MathematicsClass 10ICSE

How Two Equations With the Same Angle Can Be Squeezed Into One Without It

Prove identities with fractions by taking the LCM or rationalising, eliminate theta from a pair of equations, use complementary-angle relations, and read four-figure trigonometric tables in both directions.

What harder identity skills does Part 2 add?

Part 1 used the three basic identities on single expressions. **Part 2 applies them to fractions, to pairs of equations and to angles that add up to

9 0^{\circ}

**, and shows how to find ratios for any angle from tables.

The same identities are the tools:

sin^{2} A + cos^{2} A = 1 sec^{2} A - tan^{2} A = 1 cosec^{2} A - cot^{2} A = 1

This part covers composite identities, eliminating an angle, complementary angles and four-figure tables.

How do you prove identities with algebraic fractions by taking the LCM or rationalising?

Combine fractions over a common denominator and simplify with the identities, or multiply numerator and denominator by a conjugate so that a difference of squares becomes a single ratio.

Worked example 1 — LCM. Prove

\frac{sin A}{1 + cos A} + \frac{1 + cos A}{sin A} = 2 cosec A

\frac{sin ^{2} A + 1 + 2 cos A + cos ^{2} A}{sin A ( 1 + cos A )} = \frac{2 ( 1 + cos A )}{sin A ( 1 + cos A )} = \frac{2}{sin A}

Worked example 2 — rationalising. Prove

\frac{1 + sin A}{1 - sin A} = sec A + tan A

for acute

A

\frac{( 1 + sin A ) ^{2}}{1 - sin ^{2} A} = \frac{1 + sin A}{cos A} = sec A + tan A

Worked example 3 — conjugate. Prove

\frac{1}{sec A - tan A} = sec A + tan A

\frac{sec A + tan A}{sec ^{2} A - tan ^{2} A} = \frac{sec A + tan A}{1} = sec A + tan A

An everyday example. **Rationalising here works just like

\frac{1}{2 - 1} = 2 + 1

** from surds, because

(2 - 1) (2 + 1) = 1

.

The substance. **Choose the conjugate that turns the denominator into an identity equal to

1

.**

How do you eliminate theta from a pair of equations?

**Rearrange each equation to isolate a ratio of

θ

, then square and add or subtract so that an identity removes

θ

completely.

Worked example 1.**

x = a cos θ

and

y = b sin θ

\frac{x}{a} = cos θ \frac{y}{b} = sin θ \frac{x ^{2}}{a ^{2}} + \frac{y ^{2}}{b ^{2}} = 1

Worked example 2.

x = a sec θ

and

y = b tan θ

\frac{x ^{2}}{a ^{2}} - \frac{y ^{2}}{b ^{2}} = sec^{2} θ - tan^{2} θ = 1

Worked example 3.

sin θ + cos θ = m

and

sin θ - cos θ = n

m^{2} + n^{2} = 2 sin^{2} θ + 2 cos^{2} θ = 2

An everyday example. A seat on a giant wheel at a mela of radius

10

m has position

x = 10 cos θ

y = 10 sin θ

. Eliminating

θ

gives

x^{2} + y^{2} = 100

— the circle the seat travels round.

The substance. **Aim to create

sin^{2} θ + cos^{2} θ

sec^{2} θ - tan^{2} θ

**, since those collapse to

1

How do you use complementary-angle relations such as sin A = cos(90° - A) to evaluate expressions?

**In a right triangle the two acute angles add to

9 0^{\circ}

, so each ratio of one angle equals the co-ratio of the other:

sin A = cos (9 0^{\circ} - A)

tan A = cot (9 0^{\circ} - A)

and

sec A = cosec (9 0^{\circ} - A)

.

Worked example 1.**

\frac{sin 3 5 ^{\circ}}{cos 5 5 ^{\circ}} = \frac{sin 3 5 ^{\circ}}{sin 3 5 ^{\circ}} = 1

Worked example 2.

tan 1 0^{\circ} tan 2 0^{\circ} tan 7 0^{\circ} tan 8 0^{\circ} = (tan 1 0^{\circ} cot 1 0^{\circ}) (tan 2 0^{\circ} cot 2 0^{\circ}) = 1

Worked example 3.

cos^{2} 2 0^{\circ} + cos^{2} 7 0^{\circ} = cos^{2} 2 0^{\circ} + sin^{2} 2 0^{\circ} = 1

Worked example 4. If

sin 3 A = cos (A + 2 6^{\circ})

, with both angles acute:

3 A + A + 2 6^{\circ} = 9 0^{\circ} 4 A = 6 4^{\circ} A = 1 6^{\circ}

An everyday example. A ladder leaning against a wall makes one angle with the ground and the complementary angle with the wall, so the sine of one equals the cosine of the other.

The substance. The prefix co- means complementary — cosine is the sine of the complement.

How do you use four-figure trigonometric tables to find a ratio or an angle?

Find the row for the degrees and the column for the minutes to read a ratio; to find an angle, search the body of the table for the given value and read its row and column, remembering that cosine values decrease as the angle increases.

Reading ratios.

sin 2 5^{\circ} = 0.4226 cos 4 0^{\circ} = 0.7660 tan 3 5^{\circ} = 0.7002

sin 2 5^{\circ} 3 0^{'} = 0.4305 cos 5 2^{\circ} = 0.6157

Finding angles.

sin θ = 0.5736 \Rightarrow θ = 3 5^{\circ} tan θ = 1.1918 \Rightarrow θ = 5 0^{\circ}

Mean difference. When the angle lies between columns, the table's mean difference is added for sine and tangent, but subtracted for cosine, because cosine falls as the angle grows.

Worked example. A kite string

100

m long makes

3 5^{\circ}

with the ground.

height = 100 \times sin 3 5^{\circ} = 100 \times 0.5736 = 57.36 m

An everyday example. Flying kites on a festival afternoon, the height of a kite can be estimated from the string length and its angle using a table.

The substance. **

sin 3 0^{\circ}

reads exactly

0.5000

**, a useful check that you are reading the right column.

Exam tip

What earns full marks on harder identities and tables?

Show each algebraic step in fractions, name the identity used, and write table values to four decimal places.

- LCM: expand the numerator fully before simplifying
- Rationalising: multiply by the conjugate that gives an identity equal to

1

- **Eliminating

θ

: isolate ratios, square, then add or subtract
- Complementary angles**: pair angles that add to

9 0^{\circ}

- Tables: subtract mean differences for cosine and cotangent
- Final line: write LHS

=

RHS or the relation without

θ

The trap. Adding the mean difference for cosine. Cosine decreases as the angle increases, so the correction is subtracted.

Did you know

Why is sin 30° exactly 0.5 when most sines are endless decimals?

Take an equilateral triangle with sides

2

cm and cut it in half. **Each half is a right triangle with angles

3 0^{\circ}

6 0^{\circ}

and

9 0^{\circ}

**, hypotenuse

2

cm and shortest side

1

cm.

sin 3 0^{\circ} = \frac{1}{2} = 0.5

Most angles do not come from such neat shapes, so their sines are irrational numbers with endless, non-repeating decimals — which is why tables give them only to four figures, such as

sin 2 5^{\circ} \approx 0.4226

Exam relevance

How does eliminating theta lead into Conic Sections for JEE Main?

This is foundation work for Class 11 Conic Sections and Trigonometric Functions, both JEE Main chapters.

What gets built on. The equations

x = a cos θ

y = b sin θ

are the parametric form of an ellipse, and eliminating

θ

gives its standard equation

\frac{x ^{2}}{a ^{2}} + \frac{y ^{2}}{b ^{2}} = 1

. Likewise

x = a sec θ

y = b tan θ

gives a hyperbola. JEE Main uses these parametric forms to find tangents, normals and locus equations.

Question types. Multiple-choice questions on parametric points and on relations obtained by eliminating a parameter.

The trap that costs marks. Adding when the identity requires subtracting —

sec^{2} θ - tan^{2} θ = 1

, not a sum.

Key takeaways

What must you be able to do from this part?

- LCM proofs: expand, use

sin^{2} A + cos^{2} A = 1

, cancel common factors
- Rationalising:

\frac{1}{s e c A - t a n A} = sec A + tan A

- **Eliminating

θ

**:

x = a cos θ

y = b sin θ

give

\frac{x ^{2}}{a ^{2}} + \frac{y ^{2}}{b ^{2}} = 1

- Complementary angles:

sin A = cos (9 0^{\circ} - A)

;

tan 1 0^{\circ} tan 2 0^{\circ} tan 7 0^{\circ} tan 8 0^{\circ} = 1

- Tables:

sin 3 5^{\circ} = 0.5736

; subtract mean differences for cosine

Eliminate

θ

from

x = 3 cosec θ

and

y = 3 cot θ

, then check your relation with

θ = 3 0^{\circ}