Trigonometry

high importance~3 Q in Tier 119 formulas⚡ 15 shortcuts5 subtopics

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Fundamental identities

⏱ 3 min read🧩 5 question types🎯 17 practice Q

The idea in one minute

Three identities drive nearly every simplification: sine squared plus cosine squared equals one, and its tan-sec and cot-cosec cousins. Conjugate pairs turn sums into reciprocals, and squaring a sum removes the cross terms.

The three engines

\sin^2\theta+\cos^2\theta=1,\qquad 1+\tan^2\theta=\sec^2\theta,\qquad 1+\cot^2\theta=\cosec^2\theta

Divide the first by $\cos^2\theta$ and the second appears; divide by $\sin^2\theta$ for the third. One identity, three costumes.

Verify with $\theta=30^\circ$ : $\sin^2+\cos^2=\dfrac14+\dfrac34=1$ holds, and $1+\tan^230^\circ=1+\dfrac13=\dfrac43=\sec^230^\circ$ . Checks like this catch a misremembered formula instantly.

Rule: Spot which squared pair the question shows, then apply the matching engine and cancel.

Conjugate pairs

(\sec\theta+\tan\theta)(\sec\theta-\tan\theta)=\sec^2\theta-\tan^2\theta=1

So the two factors are reciprocals. Given $\sec+\tan=5$ , the partner is $\dfrac15$ . Adding the two: $2\sec=5+\dfrac15=\dfrac{26}{5}$ , so $\sec=\dfrac{13}{5}$ and $\tan=\dfrac{12}{5}$ .

The same trick works for $\cosec+\cot$ : the conjugate product is also $1$ .

Numbers stay friendly here. $\cosec\theta+\cot\theta=\dfrac52$ forces $\cosec\theta-\cot\theta=\dfrac25$ ; adding, $2\cosec\theta=\dfrac{29}{10}$ , so $\cosec\theta=\dfrac{29}{20}$ and $\sin\theta=\dfrac{20}{29}$ . The $20$ - $21$ - $29$ triplet was hiding inside the fractions.

Tip: A 'sum given, difference asked' question is this pair in disguise. Write the product-of-difference identity before anything else.

Squares of sums

$(a+b)^2+(a-b)^2=2(a^2+b^2)$ , so

(\sin\theta+\cos\theta)^2+(\sin\theta-\cos\theta)^2=2(\sin^2\theta+\cos^2\theta)=2

Check: each square expands to $1\pm2\sin\theta\cos\theta$ , and the cross terms cancel.

The same square bridges a given sum to a product: $\sin\theta+\cos\theta=\dfrac75$ squares to $1+2\sin\theta\cos\theta=\dfrac{49}{25}$ , so $\sin\theta\cos\theta=\dfrac{12}{25}$ . The product falls out of the square every time.

Build tan by dividing by cos

Any fraction of sines and cosines becomes a fraction of tangents after dividing top and bottom by $\cos\theta$ :

\frac{4\sin\theta-\cos\theta}{4\cos\theta+\sin\theta}=\frac{4\tan\theta-1}{4+\tan\theta}

With $\tan\theta=\dfrac34$ : numerator $3-1=2$ , denominator $4+\dfrac34=\dfrac{19}{4}$ , so the value is $\dfrac{8}{19}$ .

Watch: Divide every term, top and bottom. A missed term breaks the whole fraction.

Squared sums from a given sum

$\tan\theta+\cot\theta=5$ . Square: $\tan^2+\cot^2+2\tan\cot=25$ , and $\tan\cot=1$ , so $\tan^2+\cot^2=23$ .

The same layout gives $\sin+\cosec$ questions: square, use the product $1$ , subtract $2$ . For instance $\sin\theta+\cosec\theta=3$ squares to $\sin^2+\cosec^2+2=9$ , so $\sin^2+\cosec^2=7$ .

Remember: Products of reciprocal pairs are always $1$ : $\sin\cdot\cosec$ , $\cos\cdot\sec$ , $\tan\cdot\cot$ . That is where terms vanish.

Question types you will see

Each type: how to recognise it, the method step by step, and one question to try.

Type 1very common3 practice Q

Conjugate pairs: sec plus-minus tan, cosec plus-minus cot

How to spot it:

A sum like sec + tan given; the difference or the individuals asked.

Method

Write the conjugate product $=1$ .
Invert the given sum for the partner.
Add or subtract the pair to isolate one function.

Why it works:

The product identity makes the pair reciprocal, so one given value answers everything.

Try this

If $\sec\theta + \tan\theta = 5$ ( $\theta$ acute), the value of $\sec\theta - \tan\theta$ is:

Show solution

$(\sec+\tan)(\sec-\tan)=1$ .
$\sec-\tan=\dfrac15$ .

Answer

1/5

Type 2very common4 practice Q

Squares of sums and sin-squared plus cos-squared

How to spot it:

An expression with plus and minus twins, or an identity check.

Method

Expand both squares.
Cross terms $\\pm2\\sin\\cos$ cancel.
Use $\\sin^2+\\cos^2=1$ on what is left.

Why it works:

Twin squares always collapse to a multiple of the base identity.

Try this

The value of $(\sin\theta + \cos\theta)^2 + (\sin\theta - \cos\theta)^2$ is:

Show solution

Expand: $1+2\sin\cos+1-2\sin\cos$ .
$=2$ .

Answer

Type 3very common2 practice Q

Substituting tan via divide-by-cos

How to spot it:

A linear sin-cos fraction with tan given.

Method

Divide numerator and denominator by $\\cos\\theta$ .
Replace $\\dfrac{\\sin}{\\cos}$ by $\\tan\\theta$ .
Substitute the given value and simplify.

Why it works:

One division converts the whole fraction into tangent arithmetic.

Try this

If $\tan\theta = \frac{3}{4}$ , the value of $\frac{4\sin\theta - \cos\theta}{4\cos\theta + \sin\theta}$ is:

Show solution

$\dfrac{4\tan-1}{4+\tan}$ with $\tan=\dfrac34$ .
$=\dfrac{2}{19/4}=\dfrac{8}{19}$ .

Answer

8/19

Type 4common3 practice Q

Squared sums from a given sum

How to spot it:

A sum of reciprocal partners given; the squared sum asked.

Method

Square the given sum.
Replace the product term by $1$ .
Subtract $2$ .

Why it works:

The product of reciprocal pairs is $1$ , so squaring delivers the answer immediately.

Try this

If $\tan\theta + \cot\theta = 5$ ( $\theta$ acute), the value of $\tan^2\theta + \cot^2\theta$ is:

Show solution

$25=\tan^2+\cot^2+2\tan\cot$ .
$\tan\cot=1$ , so the sum is $25-2=23$ .

Answer

Type 5occasional

Triple-product identity checks

How to spot it:

A product of three brackets asked to simplify to a constant.

Method

Simplify each bracket using the engines.
Multiply the pieces.
Cancel to reach the constant.

Why it works:

Each bracket reduces to a squared ratio, and the product telescopes to $1$ .

Try this

The value of $(\cosec\theta-\sin\theta)(\sec\theta-\cos\theta)(\tan\theta+\cot\theta)$ is:

Show solution

$\cosec-\sin=\dfrac{\cos^2}{\sin}$ ; $\sec-\cos=\dfrac{\sin^2}{\cos}$ .
$\tan+\cot=\dfrac{1}{\sin\cos}$ .
Product $=\dfrac{\cos^2\sin^2}{\sin\cos\sin\cos}=1$ .

Answer

Formula sheet

Pythagorean identities

\sin^2+\cos^2=1,\quad 1+\tan^2=\sec^2,\quad 1+\cot^2=\cosec^2

Three engines from one identity.

Conjugate pairs

(\sec+\tan)(\sec-\tan)=1,\quad (\cosec+\cot)(\cosec-\cot)=1

Sum and difference are reciprocals.

Squares of sums

(a+b)^2+(a-b)^2=2(a^2+b^2)

Cross terms cancel in pairs.

Reciprocal products

\sin\cdot\cosec=\cos\cdot\sec=\tan\cdot\cot=1

The constant that kills cross terms.

Shortcuts that save time

⚡ Sum given, difference taken

sec + tan = 5 means sec - tan = 1/5, because their product is 1. Then add or subtract the pair.

Example

If $\sec\theta + \tan\theta = 5$ ( $\theta$ acute), the value of $\sec\theta - \tan\theta$ is:

Show solution

$(\sec+\tan)(\sec-\tan)=1$ .
$\sec-\tan=\dfrac{1}{5}$ .

Answer

1/5

⚡ Divide through by cosine

A sin-cos fraction becomes a tan fraction when every term is divided by cosine. Substitute tan and finish.

Example

If $\tan\theta = \frac{3}{4}$ , the value of $\frac{4\sin\theta - \cos\theta}{4\cos\theta + \sin\theta}$ is:

Show solution

Divide by $\cos\theta$ : $\dfrac{4\tan-1}{4+\tan}$ .
$4+\dfrac34=\dfrac{19}{4}$ .
$=\dfrac{2\times4}{19}=\dfrac{8}{19}$ .

Answer

8/19

⚡ Square and subtract two

tan + cot = 5: square it, use tan times cot = 1, and the squared sum is 25 - 2.

Example

If $\tan\theta + \cot\theta = 5$ ( $\theta$ acute), the value of $\tan^2\theta + \cot^2\theta$ is:

Show solution

$25=\tan^2+\cot^2+2$ .
$\tan^2+\cot^2=25-2=23$ .

Answer

Mistakes to avoid

Where most students lose marks on this subtopic.

Mistake 01

Writing $(\sec+\tan)(\sec-\tan)=0$ .

It equals $\sec^2-\tan^2=1$ , which makes the pair reciprocal.

Mistake 02

Forgetting the cross term when squaring a sum.

$(\tan+\cot)^2=\tan^2+\cot^2+2$ ; the $2$ must be subtracted.

Mistake 03

Dividing only the numerator by $\cos$ .

Divide every term of numerator and denominator, then substitute $\tan$ .

Mistake 04

Using $\sin^2+\cos^2=2$ .

The sum is $1$ ; the $2$ appears only after adding both squared sums.

Mistake 05

Treating $\tan\cdot\cot$ as $2$ .

Reciprocal pairs multiply to $1$ , never anything else.

Quick revision

Read this the night before the exam.

Engines: $\sin^2+\cos^2=1$ and its $\tan$ - $\sec$ , $\cot$ - $\cosec$ divisions.
Conjugates: $(\sec+\tan)(\sec-\tan)=1$ ; same for $\cosec\pm\cot$ .
Given a sum, the difference is its reciprocal; then add or subtract the pair.
$(a+b)^2+(a-b)^2=2(a^2+b^2)$ kills cross terms.
Divide sin-cos fractions by $\cos$ to build $\tan$ .
Reciprocal products are $1$ : the constant that removes cross terms.

Practice: 17 questions

Sets of 10, mixed across the question types above. Every answer has a step-by-step explanation.

Topic test · 10 questions

Suggested time 8 min · wrong answers go to your mistake notebook automatically.

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