How to Use This Section

This is your practice powerhouse for trigonometry. Below are 30+ fully solved problems covering the whole chapter — the six ratios, standard angles, identities, and complementary angles — roughly easy to hard.

How to read: Cover the solution, attempt each yourself, then check the steps. State the formula or identity you use — examiners reward it.

Keep these handy:

  • sin⁡2θ+cos⁡2θ=1\sin^2\theta + \cos^2\theta = 1; sec⁡2θ−tan⁡2θ=1\sec^2\theta - \tan^2\theta = 1; csc⁡2θ−cot⁡2θ=1\csc^2\theta - \cot^2\theta = 1.
  • Standard values for 0°,30°,45°,60°,90°0°,30°,45°,60°,90°.
  • sin⁡(90°−θ)=cos⁡θ\sin(90°-\theta)=\cos\theta, tan⁡(90°−θ)=cot⁡θ\tan(90°-\theta)=\cot\theta, sec⁡(90°−θ)=csc⁡θ\sec(90°-\theta)=\csc\theta.
  • tan⁡θ=sin⁡θ/cos⁡θ\tan\theta = \sin\theta/\cos\theta; reciprocals cosec, sec, cot.

Solved Examples

Example 1: Ratio from a triangle

If sin⁡θ=817\sin\theta = \tfrac{8}{17}, find cos⁡θ\cos\theta and tan⁡θ\tan\theta.

Solution:

  1. Adjacent =172−82=15= \sqrt{17^2 - 8^2} = 15.
  2. cos⁡θ=1517\cos\theta = \tfrac{15}{17}, tan⁡θ=815\tan\theta = \tfrac{8}{15}.

Final Answer: cos⁡θ=1517\cos\theta = \tfrac{15}{17}, tan⁡θ=815\tan\theta = \tfrac{8}{15}.

Takeaway: (8,15,17)(8, 15, 17) triple.

Example 2: Evaluate

Evaluate sin⁡60°cos⁡30°−cos⁡60°sin⁡30°\sin 60° \cos 30° - \cos 60° \sin 30°.

Solution:

  1. =32⋅32−12⋅12=34−14=12= \tfrac{\sqrt3}{2}\cdot\tfrac{\sqrt3}{2} - \tfrac12\cdot\tfrac12 = \tfrac34 - \tfrac14 = \tfrac12.

Final Answer: 12\tfrac12.

Takeaway: Equals sin⁡(60°−30°)=sin⁡30°=12\sin(60° - 30°) = \sin 30° = \tfrac12.

Example 3: Identity use

If cos⁡θ=2029\cos\theta = \tfrac{20}{29}, find sin⁡θ\sin\theta.

Solution:

  1. sin⁡2θ=1−400841=441841\sin^2\theta = 1 - \tfrac{400}{841} = \tfrac{441}{841}.
  2. sin⁡θ=2129\sin\theta = \tfrac{21}{29}.

Final Answer: sin⁡θ=2129\sin\theta = \tfrac{21}{29}.

Takeaway: (20,21,29)(20, 21, 29) triple.

Example 4: Complementary

Evaluate cos⁡37°sin⁡53°\dfrac{\cos 37°}{\sin 53°}.

Solution:

  1. sin⁡53°=sin⁡(90°−37°)=cos⁡37°\sin 53° = \sin(90° - 37°) = \cos 37°.
  2. Ratio =1= 1.

Final Answer: 1.

Takeaway: 37°+53°=90°37° + 53° = 90°.

Example 5: Standard-angle sum

Evaluate cos⁡245°+sin⁡245°\cos^2 45° + \sin^2 45°.

Solution:

  1. =12+12=1= \tfrac12 + \tfrac12 = 1.

Final Answer: 1.

Takeaway: A special case of sin⁡2+cos⁡2=1\sin^2+\cos^2=1.

Example 6: Prove

Prove cos⁡2θ(1+tan⁡2θ)=1\cos^2\theta(1 + \tan^2\theta) = 1.

Solution:

  1. 1+tan⁡2θ=sec⁡2θ=1cos⁡2θ1 + \tan^2\theta = \sec^2\theta = \dfrac{1}{\cos^2\theta}.
  2. cos⁡2θ⋅1cos⁡2θ=1\cos^2\theta\cdot\dfrac{1}{\cos^2\theta} = 1.

Final Answer: Proved.

Takeaway: Use 1+tan⁡2θ=sec⁡2θ1 + \tan^2\theta = \sec^2\theta.

Example 7: tan of a triangle

If cot⁡θ=125\cot\theta = \tfrac{12}{5}, find sin⁡θ\sin\theta and cos⁡θ\cos\theta.

Solution:

  1. Adjacent 12, Opposite 5, Hyp =13= 13.
  2. sin⁡θ=513\sin\theta = \tfrac{5}{13}, cos⁡θ=1213\cos\theta = \tfrac{12}{13}.

Final Answer: sin⁡θ=513\sin\theta = \tfrac{5}{13}, cos⁡θ=1213\cos\theta = \tfrac{12}{13}.

Takeaway: cot = adj/opp, so adj 12, opp 5.

Example 8: Evaluate expression

Evaluate tan⁡45°csc⁡30°+sec⁡60°\dfrac{\tan 45°}{\csc 30° + \sec 60°}.

Solution:

  1. tan⁡45°=1\tan 45° = 1; csc⁡30°=2\csc 30° = 2; sec⁡60°=2\sec 60° = 2.
  2. 12+2=14\dfrac{1}{2 + 2} = \dfrac14.

Final Answer: 14\tfrac14.

Takeaway: Substitute standard values, then simplify.

Example 9: Solve for angle

Find acute θ\theta if tan⁡θ=3\tan\theta = \sqrt3.

Solution:

  1. tan⁡60°=3\tan 60° = \sqrt3.

Final Answer: θ=60°\theta = 60°.

Takeaway: Read the table backwards.

Example 10: Prove

Prove cos⁡θ1−tan⁡θ+sin⁡θ1−cot⁡θ=sin⁡θ+cos⁡θ\dfrac{\cos\theta}{1 - \tan\theta} + \dfrac{\sin\theta}{1 - \cot\theta} = \sin\theta + \cos\theta.

Solution:

  1. cos⁡θ1−sin⁡θcos⁡θ=cos⁡2θcos⁡θ−sin⁡θ\dfrac{\cos\theta}{1 - \frac{\sin\theta}{\cos\theta}} = \dfrac{\cos^2\theta}{\cos\theta - \sin\theta}.
  2. sin⁡θ1−cos⁡θsin⁡θ=sin⁡2θsin⁡θ−cos⁡θ=−sin⁡2θcos⁡θ−sin⁡θ\dfrac{\sin\theta}{1 - \frac{\cos\theta}{\sin\theta}} = \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.

Takeaway: Convert tan, cot to sin/cos; factor the difference of squares.

Example 11: Given sin, evaluate

If sin⁡θ=12\sin\theta = \tfrac{1}{2}, evaluate 3cos⁡θ−4cos⁡3θ3\cos\theta - 4\cos^3\theta.

Solution:

  1. cos⁡θ=32\cos\theta = \tfrac{\sqrt3}{2}.
  2. 3⋅32−4⋅338=332−332=03\cdot\tfrac{\sqrt3}{2} - 4\cdot\tfrac{3\sqrt3}{8} = \tfrac{3\sqrt3}{2} - \tfrac{3\sqrt3}{2} = 0.

Final Answer: 0.

Takeaway: 3cos⁡θ−4cos⁡3θ=cos⁡3θ3\cos\theta - 4\cos^3\theta = \cos 3\theta; here cos⁡90°=0\cos 90° = 0.

Example 12: Complementary equation

If cos⁡2A=sin⁡(A+30°)\cos 2A = \sin(A + 30°), with 2A2A acute, find AA.

Solution:

  1. cos⁡2A=sin⁡(90°−2A)\cos 2A = \sin(90° - 2A), so sin⁡(90°−2A)=sin⁡(A+30°)\sin(90° - 2A) = \sin(A + 30°).
  2. 90°−2A=A+30°⇒60°=3A⇒A=20°90° - 2A = A + 30° \Rightarrow 60° = 3A \Rightarrow A = 20°.

Final Answer: A=20°A = 20°.

Takeaway: Convert cos to sin, equate the angles.

Example 13: Identity proof

Prove (sin⁡θ+csc⁡θ)2+(cos⁡θ+sec⁡θ)2=7+tan⁡2θ+cot⁡2θ(\sin\theta + \csc\theta)^2 + (\cos\theta + \sec\theta)^2 = 7 + \tan^2\theta + \cot^2\theta.

Solution:

  1. Expand: sin⁡2θ+2+csc⁡2θ+cos⁡2θ+2+sec⁡2θ\sin^2\theta + 2 + \csc^2\theta + \cos^2\theta + 2 + \sec^2\theta.
  2. =(sin⁡2θ+cos⁡2θ)+4+csc⁡2θ+sec⁡2θ=1+4+(1+cot⁡2θ)+(1+tan⁡2θ)= (\sin^2\theta + \cos^2\theta) + 4 + \csc^2\theta + \sec^2\theta = 1 + 4 + (1 + \cot^2\theta) + (1 + \tan^2\theta).
  3. =7+tan⁡2θ+cot⁡2θ= 7 + \tan^2\theta + \cot^2\theta.

Final Answer: Proved.

Takeaway: Cross-terms 2sin⁡θcsc⁡θ=22\sin\theta\csc\theta = 2 and 2cos⁡θsec⁡θ=22\cos\theta\sec\theta = 2.

Example 14: Evaluate ratio

If tan⁡θ=ab\tan\theta = \tfrac{a}{b}, find asin⁡θ−bcos⁡θasin⁡θ+bcos⁡θ\dfrac{a\sin\theta - b\cos\theta}{a\sin\theta + b\cos\theta}.

Solution:

  1. Divide by cos⁡θ\cos\theta: atan⁡θ−batan⁡θ+b\dfrac{a\tan\theta - b}{a\tan\theta + b}.
  2. =a⋅ab−ba⋅ab+b=a2−b2a2+b2= \dfrac{a\cdot\frac{a}{b} - b}{a\cdot\frac{a}{b} + b} = \dfrac{a^2 - b^2}{a^2 + b^2}.

Final Answer: a2−b2a2+b2\dfrac{a^2 - b^2}{a^2 + b^2}.

Takeaway: Divide through by cos⁡θ\cos\theta to introduce tan⁡θ\tan\theta.

Example 15: Standard-angle product

Evaluate tan⁡30°tan⁡60°\tan 30° \tan 60°.

Solution:

  1. =13⋅3=1= \tfrac{1}{\sqrt3}\cdot\sqrt3 = 1.

Final Answer: 1.

Takeaway: tan⁡30°\tan 30° and tan⁡60°\tan 60° are reciprocals.

Example 16: Prove

Prove 1−cos⁡θ1+cos⁡θ=csc⁡θ−cot⁡θ\sqrt{\dfrac{1 - \cos\theta}{1 + \cos\theta}} = \csc\theta - \cot\theta.

Solution:

  1. Multiply inside by 1−cos⁡θ1−cos⁡θ\dfrac{1 - \cos\theta}{1 - \cos\theta}: (1−cos⁡θ)21−cos⁡2θ=(1−cos⁡θ)2sin⁡2θ\dfrac{(1-\cos\theta)^2}{1 - \cos^2\theta} = \dfrac{(1-\cos\theta)^2}{\sin^2\theta}.
  2. Square root: 1−cos⁡θsin⁡θ=1sin⁡θ−cos⁡θsin⁡θ=csc⁡θ−cot⁡θ\dfrac{1 - \cos\theta}{\sin\theta} = \dfrac{1}{\sin\theta} - \dfrac{\cos\theta}{\sin\theta} = \csc\theta - \cot\theta.

Final Answer: Proved.

Takeaway: Rationalise inside the root using the conjugate.

Example 17: Mixed evaluation

Evaluate cos⁡0°+sin⁡90°+2sin⁡45°\cos 0° + \sin 90° + \sqrt2\sin 45°.

Solution:

  1. cos⁡0°=1\cos 0° = 1; sin⁡90°=1\sin 90° = 1; 2sin⁡45°=2⋅12=1\sqrt2\sin 45° = \sqrt2\cdot\tfrac{1}{\sqrt2} = 1.
  2. Sum =1+1+1=3= 1 + 1 + 1 = 3.

Final Answer: 3.

Takeaway: Watch the boundary values 0° and 90°.

Example 18: Given sec, evaluate

If sec⁡θ=54\sec\theta = \tfrac{5}{4}, evaluate sin⁡θ−2cos⁡θtan⁡θ−cot⁡θ\dfrac{\sin\theta - 2\cos\theta}{\tan\theta - \cot\theta}.

Solution:

  1. cos⁡θ=45\cos\theta = \tfrac45, Opp =3= 3, sin⁡θ=35\sin\theta = \tfrac35, tan⁡θ=34\tan\theta = \tfrac34, cot⁡θ=43\cot\theta = \tfrac43.
  2. Numerator =35−2⋅45=35−85=−1= \tfrac35 - 2\cdot\tfrac45 = \tfrac35 - \tfrac85 = -1.
  3. Denominator =34−43=9−1612=−712= \tfrac34 - \tfrac43 = \tfrac{9 - 16}{12} = -\tfrac{7}{12}.
  4. −1−7/12=127\dfrac{-1}{-7/12} = \tfrac{12}{7}.

Final Answer: 127\tfrac{12}{7}.

Takeaway: Build all ratios from the triangle, then substitute.

Example 19: Prove

Prove tan⁡θ1−cot⁡θ+cot⁡θ1−tan⁡θ=1+sec⁡θcsc⁡θ\dfrac{\tan\theta}{1 - \cot\theta} + \dfrac{\cot\theta}{1 - \tan\theta} = 1 + \sec\theta\csc\theta.

Solution:

  1. Write in sin/cos. tan⁡θ1−cot⁡θ=sin⁡2θcos⁡θ(sin⁡θ−cos⁡θ)\dfrac{\tan\theta}{1-\cot\theta} = \dfrac{\sin^2\theta}{\cos\theta(\sin\theta - \cos\theta)} and cot⁡θ1−tan⁡θ=cos⁡2θsin⁡θ(cos⁡θ−sin⁡θ)\dfrac{\cot\theta}{1-\tan\theta} = \dfrac{\cos^2\theta}{\sin\theta(\cos\theta - \sin\theta)}.
  2. Sum =1sin⁡θ−cos⁡θ[sin⁡2θcos⁡θ−cos⁡2θsin⁡θ]=sin⁡3θ−cos⁡3θsin⁡θcos⁡θ(sin⁡θ−cos⁡θ)= \dfrac{1}{\sin\theta - \cos\theta}\left[\dfrac{\sin^2\theta}{\cos\theta} - \dfrac{\cos^2\theta}{\sin\theta}\right] = \dfrac{\sin^3\theta - \cos^3\theta}{\sin\theta\cos\theta(\sin\theta - \cos\theta)}.
  3. sin⁡3−cos⁡3=(sin⁡θ−cos⁡θ)(sin⁡2+sin⁡θcos⁡θ+cos⁡2)\sin^3 - \cos^3 = (\sin\theta-\cos\theta)(\sin^2 + \sin\theta\cos\theta + \cos^2).
  4. =1+sin⁡θcos⁡θsin⁡θcos⁡θ=1+1sin⁡θcos⁡θ=1+sec⁡θcsc⁡θ= \dfrac{1 + \sin\theta\cos\theta}{\sin\theta\cos\theta} = 1 + \dfrac{1}{\sin\theta\cos\theta} = 1 + \sec\theta\csc\theta.

Final Answer: Proved.

Takeaway: Factor the difference of cubes after combining.

Example 20: Eliminate theta

If x=acos⁡θx = a\cos\theta and y=bsin⁡θy = b\sin\theta, show x2a2+y2b2=1\dfrac{x^2}{a^2} + \dfrac{y^2}{b^2} = 1.

Solution:

  1. x2a2=cos⁡2θ\dfrac{x^2}{a^2} = \cos^2\theta, y2b2=sin⁡2θ\dfrac{y^2}{b^2} = \sin^2\theta.
  2. Sum =cos⁡2θ+sin⁡2θ=1= \cos^2\theta + \sin^2\theta = 1.

Final Answer: Proved.

Takeaway: Eliminating θ\theta with sin⁡2+cos⁡2=1\sin^2+\cos^2=1 gives an ellipse.

Example 21: Numerical identity

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⁡30°=3\cot 30° = \sqrt3, cot⁡230°=3\cot^2 30° = 3, so 43\tfrac43.
  2. sin⁡230°=14\sin^2 30° = \tfrac14, so 11/4=4\tfrac{1}{1/4} = 4.
  3. 2cos⁡245°=2⋅12=12\cos^2 45° = 2\cdot\tfrac12 = 1; sin⁡20°=0\sin^2 0° = 0.
  4. 43+4−1−0=43+3=133\tfrac43 + 4 - 1 - 0 = \tfrac43 + 3 = \tfrac{13}{3}.

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

Takeaway: Term by term, keeping fractions exact.

Example 22: Complementary product

Evaluate cos⁡1°cos⁡2°⋯cos⁡89°cos⁡90°\cos 1° \cos 2° \cdots \cos 89° \cos 90°.

Solution:

  1. The product contains the factor cos⁡90°=0\cos 90° = 0.
  2. Anything times 0 is 0.

Final Answer: 0.

Takeaway: Always scan a long product for a zero factor like cos⁡90°\cos 90°.

Example 23: Given expression value

If tan⁡θ+cot⁡θ=2\tan\theta + \cot\theta = 2, find tan⁡2θ+cot⁡2θ\tan^2\theta + \cot^2\theta.

Solution:

  1. Square: tan⁡2θ+2+cot⁡2θ=4\tan^2\theta + 2 + \cot^2\theta = 4.
  2. tan⁡2θ+cot⁡2θ=2\tan^2\theta + \cot^2\theta = 2.

Final Answer: 2.

Takeaway: Square the sum; the cross-term 2tan⁡θcot⁡θ=22\tan\theta\cot\theta = 2.

Example 24: Prove

Prove 1+cos⁡θ−sin⁡2θsin⁡θ(1+cos⁡θ)=cot⁡θ\dfrac{1 + \cos\theta - \sin^2\theta}{\sin\theta(1 + \cos\theta)} = \cot\theta.

Solution:

  1. Numerator =1+cos⁡θ−(1−cos⁡2θ)=cos⁡θ+cos⁡2θ=cos⁡θ(1+cos⁡θ)= 1 + \cos\theta - (1 - \cos^2\theta) = \cos\theta + \cos^2\theta = \cos\theta(1 + \cos\theta).
  2. cos⁡θ(1+cos⁡θ)sin⁡θ(1+cos⁡θ)=cos⁡θsin⁡θ=cot⁡θ\dfrac{\cos\theta(1+\cos\theta)}{\sin\theta(1+\cos\theta)} = \dfrac{\cos\theta}{\sin\theta} = \cot\theta.

Final Answer: Proved.

Takeaway: Replace sin⁡2θ\sin^2\theta with 1−cos⁡2θ1 - \cos^2\theta, then factor.

Example 25: Evaluate

Evaluate 2tan⁡245°+cos⁡230°−sin⁡260°2\tan^2 45° + \cos^2 30° - \sin^2 60°.

Solution:

  1. tan⁡245°=1\tan^2 45° = 1, so 22.
  2. cos⁡230°=34\cos^2 30° = \tfrac34, sin⁡260°=34\sin^2 60° = \tfrac34.
  3. 2+34−34=22 + \tfrac34 - \tfrac34 = 2.

Final Answer: 2.

Takeaway: cos⁡30°=sin⁡60°\cos 30° = \sin 60°, so those terms cancel.

Example 26: Given relation

If 5tan⁡θ=45\tan\theta = 4, evaluate 5sin⁡θ−3cos⁡θ5sin⁡θ+2cos⁡θ\dfrac{5\sin\theta - 3\cos\theta}{5\sin\theta + 2\cos\theta}.

Solution:

  1. tan⁡θ=45\tan\theta = \tfrac45. Divide by cos⁡θ\cos\theta: 5tan⁡θ−35tan⁡θ+2\dfrac{5\tan\theta - 3}{5\tan\theta + 2}.
  2. =5⋅45−35⋅45+2=4−34+2=16= \dfrac{5\cdot\frac45 - 3}{5\cdot\frac45 + 2} = \dfrac{4 - 3}{4 + 2} = \dfrac16.

Final Answer: 16\tfrac16.

Takeaway: Divide top and bottom by cos⁡θ\cos\theta.

Example 27: Complementary simplification

Evaluate the numerator sin⁡35°cos⁡55°+cos⁡35°sin⁡55°\sin 35° \cos 55° + \cos 35° \sin 55°.

Solution:

  1. Numerator =sin⁡35°cos⁡55°+cos⁡35°sin⁡55°= \sin 35°\cos 55° + \cos 35°\sin 55°. Using cos⁡55°=sin⁡35°\cos 55° = \sin 35° and sin⁡55°=cos⁡35°\sin 55° = \cos 35°: sin⁡235°+cos⁡235°=1\sin^2 35° + \cos^2 35° = 1.

Final Answer: Numerator =1= 1.

Takeaway: Complementary substitution turns it into the master identity.

Example 28: Prove

Prove sec⁡4θ−sec⁡2θ=tan⁡4θ+tan⁡2θ\sec^4\theta - \sec^2\theta = \tan^4\theta + \tan^2\theta.

Solution:

  1. LHS =sec⁡2θ(sec⁡2θ−1)=sec⁡2θtan⁡2θ= \sec^2\theta(\sec^2\theta - 1) = \sec^2\theta\tan^2\theta.
  2. RHS =tan⁡2θ(tan⁡2θ+1)=tan⁡2θsec⁡2θ= \tan^2\theta(\tan^2\theta + 1) = \tan^2\theta\sec^2\theta.
  3. LHS = RHS.

Final Answer: Proved.

Takeaway: Factor both sides; sec⁡2−1=tan⁡2\sec^2 - 1 = \tan^2 and tan⁡2+1=sec⁡2\tan^2 + 1 = \sec^2.

Example 29: Boundary value

Is cos⁡θ=32\cos\theta = \tfrac32 possible for any angle θ\theta?

Solution:

  1. cos⁡θ\cos\theta never exceeds 1 in magnitude.
  2. 32>1\tfrac32 > 1, so impossible.

Final Answer: No.

Takeaway: −1≤cos⁡θ≤1-1 \le \cos\theta \le 1 for every angle.

Example 30: Find angle from equation

Find acute θ\theta if sin⁡θcos⁡θ=34\sin\theta\cos\theta = \tfrac{\sqrt3}{4}.

Solution:

  1. 2sin⁡θcos⁡θ=322\sin\theta\cos\theta = \tfrac{\sqrt3}{2}, i.e. sin⁡2θ=32\sin 2\theta = \tfrac{\sqrt3}{2}.
  2. 2θ=60°⇒θ=30°2\theta = 60° \Rightarrow \theta = 30°.

Final Answer: θ=30°\theta = 30°.

Takeaway: Use 2sin⁡θcos⁡θ=sin⁡2θ2\sin\theta\cos\theta = \sin 2\theta (a Class 11 preview, but the value check works).

Example 31: Combined

Evaluate cos⁡58°sin⁡32°+sin⁡22°cos⁡68°−cos⁡38°csc⁡52°tan⁡18°tan⁡35°tan⁡60°tan⁡72°tan⁡55°\dfrac{\cos 58°}{\sin 32°} + \dfrac{\sin 22°}{\cos 68°} - \dfrac{\cos 38° \csc 52°}{\tan 18° \tan 35° \tan 60° \tan 72° \tan 55°}.

Solution:

  1. sin⁡32°=cos⁡58°\sin 32° = \cos 58°, so first term =1= 1. cos⁡68°=sin⁡22°\cos 68° = \sin 22°, so second term =1= 1.
  2. csc⁡52°=sec⁡38°\csc 52° = \sec 38°, so cos⁡38°csc⁡52°=cos⁡38°sec⁡38°=1\cos 38°\csc 52° = \cos 38°\sec 38° = 1.
  3. Denominator: tan⁡18°tan⁡72°=1\tan 18°\tan 72° = 1, tan⁡35°tan⁡55°=1\tan 35°\tan 55° = 1, tan⁡60°=3\tan 60° = \sqrt3. So denominator =3= \sqrt3.
  4. Third term =13= \dfrac{1}{\sqrt3}.
  5. Total =1+1−13=2−13= 1 + 1 - \tfrac{1}{\sqrt3} = 2 - \tfrac{1}{\sqrt3}.

Final Answer: 2−132 - \tfrac{1}{\sqrt3}.

Takeaway: Pair complementary angles everywhere before computing.