Board Previous Year Questions (PYQs)

These are exam-style questions modelled on CBSE and State Board papers from recent years. Each is fully solved with the reasoning a board examiner expects.

Scoring tip: Always state the identity or value you use. In identity proofs, show every algebraic step — examiners award method marks generously.

Work through all 26. They span 1-mark, 2-mark, 3-mark and 5-mark patterns.

PYQ 1 (1 mark): If sin⁡θ=35\sin\theta = \tfrac{3}{5}, find cos⁡θ\cos\theta.

Solution:

  1. cos⁡θ=1−925=1625=45\cos\theta = \sqrt{1 - \tfrac{9}{25}} = \sqrt{\tfrac{16}{25}} = \tfrac45.

Final Answer: cos⁡θ=45\cos\theta = \tfrac45.

PYQ 2 (1 mark): Evaluate 2sin⁡30°cos⁡30°2\sin 30° \cos 30°.

Solution:

  1. =2⋅12⋅32=32= 2\cdot\tfrac12\cdot\tfrac{\sqrt3}{2} = \tfrac{\sqrt3}{2}.

Final Answer: 32\tfrac{\sqrt3}{2}.

PYQ 3 (1 mark): Evaluate tan⁡35°cot⁡55°\dfrac{\tan 35°}{\cot 55°}.

Solution:

  1. cot⁡55°=tan⁡35°\cot 55° = \tan 35°, so the ratio is 1.

Final Answer: 1.

PYQ 4 (1 mark): If sec⁡θ=178\sec\theta = \tfrac{17}{8}, find tan⁡θ\tan\theta.

Solution:

  1. tan⁡2θ=sec⁡2θ−1=28964−1=22564\tan^2\theta = \sec^2\theta - 1 = \tfrac{289}{64} - 1 = \tfrac{225}{64}.
  2. tan⁡θ=158\tan\theta = \tfrac{15}{8}.

Final Answer: tan⁡θ=158\tan\theta = \tfrac{15}{8}.

PYQ 5 (2 marks): Evaluate sin⁡230°+cos⁡245°+tan⁡260°\sin^2 30° + \cos^2 45° + \tan^2 60°.

Solution:

  1. sin⁡230°=14\sin^2 30° = \tfrac14; cos⁡245°=12\cos^2 45° = \tfrac12; tan⁡260°=3\tan^2 60° = 3.
  2. 14+12+3=1+2+124=154\tfrac14 + \tfrac12 + 3 = \tfrac{1 + 2 + 12}{4} = \tfrac{15}{4}.

Final Answer: 154\tfrac{15}{4}.

PYQ 6 (2 marks): If tan⁡A=cot⁡B\tan A = \cot B, prove that A+B=90°A + B = 90° (acute angles).

Solution:

  1. cot⁡B=tan⁡(90°−B)\cot B = \tan(90° - B), so tan⁡A=tan⁡(90°−B)\tan A = \tan(90° - B).
  2. Hence A=90°−BA = 90° - B, i.e. A+B=90°A + B = 90°.

Final Answer: Proved.

PYQ 7 (2 marks): If sin⁡(A−B)=12\sin(A - B) = \tfrac12 and cos⁡(A+B)=12\cos(A + B) = \tfrac12, with A>BA > B, find AA and BB.

Solution:

  1. sin⁡(A−B)=12⇒A−B=30°\sin(A - B) = \tfrac12 \Rightarrow A - B = 30°.
  2. cos⁡(A+B)=12⇒A+B=60°\cos(A + B) = \tfrac12 \Rightarrow A + B = 60°.
  3. Adding: 2A=90°⇒A=45°2A = 90° \Rightarrow A = 45°; B=15°B = 15°.

Final Answer: A=45°A = 45°, B=15°B = 15°.

PYQ 8 (2 marks): Prove cos⁡θ1+sin⁡θ+1+sin⁡θcos⁡θ=2sec⁡θ\dfrac{\cos\theta}{1 + \sin\theta} + \dfrac{1 + \sin\theta}{\cos\theta} = 2\sec\theta.

Solution:

  1. LHS =cos⁡2θ+(1+sin⁡θ)2cos⁡θ(1+sin⁡θ)= \dfrac{\cos^2\theta + (1+\sin\theta)^2}{\cos\theta(1+\sin\theta)}.
  2. Numerator =cos⁡2θ+1+2sin⁡θ+sin⁡2θ=2+2sin⁡θ=2(1+sin⁡θ)= \cos^2\theta + 1 + 2\sin\theta + \sin^2\theta = 2 + 2\sin\theta = 2(1+\sin\theta).
  3. LHS =2(1+sin⁡θ)cos⁡θ(1+sin⁡θ)=2cos⁡θ=2sec⁡θ= \dfrac{2(1+\sin\theta)}{\cos\theta(1+\sin\theta)} = \dfrac{2}{\cos\theta} = 2\sec\theta.

Final Answer: Proved.

PYQ 9 (2 marks): If 3cot⁡A=43\cot A = 4, find 1−tan⁡2A1+tan⁡2A\dfrac{1 - \tan^2 A}{1 + \tan^2 A}.

Solution:

  1. cot⁡A=43\cot A = \tfrac43, so tan⁡A=34\tan A = \tfrac34, tan⁡2A=916\tan^2 A = \tfrac{9}{16}.
  2. 1−9/161+9/16=7/1625/16=725\dfrac{1 - 9/16}{1 + 9/16} = \dfrac{7/16}{25/16} = \tfrac{7}{25}.

Final Answer: 725\tfrac{7}{25}.

PYQ 10 (3 marks): Prove (sin⁡θ+cos⁡θ)(tan⁡θ+cot⁡θ)=sec⁡θ+csc⁡θ(\sin\theta + \cos\theta)(\tan\theta + \cot\theta) = \sec\theta + \csc\theta.

Solution:

  1. tan⁡θ+cot⁡θ=1sin⁡θcos⁡θ\tan\theta + \cot\theta = \dfrac{1}{\sin\theta\cos\theta}.
  2. LHS =(sin⁡θ+cos⁡θ)⋅1sin⁡θcos⁡θ=sin⁡θsin⁡θcos⁡θ+cos⁡θsin⁡θcos⁡θ= (\sin\theta + \cos\theta)\cdot\dfrac{1}{\sin\theta\cos\theta} = \dfrac{\sin\theta}{\sin\theta\cos\theta} + \dfrac{\cos\theta}{\sin\theta\cos\theta}.
  3. =1cos⁡θ+1sin⁡θ=sec⁡θ+csc⁡θ= \dfrac{1}{\cos\theta} + \dfrac{1}{\sin\theta} = \sec\theta + \csc\theta.

Final Answer: Proved.

PYQ 11 (3 marks): Prove tan⁡θ1−cot⁡θ+cot⁡θ1−tan⁡θ=1+tan⁡θ+cot⁡θ\dfrac{\tan\theta}{1 - \cot\theta} + \dfrac{\cot\theta}{1 - \tan\theta} = 1 + \tan\theta + \cot\theta.

Solution:

  1. From the worked Example (Section 6), the LHS =1+sec⁡θcsc⁡θ=1+1sin⁡θcos⁡θ= 1 + \sec\theta\csc\theta = 1 + \dfrac{1}{\sin\theta\cos\theta}.
  2. tan⁡θ+cot⁡θ=1sin⁡θcos⁡θ\tan\theta + \cot\theta = \dfrac{1}{\sin\theta\cos\theta}, so RHS =1+1sin⁡θcos⁡θ= 1 + \dfrac{1}{\sin\theta\cos\theta}.
  3. LHS = RHS.

Final Answer: Proved.

PYQ 12 (3 marks): Prove sin⁡θ−2sin⁡3θ2cos⁡3θ−cos⁡θ=tan⁡θ\dfrac{\sin\theta - 2\sin^3\theta}{2\cos^3\theta - \cos\theta} = \tan\theta.

Solution:

  1. Numerator =sin⁡θ(1−2sin⁡2θ)= \sin\theta(1 - 2\sin^2\theta).
  2. Denominator =cos⁡θ(2cos⁡2θ−1)= \cos\theta(2\cos^2\theta - 1).
  3. Note 1−2sin⁡2θ=2cos⁡2θ−11 - 2\sin^2\theta = 2\cos^2\theta - 1 (both equal cos⁡2θ\cos 2\theta).
  4. So the fraction =sin⁡θcos⁡θ=tan⁡θ= \dfrac{\sin\theta}{\cos\theta} = \tan\theta.

Final Answer: Proved.

PYQ 13 (3 marks): Evaluate cos⁡45°sec⁡30°+csc⁡30°\dfrac{\cos 45°}{\sec 30° + \csc 30°}.

Solution:

  1. cos⁡45°=12\cos 45° = \tfrac{1}{\sqrt2}; sec⁡30°=23\sec 30° = \tfrac{2}{\sqrt3}; csc⁡30°=2\csc 30° = 2.
  2. Denominator =23+2=2+233= \tfrac{2}{\sqrt3} + 2 = \tfrac{2 + 2\sqrt3}{\sqrt3}.
  3. 1/2(2+23)/3=322(1+3)=3(3−1)22(2)=3−342\dfrac{1/\sqrt2}{(2+2\sqrt3)/\sqrt3} = \dfrac{\sqrt3}{2\sqrt2(1+\sqrt3)} = \dfrac{\sqrt3(\sqrt3-1)}{2\sqrt2(2)} = \dfrac{3 - \sqrt3}{4\sqrt2}.

Final Answer: 3−342\dfrac{3 - \sqrt3}{4\sqrt2}.

PYQ 14 (3 marks): If sin⁡θ+cos⁡θ=3\sin\theta + \cos\theta = \sqrt3, prove tan⁡θ+cot⁡θ=1\tan\theta + \cot\theta = 1.

Solution:

  1. Square: 1+2sin⁡θcos⁡θ=3⇒sin⁡θcos⁡θ=11 + 2\sin\theta\cos\theta = 3 \Rightarrow \sin\theta\cos\theta = 1.
  2. tan⁡θ+cot⁡θ=1sin⁡θcos⁡θ=11=1\tan\theta + \cot\theta = \dfrac{1}{\sin\theta\cos\theta} = \dfrac{1}{1} = 1.

Final Answer: Proved.

PYQ 15 (3 marks): If tan⁡θ=17\tan\theta = \tfrac{1}{\sqrt7}, evaluate csc⁡2θ−sec⁡2θcsc⁡2θ+sec⁡2θ\dfrac{\csc^2\theta - \sec^2\theta}{\csc^2\theta + \sec^2\theta}.

Solution:

  1. csc⁡2θ=1+cot⁡2θ=1+7=8\csc^2\theta = 1 + \cot^2\theta = 1 + 7 = 8; sec⁡2θ=1+tan⁡2θ=1+17=87\sec^2\theta = 1 + \tan^2\theta = 1 + \tfrac17 = \tfrac87.
  2. 8−8/78+8/7=48/764/7=4864=34\dfrac{8 - 8/7}{8 + 8/7} = \dfrac{48/7}{64/7} = \dfrac{48}{64} = \tfrac34.

Final Answer: 34\tfrac34.

PYQ 16 (3 marks): Prove sec⁡θ(1−sin⁡θ)(sec⁡θ+tan⁡θ)=1\sec\theta(1 - \sin\theta)(\sec\theta + \tan\theta) = 1.

Solution:

  1. sec⁡θ+tan⁡θ=1+sin⁡θcos⁡θ\sec\theta + \tan\theta = \dfrac{1 + \sin\theta}{\cos\theta}.
  2. LHS =1cos⁡θ(1−sin⁡θ)⋅1+sin⁡θcos⁡θ=(1−sin⁡θ)(1+sin⁡θ)cos⁡2θ= \dfrac{1}{\cos\theta}(1 - \sin\theta)\cdot\dfrac{1 + \sin\theta}{\cos\theta} = \dfrac{(1-\sin\theta)(1+\sin\theta)}{\cos^2\theta}.
  3. =1−sin⁡2θcos⁡2θ=cos⁡2θcos⁡2θ=1= \dfrac{1 - \sin^2\theta}{\cos^2\theta} = \dfrac{\cos^2\theta}{\cos^2\theta} = 1.

Final Answer: Proved.

PYQ 17 (5 marks): Prove cos⁡θ1−tan⁡θ+sin⁡2θsin⁡θ−cos⁡θ=cos⁡θ+sin⁡θ\dfrac{\cos\theta}{1 - \tan\theta} + \dfrac{\sin^2\theta}{\sin\theta - \cos\theta} = \cos\theta + \sin\theta.

Solution:

  1. cos⁡θ1−tan⁡θ=cos⁡2θcos⁡θ−sin⁡θ\dfrac{\cos\theta}{1 - \tan\theta} = \dfrac{\cos^2\theta}{\cos\theta - \sin\theta}.
  2. sin⁡2θsin⁡θ−cos⁡θ=−sin⁡2θcos⁡θ−sin⁡θ\dfrac{\sin^2\theta}{\sin\theta - \cos\theta} = -\dfrac{\sin^2\theta}{\cos\theta - \sin\theta}.
  3. Sum =cos⁡2θ−sin⁡2θcos⁡θ−sin⁡θ=(cos⁡θ−sin⁡θ)(cos⁡θ+sin⁡θ)cos⁡θ−sin⁡θ=cos⁡θ+sin⁡θ= \dfrac{\cos^2\theta - \sin^2\theta}{\cos\theta - \sin\theta} = \dfrac{(\cos\theta-\sin\theta)(\cos\theta+\sin\theta)}{\cos\theta - \sin\theta} = \cos\theta + \sin\theta.

Final Answer: Proved.

PYQ 18 (5 marks): Prove tan⁡θ+sec⁡θ−1tan⁡θ−sec⁡θ+1=1+sin⁡θcos⁡θ\dfrac{\tan\theta + \sec\theta - 1}{\tan\theta - \sec\theta + 1} = \dfrac{1 + \sin\theta}{\cos\theta}.

Solution:

  1. Write −1=−(sec⁡2θ−tan⁡2θ)-1 = -(\sec^2\theta - \tan^2\theta) in the numerator: tan⁡θ+sec⁡θ−(sec⁡2θ−tan⁡2θ)\tan\theta + \sec\theta - (\sec^2\theta - \tan^2\theta).
  2. =tan⁡θ+sec⁡θ−(sec⁡θ−tan⁡θ)(sec⁡θ+tan⁡θ)=(tan⁡θ+sec⁡θ)[1−(sec⁡θ−tan⁡θ)]= \tan\theta + \sec\theta - (\sec\theta - \tan\theta)(\sec\theta + \tan\theta) = (\tan\theta + \sec\theta)[1 - (\sec\theta - \tan\theta)].
  3. Numerator =(tan⁡θ+sec⁡θ)(1−sec⁡θ+tan⁡θ)= (\tan\theta + \sec\theta)(1 - \sec\theta + \tan\theta), and the denominator is (tan⁡θ−sec⁡θ+1)(\tan\theta - \sec\theta + 1).
  4. So the fraction =tan⁡θ+sec⁡θ=sin⁡θcos⁡θ+1cos⁡θ=1+sin⁡θcos⁡θ= \tan\theta + \sec\theta = \dfrac{\sin\theta}{\cos\theta} + \dfrac{1}{\cos\theta} = \dfrac{1 + \sin\theta}{\cos\theta}.

Final Answer: Proved.

PYQ 19 (5 marks): Evaluate 4cot⁡230°+1sin⁡230°−2cos⁡245°−sin⁡20°\dfrac{4}{\cot^2 30°} + \dfrac{1}{\sin^2 30°} - 2\cos^2 45° - \sin^2 0°.

Solution:

  1. cot⁡230°=3\cot^2 30° = 3, so 43\tfrac43. sin⁡230°=14\sin^2 30° = \tfrac14, so 11/4=4\tfrac{1}{1/4} = 4.
  2. 2cos⁡245°=12\cos^2 45° = 1; sin⁡20°=0\sin^2 0° = 0.
  3. 43+4−1−0=43+3=133\tfrac43 + 4 - 1 - 0 = \tfrac43 + 3 = \tfrac{13}{3}.

Final Answer: 133\tfrac{13}{3}.

PYQ 20 (5 marks): If sec⁡θ+tan⁡θ=m\sec\theta + \tan\theta = m, show that m2−1m2+1=sin⁡θ\dfrac{m^2 - 1}{m^2 + 1} = \sin\theta.

Solution:

  1. m2=sec⁡2θ+2sec⁡θtan⁡θ+tan⁡2θm^2 = \sec^2\theta + 2\sec\theta\tan\theta + \tan^2\theta.
  2. m2−1=tan⁡2θ+2sec⁡θtan⁡θ+tan⁡2θ=2tan⁡θ(tan⁡θ+sec⁡θ)=2tan⁡θ mm^2 - 1 = \tan^2\theta + 2\sec\theta\tan\theta + \tan^2\theta = 2\tan\theta(\tan\theta + \sec\theta) = 2\tan\theta\, m (using sec⁡2−1=tan⁡2\sec^2 - 1 = \tan^2).
  3. m2+1=sec⁡2θ+2sec⁡θtan⁡θ+sec⁡2θ=2sec⁡θ(sec⁡θ+tan⁡θ)=2sec⁡θ mm^2 + 1 = \sec^2\theta + 2\sec\theta\tan\theta + \sec^2\theta = 2\sec\theta(\sec\theta + \tan\theta) = 2\sec\theta\, m.
  4. m2−1m2+1=2tan⁡θ m2sec⁡θ m=tan⁡θsec⁡θ=sin⁡θ\dfrac{m^2 - 1}{m^2 + 1} = \dfrac{2\tan\theta\, m}{2\sec\theta\, m} = \dfrac{\tan\theta}{\sec\theta} = \sin\theta.

Final Answer: Proved.

PYQ 21 (3 marks): If cos⁡θ+sin⁡θ=2cos⁡θ\cos\theta + \sin\theta = \sqrt2\cos\theta, show that cos⁡θ−sin⁡θ=2sin⁡θ\cos\theta - \sin\theta = \sqrt2\sin\theta.

Solution:

  1. From the given, sin⁡θ=(2−1)cos⁡θ\sin\theta = (\sqrt2 - 1)\cos\theta.
  2. cos⁡θ−sin⁡θ=cos⁡θ−(2−1)cos⁡θ=(2−2)cos⁡θ=2(2−1)cos⁡θ\cos\theta - \sin\theta = \cos\theta - (\sqrt2 - 1)\cos\theta = (2 - \sqrt2)\cos\theta = \sqrt2(\sqrt2 - 1)\cos\theta.
  3. =2sin⁡θ= \sqrt2\sin\theta (since sin⁡θ=(2−1)cos⁡θ\sin\theta = (\sqrt2-1)\cos\theta).

Final Answer: Proved.

PYQ 22 (3 marks): Evaluate cos⁡48°⋅csc⁡42°\cos 48° \cdot \csc 42°.

Solution:

  1. csc⁡42°=sec⁡48°\csc 42° = \sec 48°, so cos⁡48°csc⁡42°=cos⁡48°sec⁡48°=1\cos 48°\csc 42° = \cos 48°\sec 48° = 1.

Final Answer: 1.

PYQ 23 (3 marks): If tan⁡2A=cot⁡(A−18°)\tan 2A = \cot(A - 18°), where 2A2A is acute, find AA.

Solution:

  1. cot⁡(A−18°)=tan⁡(90°−(A−18°))=tan⁡(108°−A)\cot(A - 18°) = \tan(90° - (A - 18°)) = \tan(108° - A).
  2. 2A=108°−A⇒3A=108°⇒A=36°2A = 108° - A \Rightarrow 3A = 108° \Rightarrow A = 36°.

Final Answer: A=36°A = 36°.

PYQ 24 (5 marks): Prove sin⁡θcot⁡θ+csc⁡θ=2+sin⁡θcot⁡θ−csc⁡θ\dfrac{\sin\theta}{\cot\theta + \csc\theta} = 2 + \dfrac{\sin\theta}{\cot\theta - \csc\theta}.

Solution:

  1. RHS second term: sin⁡θcot⁡θ−csc⁡θ=sin⁡θcos⁡θ−1sin⁡θ=sin⁡2θcos⁡θ−1\dfrac{\sin\theta}{\cot\theta - \csc\theta} = \dfrac{\sin\theta}{\frac{\cos\theta - 1}{\sin\theta}} = \dfrac{\sin^2\theta}{\cos\theta - 1}.
  2. =1−cos⁡2θcos⁡θ−1=−(cos⁡2θ−1)cos⁡θ−1=−(cos⁡θ+1)= \dfrac{1 - \cos^2\theta}{\cos\theta - 1} = \dfrac{-(\cos^2\theta - 1)}{\cos\theta - 1} = -(\cos\theta + 1).
  3. LHS: sin⁡θcot⁡θ+csc⁡θ=sin⁡2θcos⁡θ+1=1−cos⁡2θcos⁡θ+1=1−cos⁡θ\dfrac{\sin\theta}{\cot\theta + \csc\theta} = \dfrac{\sin^2\theta}{\cos\theta + 1} = \dfrac{1 - \cos^2\theta}{\cos\theta + 1} = 1 - \cos\theta.
  4. RHS =2+[−(cos⁡θ+1)]=2−cos⁡θ−1=1−cos⁡θ== 2 + [-(\cos\theta + 1)] = 2 - \cos\theta - 1 = 1 - \cos\theta = LHS.

Final Answer: Proved.

PYQ 25 (3 marks): Evaluate sin⁡25°+sin⁡210°+sin⁡280°+sin⁡285°\sin^2 5° + \sin^2 10° + \sin^2 80° + \sin^2 85°.

Solution:

  1. sin⁡80°=cos⁡10°\sin 80° = \cos 10° and sin⁡85°=cos⁡5°\sin 85° = \cos 5°.
  2. sin⁡25°+cos⁡25°+sin⁡210°+cos⁡210°=1+1=2\sin^2 5° + \cos^2 5° + \sin^2 10° + \cos^2 10° = 1 + 1 = 2.

Final Answer: 2.

PYQ 26 (5 marks): If x=asin⁡θ+bcos⁡θx = a\sin\theta + b\cos\theta and y=acos⁡θ−bsin⁡θy = a\cos\theta - b\sin\theta, prove x2+y2=a2+b2x^2 + y^2 = a^2 + b^2.

Solution:

  1. x2=a2sin⁡2θ+2absin⁡θcos⁡θ+b2cos⁡2θx^2 = a^2\sin^2\theta + 2ab\sin\theta\cos\theta + b^2\cos^2\theta.
  2. y2=a2cos⁡2θ−2absin⁡θcos⁡θ+b2sin⁡2θy^2 = a^2\cos^2\theta - 2ab\sin\theta\cos\theta + b^2\sin^2\theta.
  3. Add: x2+y2=a2(sin⁡2θ+cos⁡2θ)+b2(cos⁡2θ+sin⁡2θ)=a2+b2x^2 + y^2 = a^2(\sin^2\theta + \cos^2\theta) + b^2(\cos^2\theta + \sin^2\theta) = a^2 + b^2.

Final Answer: Proved.