MathsCBSE Class 10 · Introduction to Trigonometry

Trigonometric Ratios of 0° and 90°: Values and Why

Q: Why is tan 90 degrees not defined?

tan 90 degrees equals sin 90 divided by cos 90, which is 1 divided by 0. Division by zero has no value, so tan 90 degrees is not defined. In the triangle picture, the adjacent side shrinks to zero.

Q: Is tan 90 degrees equal to infinity?

In Class 10 you must write that it is not defined. Infinity is not a number and is not accepted as a value in the board exam.

Q: How do we get values at 0 and 90 degrees without a triangle?

You take a right triangle and let the acute angle get closer and closer to 0 or 90 degrees, watching the ratios settle. NCERT uses this idea to motivate the values rather than to prove them with a real triangle.

Q: Are complementary angle results like sin(90 − A) = cos A in the syllabus?

No. Trigonometric ratios of complementary angles are not in the 2026-27 syllabus.

Updated 13 October 2026 · Fact-checked

At 0 degrees, sin is 0, cos is 1 and tan is 0, while cosec and cot are not defined. At 90 degrees, sin is 1, cos is 0 and cot is 0, while tan and sec are not defined. The board asks you to evaluate expressions, spot undefined ratios and reason about why they fail.

Understand Trigonometric Ratios of 0° and 90°

Trigonometric ratios were defined for an acute angle of a right triangle. At $0^{\circ}$ or $9 0^{\circ}$ there is no genuine right triangle, so you cannot read the ratios off a figure. Instead you watch what happens to the ratios as the angle gets very close to these limits, and take the values they settle at.

Take $△ A B C$ right-angled at $B$ and keep the hypotenuse $A C$ fixed. Now shrink $∠ A$ . Point $C$ slides down towards $B$ , so $B C$ (opposite $A$ ) gets closer and closer to 0, and $A B$ gets closer to $A C$ . So $sin A = \frac{B C}{A C}$ approaches 0 and $cos A = \frac{A B}{A C}$ approaches 1. This motivates $sin 0^{\circ} = 0$ , $cos 0^{\circ} = 1$ , $tan 0^{\circ} = \frac{0}{1} = 0$ and $sec 0^{\circ} = 1$ .

Now make $∠ A$ grow towards $9 0^{\circ}$ . Then $∠ C$ shrinks towards $0^{\circ}$ , $A B$ approaches 0 and $B C$ approaches $A C$ . So $sin 9 0^{\circ} = 1$ , $cos 9 0^{\circ} = 0$ , $cot 9 0^{\circ} = \frac{0}{1} = 0$ and $cosec 9 0^{\circ} = 1$ .

Why are some ratios not defined? Every ratio is a division, and division by zero has no meaning. $cosec 0^{\circ} = \frac{1}{s i n 0 ^{\circ}}$ and $cot 0^{\circ} = \frac{c o s 0 ^{\circ}}{s i n 0 ^{\circ}}$ both divide by 0. Likewise $tan 9 0^{\circ} = \frac{s i n 9 0 ^{\circ}}{c o s 9 0 ^{\circ}}$ and $sec 9 0^{\circ} = \frac{1}{c o s 9 0 ^{\circ}}$ divide by 0. So exactly two ratios fail at each end. Write "not defined"; never write 0 or infinity.

These values connect to the rest of the chapter. They complete the table with $3 0^{\circ}$ , $4 5^{\circ}$ and $6 0^{\circ}$ : sin rises $0, \frac{1}{2}, \frac{1}{2}, \frac{3}{2}, 1$ and cos runs the same list backwards. They also extend the identities: $sin^{2} A + cos^{2} A = 1$ holds for $0^{\circ} \leq A \leq 9 0^{\circ}$ (check: $0 + 1 = 1$ at $0^{\circ}$ ), while $1 + tan^{2} A = sec^{2} A$ needs $A < 9 0^{\circ}$ and $1 + cot^{2} A = cosec^{2} A$ needs $A > 0^{\circ}$ , precisely because those ratios fail at the ends.

One last condition: an expression is defined only if no denominator becomes 0. So before substituting, check every denominator, including ones like $1 - cos θ$ at $0^{\circ}$ or $1 - sin θ$ at $9 0^{\circ}$ .

Trigonometric Ratios of 0° and 90°: formulas to remember

Ratios at 0°: $sin 0^{\circ} = 0, cos 0^{\circ} = 1, tan 0^{\circ} = 0, sec 0^{\circ} = 1$; Opposite side shrinks to 0, adjacent side becomes the hypotenuse.
Not defined at 0°: $cosec 0^{\circ}, cot 0^{\circ} not defined$; Both have sin 0° = 0 in the denominator.
Ratios at 90°: $sin 9 0^{\circ} = 1, cos 9 0^{\circ} = 0, cot 9 0^{\circ} = 0, cosec 9 0^{\circ} = 1$; Adjacent side shrinks to 0, opposite side becomes the hypotenuse.
Not defined at 90°: $tan 9 0^{\circ}, sec 9 0^{\circ} not defined$; Both have cos 90° = 0 in the denominator.
Ranges of validity of identities: $sin^{2} A + cos^{2} A = 1 (0^{\circ} \leq A \leq 9 0^{\circ}), 1 + tan^{2} A = sec^{2} A (A < 9 0^{\circ}), 1 + cot^{2} A = cosec^{2} A (A > 0^{\circ})$; An identity holds only where every ratio in it is defined.

How to solve Trigonometric Ratios of 0° and 90° questions

Use this routine for any evaluation, defined-or-not, or reasoning question involving 0° or 90°.

1Write the angle each ratio is taken at; if the angle is a sum or multiple like 2x or A + B, find that angle first.
2Check every denominator: cosec and cot fail at 0°, tan and sec fail at 90°, and expressions like 1 − cos θ can also become 0.
3If any denominator is 0, state clearly that the expression is not defined and stop.
4Otherwise substitute exact values, keeping 0s and 1s visible rather than skipping them.
5Square whole values carefully: 0² = 0 and 1² = 1, but cos²30° = 3/4.
6Simplify numerator and denominator separately, then divide.
7For an unknown angle, reduce to one ratio equal to 0 or 1 and use the fact that, for 0° to 90°, sin = 0 only at 0°, sin = 1 only at 90°, cos = 0 only at 90°, cos = 1 only at 0°.

Trigonometric Ratios of 0° and 90° in real life

Each concept, met in an everyday situation. Read the situation, try the problem, then check the solution. The one-line hook is what to remember.

Ratios at 0° and 90°

Imagine a road barrier arm at a school gate, 4 m long, hinged at one end. It rests flat (horizontal) when closed and stands straight up when fully open. When the arm makes angle θ with the ground, the height of its tip above the hinge is 4 sin θ metres and its horizontal reach is 4 cos θ metres.

Find the tip's height and reach when θ = 0°, 60° and 90°. A sensor needs the slope of the arm, tan θ. Explain for which of these angles the sensor's reading is not defined, and why.

Try it first, then show the solutionSolution

At θ = 0°: height = 4 × sin 0° = 4 × 0 = 0 m; reach = 4 × cos 0° = 4 × 1 = 4 m.
At θ = 60°: height = 4 × √3/2 = 2√3 m; reach = 4 × 1/2 = 2 m.
At θ = 90°: height = 4 × 1 = 4 m; reach = 4 × 0 = 0 m.
tan θ = height ÷ reach.
At 0°: tan 0° = 0 ÷ 4 = 0, defined.
At 60°: tan 60° = 2√3 ÷ 2 = √3, defined.
At 90°: reach is 0, so tan 90° = 4 ÷ 0, division by zero, not defined.

Answer: Heights 0 m, 2√3 m, 4 m; reaches 4 m, 2 m, 0 m. The slope reading is not defined at 90° because the horizontal reach is 0.

Memory hook: Arm flat: no height, sin 0° = 0. Arm straight up: no reach, so tan 90° breaks.

Tricks to remember Trigonometric Ratios of 0° and 90°

Short memory aids. Say each one aloud once, then use it on a question the same day.

Square-root count
Write sin as √0/2, √1/2, √2/2, √3/2, √4/2 for 0°, 30°, 45°, 60°, 90°. This gives 0, 1/2, 1/√2, √3/2, 1. Cos is the same list reversed.
Use it when: Rebuild sin 0°, sin 90°, cos 0°, cos 90° in seconds.
Zero breaks its partners
sin 0° = 0, so the ratios with sin downstairs (cosec, cot) break at 0°. cos 90° = 0, so the ratios with cos downstairs (sec, tan) break at 90°.
Use it when: Defined-or-not MCQs and assertion-reason items.
Flat ladder, standing ladder
A ladder lying flat has no height: sin 0° = 0. A ladder standing upright has no base: cos 90° = 0.
Use it when: Recover the values from a picture instead of memorising.

Trigonometric Ratios of 0° and 90°: solved examples

Ordered from board level to HOTS. Try each one before you open the solution.

Example 1: Evaluate with 0° and 90° · 2 marks · Very short answerBoard level

Evaluate: $\frac{4 s i n ^{2} 9 0 ^{\circ} + 3 c o s ^{2} 0 ^{\circ} - 2 t a n 0 ^{\circ}}{s e c ^{2} 0 ^{\circ} + c o t 9 0 ^{\circ} + cosec 9 0 ^{\circ}}$

Try it first, then show the solutionSolution

Values: $sin 9 0^{\circ} = 1$ , $cos 0^{\circ} = 1$ , $tan 0^{\circ} = 0$ , $sec 0^{\circ} = 1$ , $cot 9 0^{\circ} = 0$ , $cosec 9 0^{\circ} = 1$ . (½ mark)
Numerator $= 4 (1) + 3 (1) - 2 (0) = 7$ . (½ mark)
Denominator $= 1 + 0 + 1 = 2$ . (½ mark)
Value $= \frac{7}{2}$ . (½ mark)

Answer: $\frac{7}{2}$

Example 2: Find unknown angles from 0 and 1 · 3 marks · Short answerBoard level

If $sin (A - B) = 0$ and $cos (A + B) = 0$ , where $0^{\circ} \leq A - B$ , $A + B \leq 9 0^{\circ}$ , find $A$ and $B$ . Hence evaluate $sin^{2} A + tan B$ .

Try it first, then show the solutionSolution

$sin (A - B) = 0 = sin 0^{\circ}$ , so $A - B = 0^{\circ}$ . (½ mark)
$cos (A + B) = 0 = cos 9 0^{\circ}$ , so $A + B = 9 0^{\circ}$ . (½ mark)
Adding: $2 A = 9 0^{\circ}$ , so $A = 4 5^{\circ}$ ; then $B = 4 5^{\circ}$ . (1 mark)
$sin^{2} 4 5^{\circ} = \frac{1}{2}$ and $tan 4 5^{\circ} = 1$ . (½ mark)
$sin^{2} A + tan B = \frac{1}{2} + 1 = \frac{3}{2}$ . (½ mark)

Answer: $A = 4 5^{\circ}$ , $B = 4 5^{\circ}$ ; value $= \frac{3}{2}$

Example 3: Which expression is defined (hidden angle) · 1 mark · MCQHarder board level

For $θ = 4 5^{\circ}$ , which of the following expressions is defined?
(a) $\frac{t a n 2 θ}{1 + s i n θ}$
(b) $\frac{1 + c o s 2 θ}{1 - s i n 2 θ}$
(c) $sec 2 θ - cot 2 θ$
(d) $\frac{s i n 2 θ + c o s 2 θ}{cosec 2 θ + c o t 2 θ}$

Try it first, then show the solutionSolution

The hidden angle is $2 θ = 9 0^{\circ}$ .
(a): $tan 9 0^{\circ}$ is not defined.
(b): $1 - sin 9 0^{\circ} = 0$ in the denominator, not defined.
(c): $sec 9 0^{\circ}$ is not defined.
(d): $\frac{1 + 0}{1 + 0} = 1$ , defined.

Answer: (d), value 1

Example 4: Long answer: values, an angle and an identity check · 5 marks · Long answerHarder board level

(i) Evaluate $\frac{5 c o s ^{2} 6 0 ^{\circ} + 4 s e c ^{2} 3 0 ^{\circ} - t a n ^{2} 4 5 ^{\circ}}{s i n ^{2} 3 0 ^{\circ} + c o s ^{2} 3 0 ^{\circ}} + 2 sin 9 0^{\circ} cos 0^{\circ}$ .
(ii) Find the acute or zero angle $x$ ( $0^{\circ} \leq 3 x \leq 9 0^{\circ}$ ) if $2 cos 3 x - 2 = 0$ .
(iii) Verify $1 + tan^{2} A = sec^{2} A$ at $A = 0^{\circ}$ and explain why it cannot be checked at $A = 9 0^{\circ}$ .

Try it first, then show the solutionSolution

(i) $cos^{2} 6 0^{\circ} = \frac{1}{4}$ , $sec^{2} 3 0^{\circ} = \frac{4}{3}$ , $tan^{2} 4 5^{\circ} = 1$ . (½ mark)
Numerator $= \frac{5}{4} + \frac{16}{3} - 1 = \frac{15 + 64 - 12}{12} = \frac{67}{12}$ . (½ mark)
Denominator $= \frac{1}{4} + \frac{3}{4} = 1$ . (½ mark)
$2 sin 9 0^{\circ} cos 0^{\circ} = 2$ , so total $= \frac{67}{12} + 2 = \frac{91}{12}$ . (½ mark)
(ii) $cos 3 x = 1 = cos 0^{\circ}$ , so $3 x = 0^{\circ}$ . (½ mark)
Hence $x = 0^{\circ}$ . (½ mark)
(iii) At $0^{\circ}$ : LHS $= 1 + 0 = 1$ , RHS $= 1^{2} = 1$ , verified. (1 mark)
At $9 0^{\circ}$ , $tan 9 0^{\circ}$ and $sec 9 0^{\circ}$ are not defined since $cos 9 0^{\circ} = 0$ , so the identity holds only for $0^{\circ} \leq A < 9 0^{\circ}$ . (1 mark)

Answer: (i) $\frac{91}{12}$ (ii) $x = 0^{\circ}$ (iii) Holds at $0^{\circ}$ ; not checkable at $9 0^{\circ}$ because tan and sec are not defined there.

Example 5: Prove an identity and find where it is valid · 3 marks · Short answerHarder board level

Prove that $(sec A - tan A) (sec A + tan A) = 1$ , using $sin^{2} A + cos^{2} A = 1$ . For which values of $A$ with $0^{\circ} \leq A \leq 9 0^{\circ}$ is it valid? Evaluate both sides at $A = 0^{\circ}$ .

Try it first, then show the solutionSolution

LHS $= sec^{2} A - tan^{2} A$ , using $(a - b) (a + b) = a^{2} - b^{2}$ . (½ mark)
$= \frac{1}{c o s ^{2} A} - \frac{s i n ^{2} A}{c o s ^{2} A} = \frac{1 - s i n ^{2} A}{c o s ^{2} A}$ . (½ mark)
Since $sin^{2} A + cos^{2} A = 1$ , $1 - sin^{2} A = cos^{2} A$ , so LHS $= \frac{c o s ^{2} A}{c o s ^{2} A} = 1 =$ RHS. (1 mark)
sec and tan need $cos A \neq = 0$ ; $cos A = 0$ only at $9 0^{\circ}$ , so the identity is valid for $0^{\circ} \leq A < 9 0^{\circ}$ . (½ mark)
At $A = 0^{\circ}$ : LHS $= (1 - 0) (1 + 0) = 1$ and RHS $= 1$ , so both sides agree. (½ mark)

Answer: Proved; valid for $0^{\circ} \leq A < 9 0^{\circ}$ ; at $A = 0^{\circ}$ both sides equal 1.

Example 6: Long answer: hidden angles, evaluation and a proof · 5 marks · Long answerHarder board level

(i) If $sin (2 x + 3 0^{\circ}) = 1$ and $tan (3 y - 4 5^{\circ}) = 0$ , where $0^{\circ} \leq 2 x + 3 0^{\circ} \leq 9 0^{\circ}$ and $0^{\circ} \leq 3 y - 4 5^{\circ} < 9 0^{\circ}$ , find $x$ and $y$ .
(ii) Hence evaluate $E = \frac{t a n ^{2} ( x + y ) + 2 s i n 3 x + 4 c o s ^{2} 2 x}{c o s ( 2 x + 3 0 ^{\circ} ) + s e c 0 ^{\circ}}$ .
(iii) Using $sin^{2} A + cos^{2} A = 1$ , prove that $(sin A + cos A)^{2} + (sin A - cos A)^{2} = 2$ , and verify it at $A = x$ .

Try it first, then show the solutionSolution

(i) $sin (2 x + 3 0^{\circ}) = 1 = sin 9 0^{\circ}$ , so $2 x + 3 0^{\circ} = 9 0^{\circ}$ . (½ mark)
$2 x = 6 0^{\circ}$ , so $x = 3 0^{\circ}$ . (½ mark)
$tan (3 y - 4 5^{\circ}) = 0 = tan 0^{\circ}$ , so $3 y - 4 5^{\circ} = 0^{\circ}$ . (½ mark)
$3 y = 4 5^{\circ}$ , so $y = 1 5^{\circ}$ . (½ mark)
(ii) $x + y = 4 5^{\circ}$ , $3 x = 9 0^{\circ}$ , $2 x = 6 0^{\circ}$ , $2 x + 3 0^{\circ} = 9 0^{\circ}$ : $tan^{2} 4 5^{\circ} = 1$ , $sin 9 0^{\circ} = 1$ , $cos^{2} 6 0^{\circ} = \frac{1}{4}$ , $cos 9 0^{\circ} = 0$ , $sec 0^{\circ} = 1$ . (½ mark)
$E = \frac{1 + 2 ( 1 ) + 4 \times \frac{1}{4}}{0 + 1} = \frac{4}{1} = 4$ . (½ mark)
(iii) LHS $= (sin^{2} A + 2 sin A cos A + cos^{2} A) + (sin^{2} A - 2 sin A cos A + cos^{2} A)$ . (½ mark)
$= 2 sin^{2} A + 2 cos^{2} A = 2 (sin^{2} A + cos^{2} A)$ . (½ mark)
$= 2 \times 1 = 2 =$ RHS, using $sin^{2} A + cos^{2} A = 1$ . (½ mark)
Check at $A = 3 0^{\circ}$ : $\frac{( 1 + 3 ) ^{2}}{4} + \frac{( 1 - 3 ) ^{2}}{4} = \frac{( 4 + 2 3 ) + ( 4 - 2 3 )}{4} = \frac{8}{4} = 2$ . (½ mark)

Answer: (i) $x = 3 0^{\circ}$ , $y = 1 5^{\circ}$ (ii) $E = 4$ (iii) Proved; at $A = 3 0^{\circ}$ both sides equal 2.

Example 7: Case study: the drawbridge · 4 marks · Case-basedHOTS / competency

Imagine a small canal drawbridge with a deck 10 m long, hinged at one end. It lifts in stages from flat to vertical. When it makes angle $θ$ with the water, the deck tip is $10 sin θ$ m above the hinge level and $10 cos θ$ m horizontally from the hinge.
(i) Find the tip's height when $θ = 0^{\circ}$ and when $θ = 6 0^{\circ}$ . (1 mark)
(ii) Find the tip's horizontal distance from the hinge when $θ = 3 0^{\circ}$ and when $θ = 9 0^{\circ}$ . (1 mark)
(iii) The control panel shows the reading $R = \frac{height of tip}{10 - horizontal distance of tip}$ . Find $R$ at $θ = 6 0^{\circ}$ . For which angle $θ$ with $0^{\circ} \leq θ \leq 9 0^{\circ}$ is $R$ not defined? Give a reason. (2 marks)

Try it first, then show the solutionSolution

(i) At $0^{\circ}$ : height $= 10 \times 0 = 0$ m. At $6 0^{\circ}$ : height $= 10 \times \frac{3}{2} = 53$ m. (1 mark)
(ii) At $3 0^{\circ}$ : distance $= 10 \times \frac{3}{2} = 53$ m. At $9 0^{\circ}$ : distance $= 10 \times 0 = 0$ m. (1 mark)
(iii) $R = \frac{10 s i n θ}{10 - 10 c o s θ} = \frac{s i n θ}{1 - c o s θ}$ . (½ mark)
At $6 0^{\circ}$ : $R = \frac{3 /2}{1 - 1/2} = \frac{3 /2}{1/2} = 3$ . (½ mark)
$R$ is not defined when the denominator $1 - cos θ = 0$ , that is when $cos θ = 1$ . (½ mark)
For $0^{\circ} \leq θ \leq 9 0^{\circ}$ , $cos θ = 1$ only at $θ = 0^{\circ}$ . There the flat deck's tip is 10 m from the hinge, so the denominator is $10 - 10 = 0$ and $R = \frac{0}{0}$ has no value. At $9 0^{\circ}$ , $R = \frac{10}{10} = 1$ is fine. So $R$ is not defined only at $0^{\circ}$ . (½ mark)

Answer: (i) 0 m and $53$ m (ii) $53$ m and 0 m (iii) $R = 3$ at $6 0^{\circ}$ ; $R$ is not defined only at $θ = 0^{\circ}$ , because $1 - cos 0^{\circ} = 0$ .

Where students lose marks in Trigonometric Ratios of 0° and 90°

The slips examiners see most often, and what to write instead.

Mistake: Writing an undefined ratio as 0
Students treat 1/0 as 0.
Wrong working
$sec 9 0^{\circ} = \frac{1}{c o s 9 0 ^{\circ}} = \frac{1}{0} = 0$
Right working
$cos 9 0^{\circ} = 0$ , so $sec 9 0^{\circ} = \frac{1}{0}$ is not defined.
Do this: Whenever a denominator is 0, write "not defined" and stop.
Usually costs: The full mark of an MCQ or the answer mark of a longer question.
Mistake: Swapping sin and cos values at 0° and 90°
Both values are 0 or 1, so they get mixed up.
Wrong working
$sin 0^{\circ} = 1$ , $cos 0^{\circ} = 0$
Right working
$sin 0^{\circ} = 0$ , $cos 0^{\circ} = 1$ (flat triangle: no opposite side).
Do this: Picture the flat ladder: no height, so sin is 0.
Usually costs: Every following step of the evaluation.
Mistake: Ignoring a denominator like 1 − sin θ
Students only check the six named ratios.
Wrong working
At $9 0^{\circ}$ : $\frac{c o s θ}{1 - s i n θ} = \frac{0}{0} = 0$
Right working
$1 - sin 9 0^{\circ} = 0$ , so the expression is not defined.
Do this: Check every denominator in the whole expression before substituting.
Usually costs: 1 mark in Section A.
Mistake: Using an identity outside its range
All identities are assumed to hold for every angle.
Wrong working
$1 + tan^{2} 9 0^{\circ} = sec^{2} 9 0^{\circ}$ , so both sides equal 1.
Right working
$tan 9 0^{\circ}$ and $sec 9 0^{\circ}$ are not defined; this identity holds only for $0^{\circ} \leq A < 9 0^{\circ}$ .
Do this: Quote the range with the identity.
Usually costs: The reasoning mark in assertion-reason or proof questions.

Exam tips

In defined-or-not MCQs, test each option's denominators one by one; usually only one option survives.
Write "not defined" in words; examiners do not accept 0, infinity or a blank.
Questions often hide 0° or 90° inside A + B or 3x; find the angle first, then read the value.
When an assertion uses 0° or 90°, check whether the reason actually explains the assertion, not just whether both are true.
Show each substituted value in the first line; even if arithmetic slips, the substitution step earns marks.

Practice questions on Trigonometric Ratios of 0° and 90°

Write each answer on paper first. Open the hint only if you are stuck, and the answer only when you have finished.

Q1 · 2 marks · Very short answerBoard level
Evaluate: $\frac{s i n 9 0 ^{\circ} + 3 c o s 0 ^{\circ} - 2 c o t 9 0 ^{\circ}}{s e c 0 ^{\circ} + 3 cosec ^{2} 9 0 ^{\circ}}$
Hint
cot 90° = 0 and cosec 90° = 1.
Answer
Numerator $= 1 + 3 - 0 = 4$ ; denominator $= 1 + 3 = 4$ ; value $= 1$ .
Q2 · 1 mark · Assertion-reasonBoard level
Assertion (A): The expression $\frac{c o s θ}{1 - s i n θ}$ is defined at $θ = 0^{\circ}$ but not at $θ = 9 0^{\circ}$ .
Reason (R): $sin 0^{\circ} = 0$ and $sin 9 0^{\circ} = 1$ .
(a) Both A and R are true and R is the correct explanation of A.
(b) Both A and R are true but R is not the correct explanation of A.
(c) A is true but R is false.
(d) A is false but R is true.
Hint
Find the denominator at each angle.
Answer
(a). At $0^{\circ}$ the value is $\frac{1}{1} = 1$ ; at $9 0^{\circ}$ the denominator is $1 - 1 = 0$ . R gives exactly why.
Q3 · 3 marks · Short answerBoard level
If $tan (2 A - B) = 0$ and $sin (A + B) = 1$ , with $0^{\circ} \leq 2 A - B$ , $A + B \leq 9 0^{\circ}$ , find $A$ and $B$ , then evaluate $cos^{2} A + sec^{2} B$ .
Hint
tan = 0 only at 0°, sin = 1 only at 90°. Solve the two linear equations.
Answer
$2 A - B = 0$ , $A + B = 9 0^{\circ}$ . Adding: $3 A = 9 0^{\circ}$ , $A = 3 0^{\circ}$ , $B = 6 0^{\circ}$ . $cos^{2} 3 0^{\circ} + sec^{2} 6 0^{\circ} = \frac{3}{4} + 4 = \frac{19}{4}$ .
Q4 · 5 marks · Long answerHarder board level
(i) Evaluate $\frac{2 s i n ^{2} 9 0 ^{\circ} - c o s ^{2} 0 ^{\circ} + 4 s i n ^{2} 3 0 ^{\circ}}{t a n ^{2} 6 0 ^{\circ} - s e c 0 ^{\circ}}$ . (ii) Show that $1 + cot^{2} A = cosec^{2} A$ holds at $A = 9 0^{\circ}$ and $A = 4 5^{\circ}$ and explain why it cannot be tested at $A = 0^{\circ}$ . (iii) Find $θ$ with $0^{\circ} \leq θ \leq 9 0^{\circ}$ if $sin θ + cos 9 0^{\circ} = 1$ .
Hint
In (iii), cos 90° = 0 leaves sin θ = 1.
Answer
(i) Numerator $= 2 - 1 + 1 = 2$ , denominator $= 3 - 1 = 2$ , value $1$ . (ii) $9 0^{\circ}$ : $1 + 0 = 1 = 1^{2}$ ; $4 5^{\circ}$ : $1 + 1 = 2 = (2)^{2}$ ; at $0^{\circ}$ , cot and cosec are not defined since $sin 0^{\circ} = 0$ . (iii) $sin θ = 1$ , so $θ = 9 0^{\circ}$ .
Q5 · 3 marks · Short answerHOTS / competency
Riya writes: " $tan 9 0^{\circ} = \frac{s i n 9 0 ^{\circ}}{c o s 9 0 ^{\circ}} = \frac{1}{0} = 0$ , so $tan 9 0^{\circ} + sec 0^{\circ} = 1$ ." Find her error. Then, using a right triangle with fixed hypotenuse, explain what happens to $tan A$ as $A$ approaches $9 0^{\circ}$ .
Hint
What is 1 divided by 0? Watch the adjacent side.
Answer
Division by zero is not allowed, so $\frac{1}{0}$ is not 0; $tan 9 0^{\circ}$ is not defined and the sum is not defined. As $A \to 9 0^{\circ}$ , opposite side $\to$ hypotenuse while adjacent side $\to 0$ , so $tan A = \frac{opp}{adj}$ keeps growing without settling at any number.

Test your recall

Answer in your head or on paper first, then open the card. Pulling an answer out of memory makes it stick far better than re-reading.

1.sin 0° and cos 0°?Show answer
0 and 1
2.sin 90° and cos 90°?Show answer
1 and 0
3.Which ratios are not defined at 0°?Show answer
cosec 0° and cot 0°
4.Which ratios are not defined at 90°?Show answer
tan 90° and sec 90°
5.tan 0° and cot 90°?Show answer
Both are 0
6.sec 0° and cosec 90°?Show answer
Both are 1

Trigonometric Ratios of 0° and 90°: frequently asked questions

Why is tan 90 degrees not defined?

tan 90 degrees equals sin 90 divided by cos 90, which is 1 divided by 0. Division by zero has no value, so tan 90 degrees is not defined. In the triangle picture, the adjacent side shrinks to zero.

Is tan 90 degrees equal to infinity?

In Class 10 you must write that it is not defined. Infinity is not a number and is not accepted as a value in the board exam.

How do we get values at 0 and 90 degrees without a triangle?

You take a right triangle and let the acute angle get closer and closer to 0 or 90 degrees, watching the ratios settle. NCERT uses this idea to motivate the values rather than to prove them with a real triangle.

Are complementary angle results like sin(90 − A) = cos A in the syllabus?

No. Trigonometric ratios of complementary angles are not in the 2026-27 syllabus.