SSC CGL Preparation: Concepts and Practice

Learning outcome

Derive basic trigonometric identities, use them to recover ratios, and simplify expressions without losing their original domain restrictions.

Concepts and assumptions

Angles are in degrees. Numerical ratio-recovery problems use acute angles, 0° < θ < 90°, so all six ratios are positive. More general identities apply only where every expression involved is defined.

An identity holds for every angle in its stated domain; an equation may hold only for selected angles. Thus sin² θ + cos² θ = 1 is an identity, whereas sin θ = 1/2 is an equation to solve.

In a right triangle, opposite² + adjacent² = hypotenuse². Dividing by the positive hypotenuse squared gives: sin² θ + cos² θ = 1. The same relation holds for every real angle through unit-circle coordinates.

Divide this identity by cos² θ only when cos θ ≠ 0: 1 + tan² θ = sec² θ. Divide instead by sin² θ only when sin θ ≠ 0: 1 + cot² θ = cosec² θ.

In degrees, cosine is zero at θ = 90° + 180°k and sine is zero at θ = 180°k, where k is any integer. Exclude the relevant angles before division.

Here sin² θ means (sin θ)². When recovering a ratio from its square, its sign needs justification; acute-angle positivity supplies that justification. Cancellation removes common nonzero factors, not the original exclusions. A simplified expression may have a wider apparent domain than the expression it replaces.

Worked examples

Example 1 — Recovering ratios. For acute θ, sin θ = 7/25. Find cos θ and tan θ.

cos² θ = 1 − 49/625 = 576/625. Choose the positive root: cos θ = 24/25. Then tan θ = sin θ/cos θ = (7/25)/(24/25) = 7/24.

Example 2 — Cancellation with restrictions. Simplify (sec² θ − 1)/tan θ.

The original requires cos θ ≠ 0 and tan θ ≠ 0; together these mean sin θ ≠ 0 and cos θ ≠ 0. Using sec² θ − 1 = tan² θ gives tan² θ/tan θ = tan θ. Both original restrictions remain, even though tan θ alone can equal zero.

Example 3 — Adding fractions. Simplify 1/(1 − sin θ) + 1/(1 + sin θ).

Require sin θ ≠ 1 and sin θ ≠ −1, equivalently cos θ ≠ 0. Combine the numerators: [(1 + sin θ) + (1 − sin θ)]/(1 − sin² θ) = 2/cos² θ = 2 sec² θ, on that domain.

Common mistakes

Taking a positive square root without checking the angle range; dividing by a zero ratio; cancelling separate terms; dropping restrictions after simplification.

Practice questions

1. For acute θ, tan θ = 3/4. Find sec θ and cosec θ.
2. Simplify cos θ/(1 − sin² θ), stating the domain.
3. Simplify (cosec² θ − 1)/cot θ, stating the domain.
4. For acute θ, tan θ + cot θ = 4. Find tan² θ + cot² θ.

Worked solutions

1. sec² θ = 1 + 9/16 = 25/16, so sec θ = 5/4. cos θ = 4/5 and sin θ = tan θ cos θ = 3/5; hence cosec θ = 5/3.
2. The denominator equals cos² θ, requiring cos θ ≠ 0. Thus cos θ/cos² θ = sec θ, with that restriction.
3. Require sin θ ≠ 0 and cos θ ≠ 0. The numerator becomes cot² θ; division by nonzero cot θ gives cot θ.
4. tan θ cot θ = 1 for acute θ. Squaring the given sum gives tan² θ + 2 + cot² θ = 16. Therefore the required value is 14.