Trigonometry — SSC CGL Quantitative Aptitude

SSC trigonometry is algebra wearing angles. There are no waves, no radian calculus — just six ratios, three Pythagorean identities and a set of standard values that combine into "simplify this expression" questions. Like SSC algebra, it's a recognition game: spot which identity the expression is dressed in, or substitute a convenient angle and evaluate.

1. What SSC actually asks

Tier 1: 2–3 Q · Tier 2: 3–4 Q, in five patterns:

1. Identity simplification — reduce a trig expression to a constant or single ratio.
2. Given one ratio, find another — sin θ = 3/5 → find tan θ + cot θ.
3. Standard-value evaluation — compute expressions at 0°, 30°, 45°, 60°, 90°.
4. Complementary collapse — sin(90−θ) chains that cancel.
5. Min–max — maximum/minimum of a sin ± b cos, sin²+cos⁴ style.

2. The six ratios and the triangle habit

For a right triangle with angle θ: $\sin \theta =\frac{P}{H}$, $\cos \theta =\frac{B}{H}$, $\tan \theta =\frac{P}{B}$, and reciprocals cosec, sec, cot.

The triangle habit: given any one ratio, draw the right triangle, fill the two known sides, get the third by Pythagoras, then read off every other ratio. sin θ = 3/5 → sides 3-4-5 → cos θ = 4/5, tan θ = 3/4, sec θ = 5/4 — all in ten seconds. SSC builds these questions on Pythagorean triples (3-4-5, 5-12-13, 8-15-17, 7-24-25, 20-21-29).

3. Standard values — the table you must own

Memory hook: sin row = $\frac{\sqrt{0}}{2},\frac{\sqrt{1}}{2},\frac{\sqrt{2}}{2},\frac{\sqrt{3}}{2},\frac{\sqrt{4}}{2}$; cos row is the reverse; tan = sin/cos.

4. The identity kit

Pythagorean trio (the engine):

${\sin }^2\theta +{\cos }^2\theta =11+{\tan }^2\theta ={\sec }^2\theta 1+{\cot }^2\theta ={\text{cosec}}^2\theta$

Rearranged forms SSC loves: ${\sec }^2\theta -{\tan }^2\theta =1$ → $(\sec \theta -\tan \theta )(\sec \theta +\tan \theta )=1$ — so sec θ − tan θ and sec θ + tan θ are reciprocals (same for cosec/cot). Given sec θ + tan θ = k, immediately sec θ − tan θ = 1/k, hence:

$\sec \theta =\frac{k+1/k}{2},\tan \theta =\frac{k-1/k}{2}$

Complementary angles: $\sin (90°-\theta )=\cos \theta$, $\tan (90°-\theta )=\cot \theta$, $\sec (90°-\theta )=\text{cosec}\theta$. Chains like $\tan 1°\tan 2°\cdots \tan 89°$ collapse to 1 (pairs multiply to 1, tan 45° = 1 survives). Similarly ${\sin }^2\theta +{\sin }^2(90-\theta )=1$.

Compound-angle values worth knowing: sin 15° = $\frac{\sqrt{3}-1}{2\sqrt{2}}$, sin 75° = cos 15° = $\frac{\sqrt{3}+1}{2\sqrt{2}}$, tan 15° = 2−√3, tan 75° = 2+√3.

5. Min–max — two rules cover every question

1. $a\sin \theta \pm b\cos \theta$ ranges over $\left[-\sqrt{a^2+b^2},\sqrt{a^2+b^2}\right]$. Max of 3 sin θ + 4 cos θ = 5.
2. ${\sin }^n\theta +{\cos }^n\theta$ (n ≥ 2): maximum 1 (at axes), minimum at θ = 45°: $2\times (1/\sqrt{2}{)}^n=2^{1-n/2}$. So sin⁴θ + cos⁴θ has min 1/2; sin⁶+cos⁶ has min 1/4.

Bonus identities: ${\sin }^4\theta +{\cos }^4\theta =1-2{\sin }^2\theta {\cos }^2\theta$, and ${\sin }^6\theta +{\cos }^6\theta =1-3{\sin }^2\theta {\cos }^2\theta$.

6. The substitution shortcut

When an expression must hold "for all θ", plug a convenient angle (usually 45°, or 0°/90° when defined) and match options.

"Find the value of $\frac{1-\sin \theta }{\cos \theta }\cdot \frac{1+\sin \theta }{\cos \theta }$." — At θ = 45°: $\frac{(1-\frac{1}{\sqrt{2}})(1+\frac{1}{\sqrt{2}})}{1/2}=\frac{1-\frac{1}{2}}{\frac{1}{2}}=1$. So the expression is identically 1 (it is $\frac{1-{\sin }^2\theta }{{\cos }^2\theta }=\frac{{\cos }^2\theta }{{\cos }^2\theta }$ — but substitution got there without the algebra).

Guard-rail: substitution is valid for identity-style questions ("find the value of…for all θ") — not for conditional equations where θ is pinned by the given equation (there, solve for θ or manipulate directly).

7. Solved PYQ-style examples

Q1. If sin θ = 5/13 (θ acute), find (tan θ + sec θ). Solution. 5-12-13 triangle: tan = 5/12, sec = 13/12 → sum = 18/12 = 3/2.

Q2. If sec θ + tan θ = 4, find cos θ. Solution. sec − tan = 1/4 → sec = (4 + ¼)/2 = 17/8 → cos θ = 8/17.

Q3. Evaluate $\frac{\sin 30°+\tan 45°-\text{cosec }60°}{\cot 45°+\cos 60°-\sec 30°}$. Solution. Numerator: ½ + 1 − 2/√3; denominator: 1 + ½ − 2/√3 — identical → 1. (Spot symmetric numerator/denominator before computing.)

Q4. Value of tan 5° · tan 25° · tan 45° · tan 65° · tan 85°. Solution. tan 5°·tan 85° = 1, tan 25°·tan 65° = 1, tan 45° = 1 → 1.

Q5. Maximum value of 5 sin θ + 12 cos θ + 7. Solution. $\sqrt{25+144}=13$ → max = 13 + 7 = 20.

Q6 (Tier 2). If sin θ + sin²θ = 1, find cos²θ + cos⁴θ. Solution. sin θ = 1 − sin²θ = cos²θ. Then cos²θ + cos⁴θ = sin θ + sin²θ = 1. (The classic self-referencing identity — appears every few cycles.)

8. Training protocol

Trigonometry marks come from three memorised assets: the value table, the Pythagorean trio with rearrangements (especially the sec−tan reciprocal), and the min–max rules. Write all three daily for a week. Then drill PYQs by pattern; every question should announce its pattern (identity / triangle / values / complementary / min–max) within ten seconds. Heights & Distances — the applied version — has its own chapter; master this one first.