Trigonometry

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Value-putting and given-ratio questions

⏱ 3 min read🧩 5 question types🎯 15 practice Q

The idea in one minute

Value-putting questions hand you a ratio or a condition and ask for an expression. Two moves cover them: substitute standard angle values directly, or draw the triangle the condition describes and read the expression off it.

Put the standard values in

$2\tan^2 45^\circ+\cos^2 30^\circ-\sin^2 60^\circ$ : with $\tan45^\circ=1$ and both $30^\circ$ / $60^\circ$ squared terms equal to $\dfrac34$ :

2+\frac34-\frac34=2

Notice how $\cos^2 30^\circ$ and $\sin^2 60^\circ$ cancel by symmetry before any arithmetic.

One more: $\sin^230^\circ+\tan^245^\circ+\cos^260^\circ=\dfrac14+1+\dfrac14=\dfrac32$ . Substitute first, add the fractions second, and the answer arrives without any identity at all.

Rule: Replace every ratio first, simplify second. Half of these questions cancel themselves.

sin + cos given

Square the given sum: $(\sin\theta+\cos\theta)^2=1+2\sin\theta\cos\theta$ . So with the sum $=\sqrt2$ :

2\sin\theta\cos\theta=2-1=1\Rightarrow\sin\theta\cos\theta=\frac12

The same square gives $\sin\theta-\cos\theta$ when the sum is known (watch the sign). With $\sin+\cos=\dfrac75$ the difference squares to $1-2\sin\cos=1-\dfrac{24}{25}=\dfrac{1}{25}$ , so $|\sin-\cos|=\dfrac15$ .

Tip: $\sin+\cos=\sqrt2$ happens only at $45^\circ$ ; the product $\dfrac12$ confirms it.

tan condition into a fraction

$5\tan\theta=4$ and the fraction $\dfrac{5\sin\theta-3\cos\theta}{5\sin\theta+3\cos\theta}$ . Divide top and bottom by $\cos\theta$ :

\frac{5\tan\theta-3}{5\tan\theta+3}=\frac{4-3}{4+3}=\frac17

The condition turns the whole expression into one number.

cot or tan fraction: draw the triangle

$\cot\theta=\dfrac{21}{20}$ : adjacent $21$ , opposite $20$ , hypotenuse $\sqrt{441+400}=29$ . Then $\cos\theta=\dfrac{21}{29}$ directly.

The triplet $20$ - $21$ - $29$ was hiding inside the fraction. Drawing beats manipulating.

With $\tan\theta=\dfrac{4}{3}$ instead, the same $3$ - $4$ - $5$ drawing hands over $\sec\theta=\dfrac{5}{4}$ and $\cosec\theta=\dfrac{5}{3}$ at a glance. The triangle keeps paying rent.

Watch: For $\cot$ the first number is the adjacent side, not the opposite. That single slip flips every answer.

Sum of reciprocal partners

$\cosec\theta+\cot\theta=3$ . The conjugate is $\dfrac13$ , so $2\cosec\theta=3+\dfrac13=\dfrac{10}{3}$ , giving $\cosec\theta=\dfrac53$ and $\sin\theta=\dfrac35$ .

One reciprocal move, no quadratic.

The pair also splits a value the other way: $\sec\theta+\tan\theta=3$ gives $\sec\theta-\tan\theta=\dfrac13$ , hence $\sec\theta=\dfrac{5}{3}$ and $\tan\theta=\dfrac{4}{3}$ . Add or subtract the pair, and the individuals appear.

Remember: Value-putting never needs to solve for $\theta$ itself. The ratio travels, the angle never appears.

Question types you will see

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

Type 1very common2 practice Q

Standard angles in an expression

How to spot it:

An expression entirely in standard angles.

Method

Replace each ratio by its value.
Cancel symmetric twin terms.
Finish the arithmetic.

Why it works:

Direct substitution plus one cancellation; the fastest pattern in the subtopic.

Try this

The value of $2\tan^2 45^\circ + \cos^2 30^\circ - \sin^2 60^\circ$ is:

Show solution

$2\tan^245^\circ=2(1)^2=2$ .
$\cos^230^\circ=\sin^260^\circ=\dfrac34$ , so they cancel.
Total $=2$ .

Answer

Type 2very common4 practice Q

sin plus cos given, products and differences

How to spot it:

A sum or difference of sin and cos given; a product or square asked.

Method

Square the given relation.
Use $1+2\\sin\\cos$ or $1-2\\sin\\cos$ .
Solve for the asked expression.

Why it works:

Squaring converts sums into products through the base identity.

Try this

If $\sin\theta + \cos\theta = \sqrt{2}$ , the value of $\sin\theta\cos\theta$ is:

Show solution

$2=1+2\\sin\\cos$ .
$\\sin\\cos=\\dfrac12$ .

Answer

1/2

Type 3very common3 practice Q

tan or cot condition into a linear fraction

How to spot it:

A multiple of tan given; a linear sin-cos fraction asked.

Method

Divide the fraction by $\\cos\\theta$ throughout.
Substitute the given $\\tan\\theta$ value.
Simplify the small fraction.

Why it works:

The division aligns the fraction with the condition so only arithmetic remains.

Try this

If $5\tan\theta = 4$ , the value of $\frac{5\sin\theta - 3\cos\theta}{5\sin\theta + 3\cos\theta}$ is:

Show solution

$\\dfrac{5\\tan-3}{5\\tan+3}=\\dfrac{4-3}{4+3}$ .
$=\\dfrac17$ .

Answer

1/7

Type 4common2 practice Q

cot or tan fraction to triangle to ratio

How to spot it:

A cot or tan value; a plain ratio asked.

Method

Place the two sides in a right triangle.
Find the hypotenuse by the triplet.
Read off the asked ratio.

Why it works:

Drawing the triangle answers every ratio at once without any identity work.

Try this

If $\cot\theta = \frac{21}{20}$ ( $\theta$ acute), the value of $\cos\theta$ is:

Show solution

Adjacent $21$ , opposite $20$ ; hypotenuse $=\\sqrt{441+400}=29$ .
$\\cos\\theta=\\dfrac{21}{29}$ .

Answer

21/29

Type 5common

cosec plus cot given, find sin

How to spot it:

A sum of reciprocal partners given; a basic ratio asked.

Method

Write the conjugate difference as the reciprocal $\\dfrac{1}{\\text{sum}}$ .
Add sum and difference to get $2\\cosec\\theta$ .
Flip for $\\sin\\theta$ .

Why it works:

The conjugate identity avoids a quadratic; two additions deliver the ratio.

Try this

If $\cosec\theta + \cot\theta = 3$ ( $\theta$ acute), the value of $\sin\theta$ is:

Show solution

$\\cosec-\\cot=\\dfrac13$ .
$2\\cosec=3+\\dfrac13=\\dfrac{10}{3}$ , so $\\cosec=\\dfrac53$ .
$\\sin=\\dfrac35$ .

Answer

3/5

Formula sheet

Square of sin+cos

(\sin\theta+\cos\theta)^2=1+2\sin\theta\cos\theta

The bridge from a sum to a product.

Divide by cosine

\frac{a\sin+b\cos}{c\sin+d\cos}=\frac{a\tan+b}{c\tan+d}

After dividing every term by cosine.

Cot fraction to triangle

\cot\theta=\frac{21}{20}\Rightarrow\text{hyp}=29

Two sides given, Pythagoras gives the third.

Reciprocal pair sum

x+\frac{1}{x}\ \text{from}\ x\cdot\frac{1}{x}=1

Conjugates of cosec plus cot.

Shortcuts that save time

⚡ Symmetry cancels first

cos squared 30 and sin squared 60 are the same number; scan for such twins before computing.

Example

The value of $2\tan^2 45^\circ + \cos^2 30^\circ - \sin^2 60^\circ$ is:

Show solution

$\tan45^\circ=1$ gives $2$ .
$\cos^230^\circ=\sin^260^\circ=\dfrac34$ : they cancel.
Total $=2$ .

Answer

⚡ Square the sum

sin + cos given: square it to reach 1 + 2 sin cos, then read off the product.

Example

If $\sin\theta + \cos\theta = \sqrt{2}$ , the value of $\sin\theta\cos\theta$ is:

Show solution

$2=1+2\sin\cos$ .
$2\sin\cos=1$ .
$\sin\cos=\dfrac12$ .

Answer

1/2

⚡ Condition straight into the fraction

5 tan = 4: divide the fraction by cosine, substitute, one line of arithmetic.

Example

If $5\tan\theta = 4$ , the value of $\frac{5\sin\theta - 3\cos\theta}{5\sin\theta + 3\cos\theta}$ is:

Show solution

Divide by $\cos\theta$ : $\dfrac{5\tan-3}{5\tan+3}$ .
$=\dfrac{4-3}{4+3}$ .
$=\dfrac17$ .

Answer

1/7

Mistakes to avoid

Where most students lose marks on this subtopic.

Mistake 01

Solving for $\theta$ when only the ratio is needed.

The condition $5\tan=4$ feeds the fraction directly; the angle never appears.

Mistake 02

Forgetting the $1$ in $(\sin+\cos)^2=1+2\sin\cos$ .

Square fully: the product term sits on top of the identity $1$ .

Mistake 03

Dividing only part of the fraction by $\cos$ .

Every term, numerator and denominator, divides by $\cos\theta$ .

Mistake 04

Reading $\cot=\dfrac{21}{20}$ with $21$ opposite.

Cot is adjacent over opposite: $21$ adjacent, $20$ opposite, hypotenuse $29$ .

Mistake 05

Sign slips with $\sin-\cos$ from a known $\sin+\cos$ .

Square the difference: $(\sin-\cos)^2=1-2\sin\cos$ , then choose the sign from the angle range.

Quick revision

Read this the night before the exam.

Substitute standard values first; scan for twin terms that cancel.
$(\sin+\cos)^2=1+2\sin\cos$ : sums give products.
Divide sin-cos fractions by $\cos$ and insert the given $\tan$ .
Cot fractions: draw the triangle, complete the triplet ( $20$ - $21$ - $29$ ).
$\cosec+\cot=3$ gives $\cosec=\dfrac53$ via the reciprocal partner, so $\sin=\dfrac35$ .
The ratio travels; the angle itself never needs finding.

Practice: 15 questions

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

Topic test · 10 questions

Suggested time 6 min · wrong answers go to your mistake notebook automatically.

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