SSC CGL Preparation: Concepts and Practice

Learning outcome

Replace ratios of complementary acute angles, simplify expressions without decimal angle values, and solve angle equations using justified domain restrictions.

Concepts and assumptions

Angles are in degrees. Two angles are complementary when their sum is 90°. Throughout this lesson both are acute: 0° < θ < 90°, so 90° − θ is also acute and all six ratios are defined.

In right triangle ABC with ∠C = 90°, the angle sum gives A + B = 90°. The leg opposite A is adjacent to B, and the leg adjacent to A is opposite B. The hypotenuse does not change.

Consequently, opposite/hypotenuse for one angle equals adjacent/hypotenuse for the other: sin(90° − θ) = cos θ; cos(90° − θ) = sin θ.

Exchanging the two legs similarly gives: tan(90° − θ) = cot θ; cot(90° − θ) = tan θ. Taking the appropriate nonzero reciprocals gives: sec(90° − θ) = cosec θ; cosec(90° − θ) = sec θ.

These are angle relationships, not instructions to subtract ratio values from 90. Do not extend a quotient relationship through a zero denominator.

For acute α and β, sin α = cos β implies α + β = 90°. Replace cos β with sin(90° − β), then use that sine is strictly increasing, hence one-to-one, on the acute interval. Tangent is also strictly increasing there, so tan α = cot β gives the same conclusion. Without a suitable angle domain, periodic repetitions prevent that conclusion from giving a unique angle.

Worked examples

Example 1 — No decimal tables needed. Simplify sin 23°/cos 67° + tan 34°/cot 56°.

Because 23° + 67° = 90°, cos 67° = sin 23°. Because 34° + 56° = 90°, cot 56° = tan 34°. Both denominators are positive, so the expression equals 1 + 1 = 2.

Example 2 — Finding complementary ratios. For acute θ, sin θ = 5/13.

cos θ = √(1 − 25/169) = 12/13, using the positive root. Therefore cos(90° − θ) = 5/13 and sec(90° − θ) = 1/(5/13) = 13/5. Also tan(90° − θ) = cot θ = (12/13)/(5/13) = 12/5.

Example 3 — An angle equation. Solve sin((2x + 10)°) = cos((3x − 5)°), with both displayed angles acute.

Their degree measures must sum to 90: 2x + 10 + 3x − 5 = 90. Thus 5x + 5 = 90, giving x = 17. The angles are 44° and 46°: both are acute and their sum is 90°, verifying the original relationship.

Common mistakes

Confusing complements with angles summing to 180°; assuming complementary angles have equal sine values; ignoring domains when solving equations; replacing secant with the wrong reciprocal.

Practice questions

1. Simplify cos 19°/sin 71°.
2. For acute θ, cot θ = 7/24. Find tan(90° − θ) and sin(90° − θ).
3. Solve tan((4x − 7)°) = cot((2x + 1)°), with both angles acute.
4. Simplify tan 18° tan 72° + sec 31° sin 59°.

Worked solutions

1. Since 19° + 71° = 90°, sin 71° = cos 19° ≠ 0. The quotient is 1.
2. Use adjacent = 7k, opposite = 24k, k > 0; hypotenuse = 25k. Thus tan(90° − θ) = 7/24, and sin(90° − θ) = cos θ = 7/25.
3. Add complementary measures: 4x − 7 + 2x + 1 = 90. Hence x = 16, giving angles 57° and 33°, both valid.
4. tan 72° = cot 18°, so their product is 1. sin 59° = cos 31°, so sec 31° sin 59° = 1. Total = 2.