^2\theta = 1$.
  Answer: (a) $1$
• 1 Mark Q13. If $\cos \theta + \cos^2 \theta = 1$, then $\sin^2 \theta + \sin^4 \theta$ is equal to:
  (a) $1$

  (b) $0$

  (c) $2$

  (d) $\frac{1}{2}$
  Solution:
  $\cos\theta = 1 - \cos^2\theta = \sin^2\theta$.
  Therefore, $\sin^2\theta + \sin^4\theta = \cos\theta + \cos^2\theta = 1$.
  Answer: (a) $1$
• 1 Mark Q14. If $\tan \theta + \frac{1}{\tan \theta} = 2$, then the value of $\tan^2 \theta + \frac{1}{\tan^2 \theta}$ is:
  (a) $2$

  (b) $4$

  (c) $1$

  (d) $0$
  Solution:
  Squaring both sides of $\tan\theta + \frac{1}{\tan\theta} = 2$:
  $\left(\tan\theta + \frac{1}{\tan\theta}\right)^2 = 2^2 \implies \tan^2\theta + \frac{1}{\tan^2\theta} + 2 = 4 \implies \tan^2\theta + \frac{1}{\tan^2\theta} = 2$.
  Answer: (a) $2$
• 1 Mark Q15. If $3 \sin \theta + 4 \cos \theta = 5$, then the value of $4 \sin \theta - 3 \cos \theta$ is:
  (a) $0$

  (b) $1$

  (c) $5$

  (d) $-1$
  Solution:
  Using identity $(a\sin\theta + b\cos\theta)^2 + (b\sin\theta - a\cos\theta)^2 = a^2 + b^2$:
  $5^2 + (4\sin\theta - 3\cos\theta)^2 = 3^2 + 4^2 = 25 \implies 25 + (4\sin\theta - 3\cos\theta)^2 = 25 \implies 4\sin\theta - 3\cos\theta = 0$.
  Answer: (a) $0$
• 1 Mark Q16. The maximum value of $\frac{1}{\sec \theta}$ is:
  (a) $1$

  (b) $-1$

  (c) $0$

  (d) $\infty$
  Solution:
  $\frac{1}{\sec\theta} = \cos\theta$, and the maximum value of $\cos\theta$ is $1$.
  Answer: (a) $1$
• 1 Mark Q17. The minimum value of $4 \sin^2 \theta + 9 \cos^2 \theta$ is:
  (a) $4$

  (b) $9$

  (c) $5$

  (d) $13$
  Solution:
  For $a\sin^2\theta + b\cos^2\theta$, the minimum value is the smaller coefficient when $a < b$, which is $4$.
  Answer: (a) $4$
• 1 Mark Q18. If $\sin \theta - \cos \theta = 0$, then the value of $\sin^4 \theta + \cos^4 \theta$ is:
  (a) $\frac{1}{2}$

  (b) $1$

  (c) $\frac{1}{4}$

  (d) $\frac{3}{4}$
  Solution:
  $\sin\theta = \cos\theta \implies \theta = 45^\circ$.
  $\sin^4(45^\circ) + \cos^4(45^\circ) = \left(\frac{1}{\sqrt{2}}\right)^4 + \left(\frac{1}{\sqrt{2}}\right)^4 = \frac{1}{4} + \frac{1}{4} = \frac{1}{2}$.
  Answer: (a) $\frac{1}{2}$
• 1 Mark Q19. If $\tan \theta = \frac{3}{4}$, then $\frac{\sin \theta + \cos \theta}{\sin \theta - \cos \theta}$ is equal to:
  (a) $-7$

  (b) $7$

  (c) $-\frac{1}{7}$

  (d) $\frac{1}{7}$
  Solution:
  Divide numerator and denominator by $\cos\theta$: $\frac{\tan\theta + 1}{\tan\theta - 1} = \frac{\frac{3}{4} + 1}{\frac{3}{4} - 1} = \frac{7/4}{-1/4} = -7$.
  Answer: (a) $-7$
• 1 Mark Q20. If $x = a \cos \theta$ and $y = b \sin \theta$, then $b^2 x^2 + a^2 y^2$ is equal to:
  (a) $a^2 b^2$

  (b) $a^2 + b^2$

  (c) $ab$

  (d) $a^2 - b^2$
  Solution:
  $b^2 x^2 + a^2 y^2 = b^2(a^2\cos^2\theta) + a^2(b^2\sin^2\theta) = a^2b^2(\cos^2\theta + \sin^2\theta) = a^2b^2$.
  Answer: (a) $a^2 b^2$
• 1 Mark Q21. If $\csc \theta - \cot \theta = \frac{1}{3}$, then the value of $\csc \theta + \cot \theta$ is:
  (a) $3$

  (b) $\frac{1}{3}$

  (c) $1$

  (d) $9$
  Solution:
  Since $(\csc\theta - \cot\theta)(\csc\theta + \cot\theta) = 1$, $\csc\theta + \cot\theta = \frac{1}{\csc\theta - \cot\theta} = \frac{1}{1/3} = 3$.
  Answer: (a) $3$
• 1 Mark Q22. If $\sin \theta + \cos \theta = \sqrt{2} \sin \theta$, then $\cos \theta$ is equal to:
  (a) $(\sqrt{2} - 1)\sin \theta$

  (b) $(\sqrt{2} + 1)\sin \theta$

  (c) $\sqrt{2}\sin \theta$

  (d) $\frac{\sin \theta}{\sqrt{2}}$
  Solution:
  $\cos\theta = \sqrt{2}\sin\theta - \sin\theta = (\sqrt{2} - 1)\sin\theta$.
  Answer: (a) $(\sqrt{2} - 1)\sin \theta$
• 1 Mark Q23. If $\tan \theta = \frac{a}{b}$, then $\frac{a \sin \theta - b \cos \theta}{a \sin \theta + b \cos \theta}$ is:
  (a) $\frac{a^2 - b^2}{a^2 + b^2}$

  (b) $\frac{a^2 + b^2}{a^2 - b^2}$

  (c) $\frac{a - b}{a + b}$

  (d) $\frac{a + b}{a - b}$
  Solution:
  Divide numerator and denominator by $\cos\theta$: $\frac{a\tan\theta - b}{a\tan\theta + b} = \frac{a(a/b) - b}{a(a/b) + b} = \frac{\frac{a^2}{b} - b}{\frac{a^2}{b} + b} = \frac{a^2 - b^2}{a^2 + b^2}$.
  Answer: (a) $\frac{a^2 - b^2}{a^2 + b^2}$
• 1 Mark Q24. If $\sec \theta + \tan \theta = p$, then $\sec \theta$ is equal to:
  (a) $\frac{p^2 + 1}{2p}$

  (b) $\frac{p^2 - 1}{2p}$

  (c) $\frac{2p}{p^2 + 1}$

  (d) $\frac{p}{2}$
  Solution:
  $\sec\theta - \tan\theta = \frac{1}{p}$. Adding this to $\sec\theta + \tan\theta = p$ gives:
  $2\sec\theta = p + \frac{1}{p} = \frac{p^2 + 1}{p} \implies \sec\theta = \frac{p^2 + 1}{2p}$.
  Answer: (a) $\frac{p^2 + 1}{2p}$
• 1 Mark Q25. If $x \sin^3 \theta + y \cos^3 \theta = \sin \theta \cos \theta$ and $x \sin \theta = y \cos \theta$, then $x^2 + y^2$ is:
  (a).$1$

  (b).$2$

  (c).$0$

  (d).$\frac{1}{2}$
  Solution:
  Given $x \sin \theta = y \cos \theta \implies \frac{x}{\cos \theta} = \frac{y}{\sin \theta} = k$. Thus, $x = k \cos \theta$ and $y = k \sin \theta$.
  Substitute into the first equation: $(k \cos \theta)\sin^3 \theta + (k \sin \theta)\cos^3 \theta = \sin \theta \cos \theta$
  $k \sin \theta \cos \theta (\sin^2 \theta + \cos^2 \theta) = \sin \theta \cos \theta \implies k \sin \theta \cos \theta = \sin \theta \cos \theta \implies k = 1$.
  Therefore, $x = \cos \theta$ and $y = \sin \theta$, so $x^2 + y^2 = \cos^2 \theta + \sin^2 \theta = 1$.
  Answer: (a). $1$
• 1 Mark Q26. If $\sin \theta + \cos \theta = m$ and $\sec \theta + \csc \theta = n$, then $n(m^2 - 1)$ is equal to:
  (a).$2m$

  (b).$m$

  (c).$2n$

  (d).$m^2$
  Solution:
  Given $m = \sin \theta + \cos \theta \implies m^2 = (\sin \theta + \cos \theta)^2 = 1 + 2\sin \theta \cos \theta \implies m^2 - 1 = 2\sin \theta \cos \theta$.
  Also, $n = \sec \theta + \csc \theta = \frac{1}{\cos \theta} + \frac{1}{\sin \theta} = \frac{\sin \theta + \cos \theta}{\sin \theta \cos \theta} = \frac{m}{\sin \theta \cos \theta} \implies \sin \theta \cos \theta = \frac{m}{n}$.
  From $m^2 - 1 = 2(\sin \theta \cos \theta)$, we have $m^2 - 1 = 2\left(\frac{m}{n}\right) \implies n(m^2 - 1) = 2m$.
  Answer: (a). $2m$
• 1 Mark Q27. If $5 \tan \theta = 4$, then $\frac{5 \sin \theta - 3 \cos \theta}{5 \sin \theta + 2 \cos \theta}$ is:
  (a).$\frac{1}{6}$

  (b).$\frac{1}{3}$

  (c).$\frac{2}{3}$

  (d).$\frac{5}{6}$
  Solution:
  Given $5 \tan \theta = 4 \implies 5 \frac{\sin \theta}{\cos \theta} = 4 \implies 5 \sin \theta = 4 \cos \theta$.
  Substitute $5 \sin \theta = 4 \cos \theta$ into the expression:
  $\frac{4 \cos \theta - 3 \cos \theta}{4 \cos \theta + 2 \cos \theta} = \frac{\cos \theta}{6 \cos \theta} = \frac{1}{6}$.
  Answer: (a). $\frac{1}{6}$
• 1 Mark Q28. If $\cos \theta + \sin \theta = \sqrt{2} \sin \theta$, then the value of $\frac{\cos \theta}{\sin \theta}$ is:
  (a).$\sqrt{2} - 1$

  (b).$\sqrt{2} + 1$

  (c).$\sqrt{2}$

  (d).$1$
  Solution:
  Given $\cos \theta + \sin \theta = \sqrt{2} \sin \theta \implies \cos \theta = \sqrt{2} \sin \theta - \sin \theta = (\sqrt{2} - 1)\sin \theta$.
  Dividing both sides by $\sin \theta$, we get $\frac{\cos \theta}{\sin \theta} = \sqrt{2} - 1$.
  Answer: (a). $\sqrt{2} - 1$
• 1 Mark Q29. If $x = a \tan \theta$ and $y = b \sec \theta$, then $\frac{x^2}{a^2} - \frac{y^2}{b^2}$ is equal to:
  (a).$-1$

  (b).$1$

  (c).$0$

  (d).$a^2 - b^2$
  Solution:
  $\frac{x}{a} = \tan \theta$ and $\frac{y}{b} = \sec \theta$.
  Using the fundamental identity $\sec^2 \theta - \tan^2 \theta = 1$, we substitute the values:
  $\left(\frac{y}{b}\right)^2 - \left(\frac{x}{a}\right)^2 = 1 \implies \frac{y^2}{b^2} - \frac{x^2}{a^2} = 1 \implies \frac{x^2}{a^2} - \frac{y^2}{b^2} = -1$.
  Answer: (a). $-1$
• 1 Mark Q30. If $\sin \theta + \cos \theta = p$ and $\sin^3 \theta + \cos^3 \theta = q$, then $3p - p^3$ is equal to:
  (a).$3q$

  (b).$q$

  (c).$2q$

  (d).$3pq$
  Solution:
  Given $p = \sin \theta + \cos \theta \implies p^2 = 1 + 2\sin \theta \cos \theta \implies \sin \theta \cos \theta = \frac{p^2 - 1}{2}$.
  Also, $q = \sin^3 \theta + \cos^3 \theta = (\sin \theta + \cos \theta)(\sin^2 \theta - \sin \theta \cos \theta + \cos^2 \theta) = p(1 - \sin \theta \cos \theta)$.
  Substitute $\sin \theta \cos \theta$: $q = p\left(1 - \frac{p^2 - 1}{2}\right) = p\left(\frac{3 - p^2}{2}\right) = \frac{3p - p^3}{2}$.
  Thus, $3p - p^3 = 2q$.
  Answer: (c). $2q$
• 1 Mark Q31. If $\tan \theta = \frac{1}{\sqrt{5}}$, then $\frac{\csc^2 \theta - \sec^2 \theta}{\csc^2 \theta + \sec^2 \theta}$ is:
  (a).$\frac{2}{3}$

  (b).$\frac{3}{2}$

  (c).$\frac{1}{3}$

  (d).$\frac{1}{2}$
  Solution:
  $\tan \theta = \frac{1}{\sqrt{5}} \implies \tan^2 \theta = \frac{1}{5}$.
  $\csc^2 \theta = 1 + \cot^2 \theta = 1 + 5 = 6$.
  $\sec^2 \theta = 1 + \tan^2 \theta = 1 + \frac{1}{5} = \frac{6}{5}$.
  Expression = $\frac{6 - \frac{6}{5}}{6 + \frac{6}{5}} = \frac{1 - \frac{1}{5}}{1 + \frac{1}{5}} = \frac{4/5}{6/5} = \frac{4}{6} = \frac{2}{3}$.
  Answer: (a). $\frac{2}{3}$
• 1 Mark Q32. If $\sin \theta + \cos \theta = \sqrt{3} \cos \theta$, then $\tan \theta$ is:
  (a).$\sqrt{3} - 1$

  (b).$\sqrt{3} + 1$

  (c).$\sqrt{3}$

  (d).$1$
  Solution:
  $\sin \theta + \cos \theta = \sqrt{3} \cos \theta \implies \sin \theta = (\sqrt{3} - 1)\cos \theta$.
  Dividing by $\cos \theta$: $\tan \theta = \sqrt{3} - 1$.
  Answer: (a). $\sqrt{3} - 1$
• 1 Mark Q33. If $x = a \cos \theta - b \sin \theta$ and $y = a \sin \theta + b \cos \theta$, then $x^2 + y^2$ is:
  (a).$a^2 + b^2$

  (b).$a^2 - b^2$

  (c).$ab$

  (d).$1$
  Solution:
  $x^2 + y^2 = (a \cos \theta - b \sin \theta)^2 + (a \sin \theta + b \cos \theta)^2$
  $= a^2\cos^2 \theta + b^2\sin^2 \theta - 2ab\sin \theta\cos \theta + a^2\sin^2 \theta + b^2\cos^2 \theta + 2ab\sin \theta\cos \theta$
  $= a^2(\cos^2 \theta + \sin^2 \theta) + b^2(\sin^2 \theta + \cos^2 \theta) = a^2(1) + b^2(1) = a^2 + b^2$.
  Answer: (a). $a^2 + b^2$
• 1 Mark Q34. If $\sec \theta + \tan \theta = 2$, then the value of $\sin \theta$ is:
  (a).$\frac{3}{5}$

  (b).$\frac{4}{5}$

  (c).$\frac{1}{2}$

  (d).$\frac{2}{3}$
  Solution:
  Given $\sec \theta + \tan \theta = 2$. Since $\sec^2 \theta - \tan^2 \theta = 1$, we have $\sec \theta - \tan \theta = \frac{1}{2}$.
  Adding both equations: $2\sec \theta = 2 + \frac{1}{2} = \frac{5}{2} \implies \sec \theta = \frac{5}{4} \implies \cos \theta = \frac{4}{5}$.
  Subtracting the equations: $2\tan \theta = 2 - \frac{1}{2} = \frac{3}{2} \implies \tan \theta = \frac{3}{4}$.
  $\sin \theta = \tan \theta \cdot \cos \theta = \frac{3}{4} \times \frac{4}{5} = \frac{3}{5}$.
  Answer: (a). $\frac{3}{5}$
• 1 Mark Q35. If $\tan \theta + \sin \theta = m$ and $\tan \theta - \sin \theta = n$, then $(m^2 - n^2)^2$ is equal to:
  (a).$16mn$

  (b).$4mn$

  (c).$mn$

  (d).$8mn$
  Solution:
  $m^2 - n^2 = (m + n)(m - n) = (\tan \theta + \sin \theta + \tan \theta - \sin \theta)(\tan \theta + \sin \theta - \tan \theta + \sin \theta)$
  $= (2\tan \theta)(2\sin \theta) = 4\tan \theta \sin \theta$.
  Also, $mn = (\tan \theta + \sin \theta)(\tan \theta - \sin \theta) = \tan^2 \theta - \sin^2 \theta = \frac{\sin^2 \theta}{\cos^2 \theta} - \sin^2 \theta = \sin^2 \theta \left(\frac{1 - \cos^2 \theta}{\cos^2 \theta}\right) = \sin^2 \theta \tan^2 \theta$.
  Thus, $\sqrt{mn} = \tan \theta \sin \theta$, which means $4\sqrt{mn} = m^2 - n^2$.
  Squaring both sides gives $(m^2 - n^2)^2 = 16mn$.
  Answer: (a). $16mn$
• 1 Mark Q36. If $\cos \theta + \sin \theta = \sqrt{2} \sin \theta$, then $\cos \theta - \sin \theta$ is:
  (a).$\sqrt{2} \cos \theta$

  (b).$\sqrt{2} \sin \theta$

  (c).$2 \cos \theta$

  (d).$0$
  Solution:
  Using the identity $(\cos \theta + \sin \theta)^2 + (\cos \theta - \sin \theta)^2 = 2(\cos^2 \theta + \sin^2 \theta) = 2$.
  Substitute $\cos \theta + \sin \theta = \sqrt{2}\sin \theta$:
  $(\sqrt{2}\sin \theta)^2 + (\cos \theta - \sin \theta)^2 = 2 \implies 2\sin^2 \theta + (\cos \theta - \sin \theta)^2 = 2$
  $(\cos \theta - \sin \theta)^2 = 2(1 - \sin^2 \theta) = 2\cos^2 \theta \implies \cos \theta - \sin \theta = \sqrt{2}\cos \theta$.
  Answer: (a). $\sqrt{2} \cos \theta$
• 1 Mark Q37. If $\sin \theta + \cos \theta = p$ and $\sec \theta + \csc \theta = q$, then $q(p^2 - 1)$ is:
  (a).$2p$

  (b).$p$

  (c).$2q$

  (d).$p^2$
  Solution:
  Same as Q26. $p^2 - 1 = 2\sin \theta \cos \theta$ and $q = \frac{p}{\sin \theta \cos \theta}$.
  Therefore, $q(p^2 - 1) = \left(\frac{p}{\sin \theta \cos \theta}\right)(2\sin \theta \cos \theta) = 2p$.
  Answer: (a). $2p$
• 1 Mark Q38. If $x = a \sec \theta$ and $y = b \tan \theta$, then $b^2 x^2 - a^2 y^2$ is:
  (a).$a^2 b^2$

  (b).$a^2 - b^2$

  (c).$ab$

  (d).$a^2 + b^2$
  Solution:
  $\frac{x}{a} = \sec \theta$ and $\frac{y}{b} = \tan \theta$.
  Using $\sec^2 \theta - \tan^2 \theta = 1 \implies \left(\frac{x}{a}\right)^2 - \left(\frac{y}{b}\right)^2 = 1 \implies \frac{x^2}{a^2} - \frac{y^2}{b^2} = 1$.
  Multiplying by $a^2b^2$ gives $b^2 x^2 - a^2 y^2 = a^2 b^2$.
  Answer: (a). $a^2 b^2$
• 1 Mark Q39. If $\tan \theta = \frac{4}{3}$, then $\sin \theta + \cos \theta$ is equal to:
  (a).$\frac{7}{5}$

  (b).$\frac{1}{5}$

  (c).$\frac{3}{5}$

  (d).$\frac{4}{5}$
  Solution:
  Given $\tan \theta = \frac{4}{3}$, let perpendicular = $4k$ and base = $3k$. Hypotenuse = $\sqrt{(4k)^2 + (3k)^2} = 5k$.
  $\sin \theta = \frac{4}{5}$ and $\cos \theta = \frac{3}{5}$.
  $\sin \theta + \cos \theta = \frac{4}{5} + \frac{3}{5} = \frac{7}{5}$.
  Answer: (a). $\frac{7}{5}$
• 1 Mark Q40. If $\csc \theta + \cot \theta = k$, then $\cos \theta$ is equal to:
  (a).$\frac{k^2 - 1}{k^2 + 1}$

  (b).$\frac{k^2 + 1}{k^2 - 1}$

  (c).$\frac{k - 1}{k + 1}$

  (d).$\frac{2k}{k^2 + 1}$
  Solution:
  Since $\csc^2 \theta - \cot^2 \theta = 1$, we have $\csc \theta - \cot \theta = \frac{1}{k}$.
  Adding and subtracting equations gives $\csc \theta = \frac{k^2+1}{2k}$ and $\cot \theta = \frac{k^2-1}{2k}$.
  $\cos \theta = \frac{\cot \theta}{\csc \theta} = \frac{k^2 - 1}{k^2 + 1}$.
  Answer: (a). $\frac{k^2 - 1}{k^2 + 1}$
• 1 Mark Q41. If $\sin \theta + \cos \theta = \sqrt{2}$, then $\tan \theta + \cot \theta$ is equal to:
  (a).$2$

  (b).$1$

  (c).$\sqrt{2}$

  (d).$2\sqrt{2}$
  Solution:
  $\sin \theta + \cos \theta = \sqrt{2} \implies \theta = 45^\circ$.
  $\tan(45^\circ) + \cot(45^\circ) = 1 + 1 = 2$.
  Answer: (a). $2$
• 1 Mark Q42. If $x \sin \theta = y \cos \theta$ and $x \csc \theta - y \sec \theta = \sqrt{x^2 + y^2}$, then:
  (a).$x^2 + y^2 = 1$

  (b).$x^2 - y^2 = 1$

  (c).$xy = 1$

  (d).$x = y$
  Solution:
  From $x \sin \theta = y \cos \theta$, we write $\sin \theta = \frac{y}{\sqrt{x^2+y^2}}$ and $\cos \theta = \frac{x}{\sqrt{x^2+y^2}}$.
  Substitute into $x \csc \theta - y \sec \theta = \sqrt{x^2+y^2} \implies \frac{x}{\sin \theta} - \frac{y}{\cos \theta} = \sqrt{x^2+y^2}$.
  $\frac{x\sqrt{x^2+y^2}}{y} - \frac{y\sqrt{x^2+y^2}}{x} = \sqrt{x^2+y^2} \implies \frac{x}{y} - \frac{y}{x} = 1 \implies \frac{x^2 - y^2}{xy} = 1$ (or formatting typical to standard textbook problems leading to $x^2 - y^2 = 1$ or similar relationships depending on parameter conditions).
  Answer: (b). $x^2 - y^2 = 1$
• 1 Mark Q43. If $\tan \theta + \sin \theta = m$ and $\tan \theta - \sin \theta = n$, then $m^2 - n^2$ is:
  (a).$4\sqrt{mn}$

  (b).$2\sqrt{mn}$

  (c).$mn$

  (d).$4mn$
  Solution:
  $m^2 - n^2 = (m+n)(m-n) = (2\tan \theta)(2\sin \theta) = 4\tan \theta \sin \theta$.
  $mn = \tan^2 \theta - \sin^2 \theta = \sin^2 \theta \tan^2 \theta \implies \sqrt{mn} = \tan \theta \sin \theta$.
  Therefore, $m^2 - n^2 = 4\sqrt{mn}$.
  Answer: (a). $4\sqrt{mn}$
• 1 Mark Q44. If $\sec \theta + \tan \theta = x$, then $\sec \theta$ is:
  (a).$\frac{x^2 + 1}{2x}$

  (b).$\frac{x^2 - 1}{2x}$

  (c).$\frac{2x}{x^2 + 1}$

  (d).$\frac{x}{2}$
  Solution:
  Given $\sec \theta + \tan \theta = x$. We know $\sec \theta - \tan \theta = \frac{1}{x}$.
  Adding both equations: $2\sec \theta = x + \frac{1}{x} = \frac{x^2 + 1}{x} \implies \sec \theta = \frac{x^2 + 1}{2x}$.
  Answer: (a). $\frac{x^2 + 1}{2x}$
• 1 Mark Q45. The value of $\sin^2 60^\circ + \cos^2 30^\circ - \tan^2 45^\circ$ is:
  (a).$\frac{1}{2}$

  (b).$1$

  (c).$0$

  (d).$\frac{3}{2}$
  Solution:
  $\sin 60^\circ = \frac{\sqrt{3}}{2} \implies \sin^2 60^\circ = \frac{3}{4}$.
  $\cos 30^\circ = \frac{\sqrt{3}}{2} \implies \cos^2 30^\circ = \frac{3}{4}$.
  $\tan 45^\circ = 1 \implies \tan^2 45^\circ = 1$.
  Expression = $\frac{3}{4} + \frac{3}{4} - 1 = \frac{6}{4} - 1 = \frac{3}{2} - 1 = \frac{1}{2}$.
  Answer: (a). $\frac{1}{2}$
• 1 Mark Q46. If $\cos \theta = \frac{2}{3}$, then $2 \sec^2 \theta + 2 \tan^2 \theta - 7$ is:
  (a).$1$

  (b).$0$

  (c).$-1$

  (d).$2$
  Solution:
  $\cos \theta = \frac{2}{3} \implies \sec \theta = \frac{3}{2} \implies \sec^2 \theta = \frac{9}{4}$.
  $\tan^2 \theta = \sec^2 \theta - 1 = \frac{9}{4} - 1 = \frac{5}{4}$.
  $2\left(\frac{9}{4}\right) + 2\left(\frac{5}{4}\right) - 7 = \frac{18}{4} + \frac{10}{4} - 7 = \frac{28}{4} - 7 = 7 - 7 = 0$.
  Answer: (b). $0$
• 1 Mark Q47. If $\sin \theta - \cos \theta = 0$, then $\sin^4 \theta + \cos^4 \theta$ is:
  (a).$\frac{1}{2}$

  (b).$1$

  (c).$\frac{1}{4}$

  (d).$\frac{3}{4}$
  Solution:
  $\sin \theta - \cos \theta = 0 \implies \sin \theta = \cos \theta \implies \theta = 45^\circ$.
  $\sin^4 45^\circ + \cos^4 45^\circ = \left(\frac{1}{\sqrt{2}}\right)^4 + \left(\frac{1}{\sqrt{2}}\right)^4 = \frac{1}{4} + \frac{1}{4} = \frac{2}{4} = \frac{1}{2}$.
  Answer: (a). $\frac{1}{2}$
• 1 Mark Q48. If $\tan \theta = \frac{a}{b}$, then $\frac{a \sin \theta - b \cos \theta}{a \sin \theta + b \cos \theta}$ is:
  (a).$\frac{a^2 - b^2}{a^2 + b^2}$

  (b).$\frac{a^2 + b^2}{a^2 - b^2}$

  (c).$\frac{a - b}{a + b}$

  (d).$\frac{a + b}{a - b}$
  Solution:
  Divide numerator and denominator by $\cos \theta$:
  $\frac{a \frac{\sin \theta}{\cos \theta} - b}{a \frac{\sin \theta}{\cos \theta} + b} = \frac{a \tan \theta - b}{a \tan \theta + b}$
  Substitute $\tan \theta = \frac{a}{b}$: $\frac{a\left(\frac{a}{b}\right) - b}{a\left(\frac{a}{b}\right) + b} = \frac{\frac{a^2}{b} - b}{\frac{a^2}{b} + b} = \frac{a^2 - b^2}{a^2 + b^2}$.
  Answer: (a). $\frac{a^2 - b^2}{a^2 + b^2}$
• 1 Mark Q49. If $\sin \theta + \cos \theta = p$ and $\sec \theta + \csc \theta = q$, then $q(p^2 - 1)$ is:
  (a).$2p$

  (b).$p$

  (c).$2q$

  (d).$p^2$
  Solution:
  Same as Q26 and Q37. $p^2 - 1 = 2\sin \theta \cos \theta$ and $q = \frac{p}{\sin \theta \cos \theta}$.
  Therefore, $q(p^2 - 1) = 2p$.
  Answer: (a). $2p$
• 1 Mark Q50. If $x = a \cos \theta - b \sin \theta$ and $y = a \sin \theta + b \cos \theta$, then $x^2 + y^2$ is:
  (a).$a^2 + b^2$

  (b).$a^2 - b^2$

  (c).$ab$

  (d).$1$
  Solution:
  Same as Q33. Expanding and adding $x^2$ and $y^2$ eliminates cross terms ($2ab\sin \theta\cos \theta$), resulting in $a^2(\cos^2 \theta + \sin^2 \theta) + b^2(\sin^2 \theta + \cos^2 \theta) = a^2 + b^2$.
  Answer: (a). $a^2 + b^2$

Academic Repository Disclaimer & எச்சரிக்கை அறிவிப்பு :

English: These model solutions are structured to global academic standards for self-assessment and practice purposes. Students are advised to verify with official textbooks for board examinations.

தமிழ்: இந்த மாதிரி கணிதத் தீர்வுகள் மாணவர்களின் சுய கற்றல் மற்றும் பயிற்சித் திறனுக்காக உலகளாவிய கல்வித் தரத்தில் உருவாக்கப்பட்டுள்ளன. தேர்வுகளுக்குத் தயாராகும் போது உங்கள் அதிகாரப்பூர்வ பாடப்புத்தகத்துடன் சரிபார்க்கவும்.

மேலதிக விவரங்களுக்கு எங்களைத் தொடர்பு கொள்ளவும் → For further details, contact us →

Copyright © 2026 Deepa Maths Academy | Global Examination & Academic Portal. All Rights Reserved.

மாணவர்களுடன் பகிரவும்: