A conditional question gives you one trig equation as a “condition” and asks for the value of some other expression. The trap is trying to find the angle θ first. You almost never need it — one algebraic move on the given equation usually hands you the exact piece the target expression is hungry for.
The master move: square the given
Most conditionals give a sum and ask for a square or a product. Square both sides and the identity sin2θ+cos2θ=1 drops a number into your lap.
From sinθ+cosθ=k, squaring gives 1+2sinθcosθ=k2, so sinθcosθ=2k2−1.
For the tan–cot family, remember tanθcotθ=1, so the pair behaves like x+x1: tan2θ+cot2θ=k2−2. The bridge back to sine–cosine is tanθ+cotθ=sinθcosθ1.
Conjugate pairs like secθ−tanθ and cscθ+cotθ obey sec2θ−tan2θ=1 and csc2θ−cot2θ=1, so the partner is just the reciprocal of the given.
📐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
Quick drill: if sinθ+cosθ=2, squaring gives 1+2sinθcosθ=2, so sinθcosθ= .
📝Practice Questions
Q1If sinθ + cosθ = √2, then sinθ·cosθ = ?
Q2If tanθ + cotθ = 4, then tan²θ + cot²θ = ?
Q3If cosθ = 3/5 and θ is acute, then sinθ = ?
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Real exam questions — Conditional Equations
5 question types · 20 solved examples from real SSC papers
Conditional questions hand you one trig equation and ask for a different expression, and the whole game is to square the given (or use a conjugate-pair identity) so a known value drops out without ever finding the angle.
How to solve this type
If you are given one ratio, draw a right triangle: that ratio fixes two sides, and Pythagoras gives the third, so EVERY other ratio is now readable. If you are instead given a ± relation, square it or use a conjugate identity. The golden one is sec2θ−tan2θ=1=(secθ−tanθ)(secθ+tanθ), so the partner is just the reciprocal of the given. Never solve for the angle itself unless the numbers are a standard 30/45/60.
