MathematicsClass 10CBSE

One Equation Connects Sine and Cosine at Every Angle at Once

Prove that sin squared plus cos squared is always one, derive the two identities that follow from it, simplify expressions that collapse to a single number, prove identities by converting everything to sine and cosine, and get all six ratios from one.

Why does one relation between sine and cosine make the whole chapter easier?

In the previous part you computed six ratios for an angle and noticed something every time you checked your work: **the square of the sine plus the square of the cosine always came to exactly

1

.**

For the triangle with sides

7

24

and

25

it was

\frac{49}{625} + \frac{576}{625} = 1

. For

3

4

5

it was

\frac{9}{25} + \frac{16}{25} = 1

. It was not luck — it is the Pythagoras theorem in disguise, and it holds for every angle without exception.

An equation that is true for every value of the variable is called an identity, and this one is the most useful in the subject:

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

Two more identities drop out of it for free, and between them the three do an enormous amount of work:

- They collapse long expressions into a single number, so an expression that looks like half a page turns out to equal

1

- They let you prove other relations by converting everything to sine and cosine
- They give you all six ratios from any one of them, without drawing a triangle or using the Pythagoras theorem again

That third use is the biggest practical gain. In Part 1, given

tan A = \frac{4}{3}

, you built a triangle, found the hypotenuse, and read off the rest. With identities you never leave the algebra — and when the given value is something like

\frac{7}{8}

, whose triangle has an ugly hypotenuse, that matters a great deal.

This page covers the second part of the CBSE Class 10 Maths chapter on trigonometry: the three fundamental identities and their proofs, simplifying expressions, proving identities, and expressing every ratio in terms of one.

Formula

What are the three trigonometric identities and how are they proved?

All three come from the Pythagoras theorem, divided by a different side each time.

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

1 + tan^{2} A = sec^{2} A

1 + cot^{2} A = cosec^{2} A

The proof of the first one. Take

△ A B C

right-angled at

B

, with the angle

A

at the vertex

A

. By the Pythagoras theorem,

A B^{2} + B C^{2} = A C^{2}

Now divide every term by

A C^{2}

\frac{A B ^{2}}{A C ^{2}} + \frac{B C ^{2}}{A C ^{2}} = 1

But

\frac{A B}{A C}

is the adjacent over the hypotenuse, which is

cos A

, and

\frac{B C}{A C}

is the opposite over the hypotenuse, which is

sin A

. So

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

The whole proof is one division. That is why it is worth writing out rather than memorising — the division by

A C^{2}

is the only idea in it.

**The second identity, from dividing by

A B^{2}

instead:**

1 + \frac{B C ^{2}}{A B ^{2}} = \frac{A C ^{2}}{A B ^{2}} ⟹ 1 + tan^{2} A = sec^{2} A

since

\frac{B C}{A B}

is opposite over adjacent, which is

tan A

, and

\frac{A C}{A B}

is hypotenuse over adjacent, which is

sec A

.

**The third, from dividing by

B C^{2}

:**

\frac{A B ^{2}}{B C ^{2}} + 1 = \frac{A C ^{2}}{B C ^{2}} ⟹ cot^{2} A + 1 = cosec^{2} A

An alternative route to the second and third, which some questions ask for. Divide the first identity by

cos^{2} A

\frac{sin ^{2} A}{cos ^{2} A} + 1 = \frac{1}{cos ^{2} A} ⟹ tan^{2} A + 1 = sec^{2} A

and dividing it by

sin^{2} A

gives the third. Either derivation earns full marks, and this second route makes clear that there is really only one identity here, written three ways.

The rearranged forms you will use constantly, and it is worth writing them out once:

-

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

and

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

sec^{2} A - tan^{2} A = 1

cosec^{2} A - cot^{2} A = 1

Notice the pattern in the last two. Each is a difference of two squares equal to

1

, which means

(sec A - tan A) (sec A + tan A) = 1

— so those two brackets are reciprocals of each other. That factorisation solves a whole family of questions in one line.

The restriction that examiners ask about. The first identity holds for every angle from

0^{\circ}

9 0^{\circ}

inclusive. **The second fails at

A = 9 0^{\circ}

**, because

tan 9 0^{\circ}

and

sec 9 0^{\circ}

are not defined; **the third fails at

A = 0^{\circ}

**, for the same reason about

cot 0^{\circ}

and

cosec 0^{\circ}

. So the correct statements carry the conditions

0^{\circ} \leq A < 9 0^{\circ}

and

0^{\circ} < A \leq 9 0^{\circ}

respectively, and a complete answer mentions them.

How do you simplify a trigonometric expression down to a number?

Look for a bracket or a pair of terms that matches one of the identities, replace it, and watch the expression collapse. Most such questions are built so that the answer is a small whole number.

Worked example 1. Simplify

(1 - cos^{2} A) cosec^{2} A

.

The first bracket is

sin^{2} A

by the first identity, and

cosec^{2} A

\frac{1}{s i n ^{2} A}

= sin^{2} A \times \frac{1}{sin ^{2} A} = 1

**Check at

A = 3 0^{\circ}

**:

(1 - \frac{3}{4}) \times 4 = \frac{1}{4} \times 4 = 1

. Correct, and checking at a standard angle is the fastest way to confirm a simplification.

Worked example 2. Simplify

(sec^{2} A - 1) cot^{2} A

.

The first bracket is

tan^{2} A

by the second identity, and

cot^{2} A

is its reciprocal:

= tan^{2} A \times cot^{2} A = 1

**Check at

A = 6 0^{\circ}

**:

sec 6 0^{\circ} = 2

, so the bracket is

4 - 1 = 3

; and

cot 6 0^{\circ} = \frac{1}{3}

, so

cot^{2} 6 0^{\circ} = \frac{1}{3}

. The product is

3 \times \frac{1}{3} = 1

. Correct.

Worked example 3. Simplify

(1 + tan^{2} θ) cos^{2} θ

= sec^{2} θ \times cos^{2} θ = \frac{1}{cos ^{2} θ} \times cos^{2} θ = 1

Worked example 4 — with a coefficient. Evaluate

9 sec^{2} A - 9 tan^{2} A

.

Take out the common factor first:

= 9 (sec^{2} A - tan^{2} A) = 9 \times 1 = 9

Factorising before substituting is the whole trick, and a student who converts

sec^{2} A

and

tan^{2} A

separately into sines and cosines will produce three lines of fractions and reach the same

9

much more slowly.

Worked example 5 — three brackets. Evaluate

(1 + tan^{2} θ) (1 - sin θ) (1 + sin θ)

.

Multiply the last two brackets first, since they are a difference of two squares:

(1 - sin θ) (1 + sin θ) = 1 - sin^{2} θ = cos^{2} θ

and the first bracket is

sec^{2} θ

, so

= sec^{2} θ \times cos^{2} θ = 1

**Check at

θ = 4 5^{\circ}

**:

tan 4 5^{\circ} = 1

so the first bracket is

2

;

sin 4 5^{\circ} = \frac{1}{2}

so the other two multiply to

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

. The product is

2 \times \frac{1}{2} = 1

. Correct.

Worked example 6 — a fraction. Simplify

\frac{s i n ^{2} A}{1 - c o s A}

for

A \neq = 0^{\circ}

.

Replace the numerator using the first identity and then factorise it:

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

**Check at

A = 6 0^{\circ}

**: the original is

\frac{3/4}{1/2} = \frac{3}{2}

, and

1 + cos 6 0^{\circ} = 1 + \frac{1}{2} = \frac{3}{2}

. Correct.

Notice the pattern across all six. Whenever you see

1 - cos^{2}

1 - sin^{2}

sec^{2} - 1

cosec^{2} - 1

, or a pair of brackets like

(1 - sin) (1 + sin)

, an identity is waiting. Spotting them is the entire skill, and the give-away is always a square or a difference of two squares.

How do you prove a trigonometric identity from scratch?

Work on one side only — usually the messier one — convert every ratio to sines and cosines, simplify, and arrive at the other side. Never operate on both sides at once.

Why you must not touch both sides. An identity is something you are asked to establish, so you are not yet allowed to assume it is true. Cross-multiplying across the equals sign assumes exactly what you are proving, and it is a marked error even when the algebra is otherwise perfect.

The standard opening move. Replace

tan

cot

sec

and

cosec

by their sine and cosine forms:

-

tan A = \frac{s i n A}{c o s A}

and

cot A = \frac{c o s A}{s i n A}

sec A = \frac{1}{c o s A}

and

cosec A = \frac{1}{s i n A}

Worked example 1. Prove that

sec A (1 - sin A) (sec A + tan A) = 1

.

Start with the left side and convert everything:

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

= \frac{1}{cos A} (1 - sin A) \times \frac{1 + sin A}{cos A}

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

which is the right side, so the identity is proved.

**Check numerically at

A = 3 0^{\circ}

.** Then

sec 3 0^{\circ} = \frac{2}{3} \approx 1.1547

1 - sin 3 0^{\circ} = 0.5

, and

sec 3 0^{\circ} + tan 3 0^{\circ} \approx 1.1547 + 0.5774 = 1.7321

. The product is

1.1547 \times 0.5 \times 1.7321 = 1.000

. Correct.

Worked example 2. Prove that

\frac{1 - c o s θ}{1 + c o s θ} = (cosec θ - cot θ)^{2}

.

Here the right side is the messier one, so work on it:

(cosec θ - cot θ)^{2} = (\frac{1}{sin θ} - \frac{cos θ}{sin θ})^{2} = (\frac{1 - cos θ}{sin θ})^{2} = \frac{( 1 - cos θ ) ^{2}}{sin ^{2} θ}

Now replace

sin^{2} θ

1 - cos^{2} θ

and factorise it:

= \frac{( 1 - cos θ ) ^{2}}{( 1 - cos θ ) ( 1 + cos θ )} = \frac{1 - cos θ}{1 + cos θ}

which is the left side.

**Check at

θ = 6 0^{\circ}

**: the left side is

\frac{1 - 0.5}{1 + 0.5} = \frac{1}{3}

; the right side is

(\frac{2}{3} - \frac{1}{3})^{2} = (\frac{1}{3})^{2} = \frac{1}{3}

. Correct.

**The decisive step there was replacing

sin^{2}

1 - cos^{2}

, which turned the denominator into something sharing a factor with the numerator. Whenever a proof stalls, look for a squared sine or cosine to swap — it is almost always the move that unlocks a cancellation.

Worked example 3.** Prove that

\frac{s i n A - 2 s i n ^{3} A}{2 c o s ^{3} A - c o s A} = tan A

.

Factorise the numerator and the denominator separately:

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

Now show the two brackets are the same thing. Using

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

1 - 2 sin^{2} A = 1 - 2 (1 - cos^{2} A) = 2 cos^{2} A - 1

So the brackets cancel, leaving

= \frac{sin A}{cos A} = tan A

**Check at

A = 3 0^{\circ}

**: the numerator is

0.5 - 2 (0.125) = 0.25

and the denominator is

2 (0.75) (0.866) - 0.866 = 1.299 - 0.866 = 0.433

, giving

\frac{0.25}{0.433} = 0.577 = tan 3 0^{\circ}

. Correct.

The three habits that make these proofs routine. Convert to sine and cosine; look for a difference of two squares to factorise; and swap a squared sine or cosine using the first identity whenever you need a common factor. And always finish by stating that the two sides are now equal, because the final sentence is itself worth a mark.

How do you get every ratio from one given value without a triangle?

Use the identity that contains the given ratio, solve for the one you want, and then read the rest off the definitions. No triangle, no square roots of awkward numbers.

Worked example 1. If

tan A = \frac{3}{4}

, find all the other trigonometric ratios of

A

.

**Start with the identity containing

tan

:**

sec^{2} A = 1 + tan^{2} A = 1 + \frac{9}{16} = \frac{25}{16}

sec A = \frac{5}{4}

, taking the positive root because

A

is acute. Then

cos A = \frac{1}{sec A} = \frac{4}{5}

sin A = tan A \times cos A = \frac{3}{4} \times \frac{4}{5} = \frac{3}{5}

and the reciprocals give

cosec A = \frac{5}{3}

and

cot A = \frac{4}{3}

.

Check with the first identity:

\frac{9}{25} + \frac{16}{25} = 1

. Correct.

**Notice that

sin A = tan A cos A

** — a rearrangement of

tan A = \frac{s i n A}{c o s A}

, and the quickest way to get from the cosine to the sine once you have one of them.

Worked example 2 — where the identity route clearly wins. If

cot θ = \frac{7}{8}

, evaluate

\frac{( 1 + cos θ ) ( 1 - cos θ )}{( 1 + sin θ ) ( 1 - sin θ )}

**Do not find

θ

and do not build a triangle.** Multiply out each pair of brackets:

= \frac{1 - cos ^{2} θ}{1 - sin ^{2} θ} = \frac{sin ^{2} θ}{cos ^{2} θ} = tan^{2} θ

and since

cot θ = \frac{7}{8}

, we have

tan θ = \frac{8}{7}

, so

= (\frac{8}{7})^{2} = \frac{64}{49}

**The whole expression was

tan^{2} θ

all along.** Building a triangle would have given a hypotenuse of

113

and a page of surds for the same answer.

Worked example 3 — verifying a stated relation. If

3 cot A = 4

, check whether

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

From

3 cot A = 4

we get

cot A = \frac{4}{3}

and so

tan A = \frac{3}{4}

. Using the ratios found in worked example 1,

cos A = \frac{4}{5}

and

sin A = \frac{3}{5}

.

Left side:

\frac{1 - \frac{9}{16}}{1 + \frac{9}{16}} = \frac{\frac{7}{16}}{\frac{25}{16}} = \frac{7}{25}

Right side:

\frac{16}{25} - \frac{9}{25} = \frac{7}{25}

**The two agree, so the relation is true for this value of

A

.** In fact it is an identity for every acute

A

, since dividing the numerator and denominator of the left side by

sec^{2} A

converts it into

cos^{2} A - sin^{2} A

directly.

Worked example 4 — an evaluation. If

sin A = \frac{3}{5}

, find the value of

tan A + sec A

cos^{2} A = 1 - \frac{9}{25} = \frac{16}{25} ⟹ cos A = \frac{4}{5}

tan A = \frac{3/5}{4/5} = \frac{3}{4}, sec A = \frac{5}{4}

tan A + sec A = \frac{3}{4} + \frac{5}{4} = 2

Check with the factorisation noted earlier. Since

(sec A - tan A) (sec A + tan A) = 1

, we must have

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

— and indeed

\frac{5}{4} - \frac{3}{4} = \frac{1}{2}

. Both agree, and that reciprocal pair is worth remembering because questions of the form "given

sec A + tan A

, find

sec A - tan A

" are answered by taking the reciprocal and nothing else.

The one decision to make consciously. Every identity involves squares, so solving one gives

\pm

a value. For an acute angle every ratio is positive, so you always take the positive root at this level — and remembering that this is a choice rather than an automatic step is what will save you in Class 11, where the sign depends on the quadrant.

Exam tip

What does a full-mark identity proof look like on paper?

Write "LHS =" at the start, work down one side only, and finish with "= RHS, hence proved". The structure is marked separately from the algebra.

- Choose the more complicated side to start from, whichever side of the equals sign it happens to be
- Never cross-multiply across the equals sign — that assumes the result you are proving
- Convert every ratio to sine and cosine as the opening move if nothing else suggests itself
- Look for a difference of two squares:

1 - cos^{2}

sec^{2} - 1

(1 - sin) (1 + sin)

- **Swap

sin^{2}

for

1 - cos^{2}

when you need a factor that cancels
- Take out common factors before substituting**, so

9 sec^{2} A - 9 tan^{2} A

becomes

9 (sec^{2} A - tan^{2} A)

- **Remember

(sec A - tan A) (sec A + tan A) = 1

and the matching cosec-and-cot version
- Use

sin A = tan A cos A

to jump between ratios
- Take the positive root for an acute angle, and say that you are doing so
- Verify at a standard angle** such as

3 0^{\circ}

6 0^{\circ}

before moving on
- State the conditions where a question asks for them: the second identity fails at

9 0^{\circ}

and the third at

0^{\circ}

The misconception to name.

sin^{2} A

means

(sin A)^{2}

, not

sin (A^{2})

. The square applies to the value of the ratio, not to the angle — so

sin^{2} 3 0^{\circ} = (0.5)^{2} = 0.25

, and

sin (3 0^{2})^{\circ}

is a completely different and irrelevant quantity. The notation is compressed, and reading it wrongly wrecks every substitution that follows.

A second trap. Treating

1 - cos^{2} A

1 - cos A

. A square root does not distribute over a sum or a difference, and

1 - cos^{2} A

sin A

. Testing at

A = 6 0^{\circ}

settles it:

1 - 0.25 = 0.866

, while

1 - 0.5 = 0.5

. Different numbers, and a one-line check catches the error every time.

Did you know

Why is this identity really the Pythagoras theorem wearing different clothes?

Divide the Pythagoras theorem by the square of the hypotenuse and you get the first identity. Nothing else happens. So every time you use

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

, you are using the Pythagoras theorem — just with the sides expressed as fractions of the hypotenuse instead of as lengths.

That explains something students find odd: why the identity has no units. The Pythagoras theorem relates areas, so

A B^{2} + B C^{2} = A C^{2}

has square centimetres on both sides. **Dividing by

A C^{2}

cancels them, leaving pure numbers — and pure numbers are exactly what a ratio is. The identity is the Pythagoras theorem with the size scaled out, which is why it holds for a triangle of any size and so for the angle alone.

It also explains the shape of the second and third identities.** Dividing by

A B^{2}

B C^{2}

instead of

A C^{2}

gives the same theorem measured against a different side, and since neither of those sides is the longest one, **the results come out with a

1

on the left rather than on the right. Three divisions, three identities, one theorem.

And it predicts the restrictions without any extra thought.** Dividing by

A B^{2}

requires

A B \neq = 0

, and

A B

shrinks to nothing as

A

approaches

9 0^{\circ}

. That is exactly the angle at which the second identity fails, and the algebraic reason —

tan 9 0^{\circ}

is not defined — and the geometric reason are the same reason.

One more thing follows from the picture. Because

sin^{2} A

and

cos^{2} A

are both non-negative and add to

1

, **neither can ever exceed

1

** — so

sin A

and

cos A

are trapped between

0

and

1

for an acute angle. That bound was stated in Part 1 by looking at the hypotenuse, and here it falls straight out of the identity, which is a useful sign that the two halves of the chapter agree.

In Class 11 the same identity is drawn as a circle, and the picture becomes even clearer. Mark the point at a distance

1

from the origin in the direction of the angle

A

; its coordinates are exactly

(cos A, sin A)

, and the statement that it lies on a circle of radius

1

is precisely

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

. The identity is the equation of that circle — which is why it is called the unit circle, and why the identity survives unchanged when the angle is allowed to grow past

9 0^{\circ}

Exam relevance

How are trigonometric identities tested in JEE?

This is foundation work for Class 11 Trigonometric Functions and Trigonometric Equations, and identity manipulation runs through a large part of JEE Main and JEE Advanced mathematics.

Where the three identities lead. Class 11 keeps all three unchanged and adds the compound-angle, multiple-angle, half-angle and sum-to-product formulas on top of them. **Every one of those derivations uses

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

at some point, and the habit of converting everything to sine and cosine when stuck remains the standard fallback for the rest of the subject.

Where the proof technique leads. JEE questions rarely say "prove this identity", but they constantly require an expression to be simplified before it can be evaluated or maximised. The skill being tested is the same one — spot the square, spot the difference of two squares, cancel — applied inside a longer problem. A candidate who cannot reduce an expression quickly runs out of time even when the underlying idea is easy.

Where the reciprocal pair leads.**

(sec A - tan A) (sec A + tan A) = 1

generalises: JEE sets problems giving

sec θ + tan θ

and asking for

sec θ - tan θ

, or giving

sin θ + cos θ

and asking for

sin θ cos θ

, which comes from squaring and using the first identity. Recognising that squaring a sum lets the identity in is a standard competitive move.

Where the all-ratios-from-one method leads. Class 11 adds the quadrant, so the answer carries a sign that must be decided from the given information. **The Class 10 habit of writing "taking the positive root since

A

is acute" is exactly the sentence that becomes a real decision there, and candidates who never noticed it was a choice lose marks systematically.

Where the unit-circle reading leads. It is the definition Class 11 uses for the trigonometric functions of any angle, and it underlies the graphs, the periodicity and the general solution of a trigonometric equation.

Question types to expect.** At this level: prove the identities, prove a given identity, simplify to a number, and evaluate from one given ratio. In competitive papers: simplification inside longer problems, conditional identities, maximum and minimum values of expressions such as

a sin θ + b cos θ

, and trigonometric equations.

The single trap that costs marks. Operating on both sides of an identity at once. Cross-multiplying across the equals sign assumes the very thing being proved, and in a competitive setting the same error appears as squaring an equation and then keeping a root that the original equation rejects.

A second trap. Reading

sin^{2} A

sin (A^{2})

, or distributing a square root over a difference. **

1 - cos^{2} A

sin A

, not

1 - cos A

** — and a single numerical check at

6 0^{\circ}

exposes it, which is why substituting a standard angle is worth doing even in an examination.

Board versus competitive emphasis. The CBSE paper marks the proof of the identity, the labelled triangle, the stated conditions and the one-sided working; a competitive paper marks the simplified value at the end of a longer chain. The transferable habit is scanning an expression for squares before touching it — the identities only apply to squares, and spotting one immediately is the difference between a two-line solution and a two-page one.

Key takeaways

What must you be able to do from this part?

Three identities, one proof habit and one warning about signs.

- **

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

** for every angle from

0^{\circ}

9 0^{\circ}

, proved by dividing the Pythagoras theorem by

A C^{2}

- **

1 + tan^{2} A = sec^{2} A

**, from dividing by

A B^{2}

, valid for

0^{\circ} \leq A < 9 0^{\circ}

- **

1 + cot^{2} A = cosec^{2} A

**, from dividing by

B C^{2}

, valid for

0^{\circ} < A \leq 9 0^{\circ}

- The second and third also follow from dividing the first by

cos^{2} A

and by

sin^{2} A

- Rearranged forms:

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

sec^{2} A - tan^{2} A = 1

cosec^{2} A - cot^{2} A = 1

- **

(sec A - tan A) (sec A + tan A) = 1

, so those brackets are reciprocals
-

(1 - cos^{2} A) cosec^{2} A = 1

** and

(sec^{2} A - 1) cot^{2} A = 1

and

(1 + tan^{2} θ) cos^{2} θ = 1

- Factorise before substituting:

9 sec^{2} A - 9 tan^{2} A = 9

- **

(1 + tan^{2} θ) (1 - sin θ) (1 + sin θ) = 1

**, and

\frac{s i n ^{2} A}{1 - c o s A} = 1 + cos A

- To prove an identity, work one side only, convert to sine and cosine, and never cross-multiply across the equals sign
- **

sec A (1 - sin A) (sec A + tan A) = 1

** and

\frac{1 - c o s θ}{1 + c o s θ} = (cosec θ - cot θ)^{2}

and

\frac{s i n A - 2 s i n ^{3} A}{2 c o s ^{3} A - c o s A} = tan A

- **From

tan A = \frac{3}{4}

**:

sec A = \frac{5}{4}

cos A = \frac{4}{5}

sin A = \frac{3}{5}

cot A = \frac{4}{3}

cosec A = \frac{5}{3}

- **

\frac{( 1 + c o s θ ) ( 1 - c o s θ )}{( 1 + s i n θ ) ( 1 - s i n θ )} = tan^{2} θ

**, so

cot θ = \frac{7}{8}

gives

\frac{64}{49}

- **If

sin A = \frac{3}{5}

then

tan A + sec A = 2

**, and

sec A - tan A

is its reciprocal,

\frac{1}{2}

- **

sin^{2} A

means

(sin A)^{2}

**, and

1 - cos^{2} A

sin A

, not

1 - cos A

- Take the positive root for an acute angle, and say so

The best self-test is a proof you check twice. Prove that

\frac{1 - c o s θ}{1 + c o s θ} = (cosec θ - cot θ)^{2}

working from the right-hand side only, then substitute

θ = 6 0^{\circ}

into both sides and see whether your algebra and your arithmetic tell the same story.

Why does one relation between sine and cosine make the whole chapter easier?

What are the three trigonometric identities and how are they proved?

How do you simplify a trigonometric expression down to a number?

How do you prove a trigonometric identity from scratch?

How do you get every ratio from one given value without a triangle?

What does a full-mark identity proof look like on paper?

Why is this identity really the Pythagoras theorem wearing different clothes?

How are trigonometric identities tested in JEE?

What must you be able to do from this part?

Related study notes

Ready to put this into practice?