A trigonometric identity is an equation that stays true for every angle. SSC leans on them heavily because, once you recognise the pattern, an expression stuffed with sin,cos,sec and cot collapses to a plain number — no slow side-hunting needed.
The three Pythagorean identities
• Master:sin2θ+cos2θ=1 — divide it by cos2θ and by sin2θ to get the other two.
• With tan:1+tan2θ=sec2θ.
• With cot:1+cot2θ=csc2θ.
The rearrangements are what you actually plug in: 1−cos2θ=sin2θ, sec2θ−tan2θ=1 and csc2θ−cot2θ=1. Whenever you see sec2θ−1, read it as tan2θ; whenever you see csc2θ−1, read it as cot2θ.
For high powers, raise the master identity. Squaring gives sin4θ+cos4θ=1−2sin2θcos2θ; cubing gives sin6θ+cos6θ=1−3sin2θcos2θ. Setting k=sinθcosθ, these are simply 1−2k2 and 1−3k2.
📐Right Triangle ExplorerPick an angle — see exact fraction values
Select angle θ
sin θP / H1/2
cos θB / H√3/2
tan θP / B1/√3
cosec θH / P2
sec θH / B2/√3
cot θB / P√3
Using sin2θ+cos2θ=1: if sinθ=53 and θ is acute, then cosθ= — the 3-4-5 triangle gives the base for free.
📝Practice Questions
Q1sin²40° + cos²40° = ?
Q2sec²θ − tan²θ = ?
Q3If sinθ cosθ = ½, then sin⁴θ + cos⁴θ = ?
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Real exam questions — Trig Identities
3 question types · 15 solved examples from real SSC papers
Trig identities turn a scary mix of sin, cos, sec and cot into a single number: spot sin2θ+cos2θ=1 (and its sec/csc cousins) and most SSC questions fall in one line.
How to solve this type
Three identities cover this whole family: sin2θ+cos2θ=1, 1+tan2θ=sec2θ and 1+cot2θ=csc2θ. Their rearrangements (sec2θ−tan2θ=1, 1−cos2θ=sin2θ, etc.) are what you actually plug in. When you are given one ratio, draw the right triangle and read the missing side with the Pythagoras theorem — the 3-4-5 and 5-12-13 triplets show up again and again.
