Introduction to Trigonometry

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CLASS X Mathematics ~6 marks/year Ch 8 of 14
Introduction to Trigonometry

Class 10 · Mathematics · NCERT chapter notes · Akanksha Classes

Snapshot
  • Trigonometry studies the relationship between the sides and acute angles of a right triangle through six ratios: $\sin,\cos,\tan,\csc,\sec,\cot$.
  • For an acute angle $A$: $\sin A=\dfrac{\text{opposite}}{\text{hypotenuse}}$, $\cos A=\dfrac{\text{adjacent}}{\text{hypotenuse}}$, $\tan A=\dfrac{\text{opposite}}{\text{adjacent}}$ — the other three are their reciprocals.
  • The values at $0^\circ,30^\circ,45^\circ,60^\circ,90^\circ$ come from simple triangles and must be memorised (Table 8.1).
  • Three identities tie everything together: $\sin^{2}\theta+\cos^{2}\theta=1$, $\;1+\tan^{2}\theta=\sec^{2}\theta$, $\;1+\cot^{2}\theta=\csc^{2}\theta$.
  • Board weightage: ~6 marks/year — usually one "evaluate / find ratios" sum (2 marks), one prove-the-identity question (3 marks) and one MCQ.
Detailed notes

1. What trigonometry is — and why

The word trigonometry comes from three Greek words: tri (three), gon (sides) and metron (measure) — literally "measuring three-sided figures". It is the study of the relationships between the sides and angles of a triangle, and in Class 10 we restrict ourselves to right triangles and their acute angles.

Why bother? Because it lets us find heights and distances we cannot measure directly. If you stand on the ground and look up at the top of the Qutub Minar, a right triangle is formed — and knowing one angle plus one length is enough to compute the tower's height without climbing it. The same idea finds the width of a river from a balcony, or the altitude of a hot-air balloon. The earliest work was recorded in Egypt and Babylon, and early astronomers used it to measure distances to stars and planets. Even today, engineering and the physical sciences rest on these concepts.

2. Naming the sides of a right triangle

Take a right triangle $ABC$ with the right angle at $B$, and focus on the acute angle $A$. Three labels matter, and they are always named relative to the angle you are looking at:

  • Hypotenuse — the side opposite the right angle (here $AC$). It is the longest side and never changes.
  • Opposite side — the side facing angle $A$ (here $BC$). It is the side "across from" $A$.
  • Adjacent side — the side next to angle $A$ that is not the hypotenuse (here $AB$).

Key warning: "opposite" and "adjacent" swap when you switch attention to the other acute angle $C$. For angle $C$, side $AB$ becomes the opposite and $BC$ becomes the adjacent. The hypotenuse $AC$ stays the hypotenuse for both. Always fix which angle you are working with first.

3. The six trigonometric ratios

For the acute angle $A$ in right triangle $ABC$ (right-angled at $B$), the six trigonometric ratios are defined as:

$$\sin A=\frac{\text{opposite}}{\text{hyp}}=\frac{BC}{AC},\qquad \cos A=\frac{\text{adjacent}}{\text{hyp}}=\frac{AB}{AC},\qquad \tan A=\frac{\text{opposite}}{\text{adjacent}}=\frac{BC}{AB}$$
$$\csc A=\frac{1}{\sin A}=\frac{AC}{BC},\qquad \sec A=\frac{1}{\cos A}=\frac{AC}{AB},\qquad \cot A=\frac{1}{\tan A}=\frac{AB}{BC}$$

So $\csc A,\sec A,\cot A$ are simply the reciprocals of $\sin A,\cos A,\tan A$ respectively. Two more relations drop out instantly by dividing:

$$\tan A=\frac{\sin A}{\cos A},\qquad \cot A=\frac{\cos A}{\sin A}$$

Memory aid: "Some People Have / Curly Brown Hair / Through Proper Brushing" gives $\sin=\tfrac{P}{H}$, $\cos=\tfrac{B}{H}$, $\tan=\tfrac{P}{B}$ (Perpendicular, Base, Hypotenuse).

Crucial note on notation: $\sin A$ is a single symbol — it is NOT "$\sin$ multiplied by $A$". "$\sin$" alone has no meaning, and you can never separate it from its angle. We also write $\sin^{2}A$ to mean $(\sin A)^{2}$. But $\csc A=(\sin A)^{-1}$ is not the same as $\sin^{-1}A$ (which means something different — studied later). The Greek letter $\theta$ (theta) is often used in place of $A$ for an angle.

History: the idea of "sine" first appears in Aryabhata's Aryabhatiyam (A.D. 500), where he used ardha-jya (half-chord), later shortened to jya. Through Arabic and Latin translation this became sinus, then "sin". The notations "cos" and "tan" came much later.

4. The ratios depend only on the angle, not the triangle's size

This is the idea that makes trigonometry work. If you take a larger or smaller right triangle that keeps the same angle $A$, every ratio comes out identical. The reason is similar triangles: enlarging the triangle multiplies all three sides by the same factor, which cancels in every ratio.

Formally, if $P$ is any point on the hypotenuse $AC$ and $PM\perp AB$, then $\triangle PAM\sim\triangle CAB$ (AA criterion). Corresponding sides are proportional, so $\dfrac{MP}{AP}=\dfrac{BC}{AC}=\sin A$ regardless of where $P$ sits. Conclusion: the value of $\sin A$ (and every other ratio) depends only on the size of the angle $A$, not on the lengths of the triangle's sides.

Remark: since the hypotenuse is the longest side, $\sin A$ and $\cos A$ are always $\le 1$. So a value like $\sin\theta=\tfrac{4}{3}$ is impossible.

5. Finding all ratios from one given ratio

If one ratio is known you can recover the others. Draw the right triangle, use the given ratio to label two sides (introducing a positive multiplier $k$), find the third side by Pythagoras, then read off the rest.

Worked illustration: if $\sin A=\dfrac13$, then $\dfrac{BC}{AC}=\dfrac13$, so take $BC=k,\ AC=3k$. By Pythagoras $AB^{2}=AC^{2}-BC^{2}=9k^{2}-k^{2}=8k^{2}$, giving $AB=2\sqrt2\,k$. Therefore $\cos A=\dfrac{AB}{AC}=\dfrac{2\sqrt2\,k}{3k}=\dfrac{2\sqrt2}{3}$, and the others follow.

NCERT Example 1 — given $\tan A=\dfrac43$, find the other ratios

Draw $\triangle ABC$ right-angled at $B$. Since $\tan A=\dfrac{BC}{AB}=\dfrac43$, take $BC=4k,\ AB=3k$.

By Pythagoras: $AC^{2}=AB^{2}+BC^{2}=(3k)^{2}+(4k)^{2}=25k^{2}$, so $AC=5k$.

Hence $\sin A=\dfrac{BC}{AC}=\dfrac{4k}{5k}=\dfrac45$ and $\cos A=\dfrac{AB}{AC}=\dfrac{3k}{5k}=\dfrac35$.

The reciprocals: $\cot A=\dfrac{1}{\tan A}=\dfrac34$, $\;\csc A=\dfrac{1}{\sin A}=\dfrac54$, $\;\sec A=\dfrac{1}{\cos A}=\dfrac53$.

NCERT Example 2 — if $\sin B=\sin Q$, prove $\angle B=\angle Q$

Take right triangles $ABC$ (right angle at $C$) and $PQR$ (right angle at $R$) with $\sin B=\sin Q$. Then $\dfrac{AC}{AB}=\dfrac{PR}{PQ}$, so $\dfrac{AC}{PR}=\dfrac{AB}{PQ}=k$ (say). By Pythagoras $BC=\sqrt{AB^{2}-AC^{2}}$ and $QR=\sqrt{PQ^{2}-PR^{2}}$, and substituting $AB=k\,PQ,\ AC=k\,PR$ gives $\dfrac{BC}{QR}=k$ too. So all three pairs of sides are in the ratio $k$, hence $\triangle ACB\sim\triangle PRQ$ and therefore $\angle B=\angle Q$.

NCERT Example 3 — $\triangle ACB$ right-angled at $C$, $AB=29$, $BC=21$, $\angle ABC=\theta$

First $AC=\sqrt{AB^{2}-BC^{2}}=\sqrt{29^{2}-21^{2}}=\sqrt{(29-21)(29+21)}=\sqrt{8\times50}=\sqrt{400}=20$ units.

So $\sin\theta=\dfrac{AC}{AB}=\dfrac{20}{29}$ and $\cos\theta=\dfrac{BC}{AB}=\dfrac{21}{29}$.

(i) $\cos^{2}\theta+\sin^{2}\theta=\left(\dfrac{21}{29}\right)^{2}+\left(\dfrac{20}{29}\right)^{2}=\dfrac{441+400}{841}=\dfrac{841}{841}=1.$

(ii) $\cos^{2}\theta-\sin^{2}\theta=\dfrac{21^{2}-20^{2}}{29^{2}}=\dfrac{(21+20)(21-20)}{841}=\dfrac{41}{841}.$

NCERT Example 4 — $\triangle ABC$ right-angled at $B$, $\tan A=1$; verify $2\sin A\cos A=1$

$\tan A=\dfrac{BC}{AB}=1\Rightarrow BC=AB$. Let $AB=BC=k$. Then $AC=\sqrt{k^{2}+k^{2}}=k\sqrt2$.

So $\sin A=\dfrac{BC}{AC}=\dfrac{1}{\sqrt2}$ and $\cos A=\dfrac{AB}{AC}=\dfrac{1}{\sqrt2}$.

Hence $2\sin A\cos A=2\cdot\dfrac{1}{\sqrt2}\cdot\dfrac{1}{\sqrt2}=2\cdot\dfrac12=1.$ Verified.

NCERT Example 5 — $\triangle OPQ$ right-angled at $P$, $OP=7$ cm, $OQ-PQ=1$ cm; find $\sin Q,\cos Q$

By Pythagoras $OQ^{2}=OP^{2}+PQ^{2}$. Since $OQ=1+PQ$: $(1+PQ)^{2}=7^{2}+PQ^{2}$, i.e. $1+2\,PQ+PQ^{2}=49+PQ^{2}$, so $2\,PQ=48\Rightarrow PQ=24$ cm and $OQ=1+24=25$ cm.

Therefore $\sin Q=\dfrac{OP}{OQ}=\dfrac{7}{25}$ and $\cos Q=\dfrac{PQ}{OQ}=\dfrac{24}{25}.$

6. Ratios of specific angles — where the values come from

You do not memorise these blindly; each comes from a simple triangle.

For $45^\circ$: in a right triangle with one angle $45^\circ$, the other acute angle is also $45^\circ$, so the two legs are equal. Let $AB=BC=a$; then $AC=a\sqrt2$. Hence $\sin45^\circ=\dfrac{a}{a\sqrt2}=\dfrac{1}{\sqrt2}$, $\cos45^\circ=\dfrac{1}{\sqrt2}$, $\tan45^\circ=\dfrac{a}{a}=1$.

For $30^\circ$ and $60^\circ$: take an equilateral triangle (all angles $60^\circ$) and drop a perpendicular from a vertex. This makes a right triangle with angles $30^\circ$ and $60^\circ$, splitting the base in half. With $AB=2a$, $BD=a$ and $AD=\sqrt{(2a)^{2}-a^{2}}=a\sqrt3$. So $\sin30^\circ=\dfrac{a}{2a}=\dfrac12$, $\cos30^\circ=\dfrac{a\sqrt3}{2a}=\dfrac{\sqrt3}{2}$, $\tan30^\circ=\dfrac{1}{\sqrt3}$; and $\sin60^\circ=\dfrac{\sqrt3}{2}$, $\cos60^\circ=\dfrac12$, $\tan60^\circ=\sqrt3$.

For $0^\circ$ and $90^\circ$: imagine angle $A$ shrinking to $0$ — the opposite side $BC\to0$, so $\sin0^\circ=0$ and $\cos0^\circ=1$. As $A\to90^\circ$, the roles reverse: $\sin90^\circ=1$, $\cos90^\circ=0$. From these, $\tan0^\circ=0$, while $\tan90^\circ$, $\sec90^\circ$, $\cot0^\circ$ and $\csc0^\circ$ are not defined (division by zero).

7. The standard-angle table (Table 8.1)

This table must be at your fingertips. Note the elegant pattern in $\sin$: the values are $\sqrt{0}/2,\sqrt1/2,\sqrt2/2,\sqrt3/2,\sqrt4/2$. As the angle rises $0^\circ\to90^\circ$, $\sin$ increases $0\to1$ and $\cos$ decreases $1\to0$.

$\angle A$ $0^\circ$ $30^\circ$ $45^\circ$ $60^\circ$ $90^\circ$
$\sin A$$0$$\dfrac12$$\dfrac{1}{\sqrt2}$$\dfrac{\sqrt3}{2}$$1$
$\cos A$$1$$\dfrac{\sqrt3}{2}$$\dfrac{1}{\sqrt2}$$\dfrac12$$0$
$\tan A$$0$$\dfrac{1}{\sqrt3}$$1$$\sqrt3$Not defined
$\csc A$Not defined$2$$\sqrt2$$\dfrac{2}{\sqrt3}$$1$
$\sec A$$1$$\dfrac{2}{\sqrt3}$$\sqrt2$$2$Not defined
$\cot A$Not defined$\sqrt3$$1$$\dfrac{1}{\sqrt3}$$0$
NCERT Example 6 — $\triangle ABC$ right-angled at $B$, $AB=5$ cm, $\angle ACB=30^\circ$; find $BC,AC$

For $BC$: $AB$ is opposite to $C$ and $BC$ is adjacent, so $\dfrac{AB}{BC}=\tan C=\tan30^\circ=\dfrac{1}{\sqrt3}$, giving $\dfrac{5}{BC}=\dfrac{1}{\sqrt3}$, so $BC=5\sqrt3$ cm.

For $AC$: $\sin30^\circ=\dfrac{AB}{AC}\Rightarrow\dfrac12=\dfrac{5}{AC}\Rightarrow AC=10$ cm. (Check by Pythagoras: $\sqrt{5^{2}+(5\sqrt3)^{2}}=\sqrt{25+75}=10$ cm.)

NCERT Example 7 — $\triangle PQR$ right-angled at $Q$, $PQ=3$ cm, $PR=6$ cm; find $\angle QPR,\angle PRQ$

$\sin R=\dfrac{PQ}{PR}=\dfrac36=\dfrac12$, so $\angle PRQ=30^\circ$. The angles of a triangle sum to $180^\circ$, so $\angle QPR=180^\circ-90^\circ-30^\circ=60^\circ$.

NCERT Example 8 — $\sin(A-B)=\dfrac12$, $\cos(A+B)=\dfrac12$, with $0^\circ<A+B\le90^\circ$, $A>B$; find $A,B$

$\sin(A-B)=\dfrac12\Rightarrow A-B=30^\circ$ … (1). $\;\cos(A+B)=\dfrac12\Rightarrow A+B=60^\circ$ … (2).

Adding (1)+(2): $2A=90^\circ\Rightarrow A=45^\circ$. Substituting back: $B=15^\circ$. So $A=45^\circ,\ B=15^\circ$.

8. NCERT Exercise 8.1 — fully solved

Q1. $\triangle ABC$ right-angled at $B$, $AB=24$ cm, $BC=7$ cm. First $AC=\sqrt{24^{2}+7^{2}}=\sqrt{576+49}=\sqrt{625}=25$ cm.

  • (i) For angle $A$: opposite $=BC=7$, adjacent $=AB=24$. So $\sin A=\dfrac{7}{25}$, $\cos A=\dfrac{24}{25}$.
  • (ii) For angle $C$: opposite $=AB=24$, adjacent $=BC=7$. So $\sin C=\dfrac{24}{25}$, $\cos C=\dfrac{7}{25}$.

Q2. In Fig. 8.13 (right-angled at $Q$, $PR=13$, $PQ=12$), find $\tan P-\cot R$. $QR=\sqrt{13^{2}-12^{2}}=\sqrt{25}=5$. Now $\tan P=\dfrac{QR}{PQ}=\dfrac{5}{12}$ and $\cot R=\dfrac{QR}{PQ}=\dfrac{5}{12}$ (for angle $R$, opposite is $PQ$, adjacent is $QR$, so $\cot R=\dfrac{\text{adj}}{\text{opp}}=\dfrac{QR}{PQ}$). Hence $\tan P-\cot R=\dfrac{5}{12}-\dfrac{5}{12}=0.$

Q3. If $\sin A=\dfrac34$, find $\cos A$ and $\tan A$. Take opposite $=3k$, hyp $=4k$; adjacent $=\sqrt{16k^{2}-9k^{2}}=k\sqrt7$. So $\cos A=\dfrac{\sqrt7}{4}$ and $\tan A=\dfrac{3}{\sqrt7}$.

Q4. Given $15\cot A=8$, find $\sin A$ and $\sec A$. $\cot A=\dfrac{8}{15}=\dfrac{\text{adj}}{\text{opp}}$, so adjacent $=8k$, opposite $=15k$, hyp $=\sqrt{64k^{2}+225k^{2}}=17k$. Thus $\sin A=\dfrac{15}{17}$ and $\sec A=\dfrac{\text{hyp}}{\text{adj}}=\dfrac{17}{8}$.

Q5. Given $\sec\theta=\dfrac{13}{12}$, find all other ratios. hyp $=13k$, adjacent $=12k$, opposite $=\sqrt{169k^{2}-144k^{2}}=5k$. So $\sin\theta=\dfrac{5}{13}$, $\cos\theta=\dfrac{12}{13}$, $\tan\theta=\dfrac{5}{12}$, $\csc\theta=\dfrac{13}{5}$, $\cot\theta=\dfrac{12}{5}$.

Q6. If $\cos A=\cos B$ for acute $A,B$, show $\angle A=\angle B$. In a right triangle, $\cos A=\dfrac{AB}{AC}$ and $\cos B=\dfrac{\text{adj to }B}{\text{hyp}}$. Building two right triangles as in Example 2, $\cos A=\cos B$ forces the side ratios to match, so the triangles are similar and $\angle A=\angle B$.

Q7. If $\cot\theta=\dfrac78$, evaluate:

  • (i) $\dfrac{(1+\sin\theta)(1-\sin\theta)}{(1+\cos\theta)(1-\cos\theta)}=\dfrac{1-\sin^{2}\theta}{1-\cos^{2}\theta}=\dfrac{\cos^{2}\theta}{\sin^{2}\theta}=\cot^{2}\theta=\left(\dfrac78\right)^{2}=\dfrac{49}{64}.$
  • (ii) $\cot^{2}\theta=\dfrac{49}{64}.$

Q8. If $3\cot A=4$, check whether $\dfrac{1-\tan^{2}A}{1+\tan^{2}A}=\cos^{2}A-\sin^{2}A$. Here $\cot A=\dfrac43$, so $\tan A=\dfrac34$. Take opposite $=3,$ adjacent $=4,$ hyp $=5$: $\sin A=\dfrac35,\cos A=\dfrac45$. LHS $=\dfrac{1-\frac{9}{16}}{1+\frac{9}{16}}=\dfrac{7/16}{25/16}=\dfrac{7}{25}$. RHS $=\dfrac{16}{25}-\dfrac{9}{25}=\dfrac{7}{25}$. They are equal.

Q9. $\triangle ABC$ right-angled at $B$, $\tan A=\dfrac{1}{\sqrt3}$. Find: Here $A=30^\circ$ and $C=60^\circ$, so $\sin A=\tfrac12,\cos A=\tfrac{\sqrt3}{2},\sin C=\tfrac{\sqrt3}{2},\cos C=\tfrac12$.

  • (i) $\sin A\cos C+\cos A\sin C=\dfrac12\cdot\dfrac12+\dfrac{\sqrt3}{2}\cdot\dfrac{\sqrt3}{2}=\dfrac14+\dfrac34=1.$
  • (ii) $\cos A\cos C-\sin A\sin C=\dfrac{\sqrt3}{2}\cdot\dfrac12-\dfrac12\cdot\dfrac{\sqrt3}{2}=\dfrac{\sqrt3}{4}-\dfrac{\sqrt3}{4}=0.$

Q10. $\triangle PQR$ right-angled at $Q$, $PR+QR=25$ cm, $PQ=5$ cm. Find $\sin P,\cos P,\tan P$. By Pythagoras $PR^{2}=PQ^{2}+QR^{2}$, and $PR=25-QR$: $(25-QR)^{2}=5^{2}+QR^{2}\Rightarrow 625-50\,QR=25\Rightarrow QR=12$ cm, so $PR=13$ cm. Then $\sin P=\dfrac{QR}{PR}=\dfrac{12}{13}$, $\cos P=\dfrac{PQ}{PR}=\dfrac{5}{13}$, $\tan P=\dfrac{QR}{PQ}=\dfrac{12}{5}$.

Q11. True or false (justify):

  • (i) "$\tan A$ is always $<1$." False — e.g. $\tan60^\circ=\sqrt3>1$.
  • (ii) "$\sec A=\dfrac{12}{5}$ for some $A$." True — $\sec A\ge1$ is allowed, and $\dfrac{12}{5}=2.4>1$.
  • (iii) "$\cos A$ is the abbreviation for cosecant of $A$." False — $\cos A$ is cosine; cosecant is $\csc A$.
  • (iv) "$\cot A$ is the product of cot and $A$." False — $\cot A$ is a single symbol, not a product.
  • (v) "$\sin\theta=\dfrac43$ for some $\theta$." False — $\sin\theta\le1$ always, and $\dfrac43>1$.

9. NCERT Exercise 8.2 — fully solved

Q1. Evaluate:

  • (i) $\sin60^\circ\cos30^\circ+\sin30^\circ\cos60^\circ=\dfrac{\sqrt3}{2}\cdot\dfrac{\sqrt3}{2}+\dfrac12\cdot\dfrac12=\dfrac34+\dfrac14=1.$
  • (ii) $2\tan^{2}45^\circ+\cos^{2}30^\circ-\sin^{2}60^\circ=2(1)^{2}+\left(\dfrac{\sqrt3}{2}\right)^{2}-\left(\dfrac{\sqrt3}{2}\right)^{2}=2+\dfrac34-\dfrac34=2.$
  • (iii) $\dfrac{\cos45^\circ}{\sec30^\circ+\csc30^\circ}=\dfrac{1/\sqrt2}{\frac{2}{\sqrt3}+2}=\dfrac{1/\sqrt2}{\frac{2+2\sqrt3}{\sqrt3}}=\dfrac{\sqrt3}{\sqrt2\,(2+2\sqrt3)}=\dfrac{\sqrt3}{2\sqrt2(1+\sqrt3)}$. Rationalising, this equals $\dfrac{3\sqrt2-\sqrt6}{8}.$
  • (iv) $\dfrac{\sin30^\circ+\tan45^\circ-\csc60^\circ}{\sec30^\circ+\cos60^\circ+\cot45^\circ}=\dfrac{\frac12+1-\frac{2}{\sqrt3}}{\frac{2}{\sqrt3}+\frac12+1}=\dfrac{\frac32-\frac{2}{\sqrt3}}{\frac32+\frac{2}{\sqrt3}}$. Multiplying through by $2\sqrt3$: $=\dfrac{3\sqrt3-4}{3\sqrt3+4}$. Rationalising gives $\dfrac{43-24\sqrt3}{11}.$
  • (v) $\dfrac{5\cos^{2}60^\circ+4\sec^{2}30^\circ-\tan^{2}45^\circ}{\sin^{2}30^\circ+\cos^{2}30^\circ}$. Numerator $=5\left(\dfrac12\right)^{2}+4\left(\dfrac{2}{\sqrt3}\right)^{2}-1^{2}=\dfrac54+\dfrac{16}{3}-1=\dfrac{15+64-12}{12}=\dfrac{67}{12}$. Denominator $=1$. So the value is $\dfrac{67}{12}.$

Q2. Choose the correct option:

  • (i) $\dfrac{2\tan30^\circ}{1+\tan^{2}30^\circ}=\dfrac{2/\sqrt3}{1+\frac13}=\dfrac{2/\sqrt3}{4/3}=\dfrac{2}{\sqrt3}\cdot\dfrac34=\dfrac{\sqrt3}{2}=\sin60^\circ$. (A).
  • (ii) $\dfrac{1-\tan^{2}45^\circ}{1+\tan^{2}45^\circ}=\dfrac{1-1}{1+1}=0$. (D).
  • (iii) "$\sin2A=2\sin A$" holds when $A=$ (A) $0^\circ$ (check: $\sin0^\circ=0=2\sin0^\circ$).
  • (iv) $\dfrac{2\tan30^\circ}{1-\tan^{2}30^\circ}=\dfrac{2/\sqrt3}{1-\frac13}=\dfrac{2/\sqrt3}{2/3}=\sqrt3=\tan60^\circ$. (C).

Q3. $\tan(A+B)=\sqrt3$, $\tan(A-B)=\dfrac{1}{\sqrt3}$, with $0^\circ<A+B\le90^\circ$, $A>B$. Find $A,B$. $\tan(A+B)=\sqrt3\Rightarrow A+B=60^\circ$; $\tan(A-B)=\dfrac{1}{\sqrt3}\Rightarrow A-B=30^\circ$. Adding: $2A=90^\circ\Rightarrow A=45^\circ$; then $B=15^\circ$.

Q4. True or false (justify):

  • (i) "$\sin(A+B)=\sin A+\sin B$." False — take $A=B=30^\circ$: $\sin60^\circ=\dfrac{\sqrt3}{2}$ but $\sin30^\circ+\sin30^\circ=1$.
  • (ii) "$\sin\theta$ increases as $\theta$ increases." True for $0^\circ$ to $90^\circ$ (values rise $0\to1$).
  • (iii) "$\cos\theta$ increases as $\theta$ increases." False — $\cos$ decreases from $1$ to $0$.
  • (iv) "$\sin\theta=\cos\theta$ for all $\theta$." False — true only at $\theta=45^\circ$.
  • (v) "$\cot A$ is not defined for $A=0^\circ$." True — $\cot0^\circ=\dfrac{\cos0^\circ}{\sin0^\circ}=\dfrac10$, undefined.

10. The three trigonometric identities

An equation true for all values of the angle is an identity. Start from Pythagoras in $\triangle ABC$ (right-angled at $B$): $AB^{2}+BC^{2}=AC^{2}$. Dividing each term by a different square produces the three identities.

Dividing by $AC^{2}$: $\left(\dfrac{AB}{AC}\right)^{2}+\left(\dfrac{BC}{AC}\right)^{2}=1$, i.e. $(\cos A)^{2}+(\sin A)^{2}=1$:

$$\sin^{2}\theta+\cos^{2}\theta=1\qquad(\text{true for all }0^\circ\le\theta\le90^\circ)$$

Dividing by $AB^{2}$ gives $1+\tan^{2}A=\sec^{2}A$:

$$1+\tan^{2}\theta=\sec^{2}\theta\qquad(0^\circ\le\theta<90^\circ)$$

Dividing by $BC^{2}$ gives $\cot^{2}A+1=\csc^{2}A$:

$$1+\cot^{2}\theta=\csc^{2}\theta\qquad(0^\circ<\theta\le90^\circ)$$

With these, knowing one ratio gives all the rest. E.g. if $\tan A=\dfrac{1}{\sqrt3}$, then $\sec^{2}A=1+\dfrac13=\dfrac43$, so $\sec A=\dfrac{2}{\sqrt3}$, $\cos A=\dfrac{\sqrt3}{2}$, and $\sin A=\sqrt{1-\cos^{2}A}=\sqrt{1-\tfrac34}=\dfrac12$, giving $\csc A=2$.

NCERT Example 9 — express $\cos A,\tan A,\sec A$ in terms of $\sin A$

From $\cos^{2}A+\sin^{2}A=1$: $\cos A=\sqrt{1-\sin^{2}A}$ (positive for acute $A$). Then $\tan A=\dfrac{\sin A}{\cos A}=\dfrac{\sin A}{\sqrt{1-\sin^{2}A}}$ and $\sec A=\dfrac{1}{\cos A}=\dfrac{1}{\sqrt{1-\sin^{2}A}}.$

NCERT Example 10 — prove $\sec A(1-\sin A)(\sec A+\tan A)=1$

LHS $=\dfrac{1}{\cos A}(1-\sin A)\left(\dfrac{1}{\cos A}+\dfrac{\sin A}{\cos A}\right)=\dfrac{(1-\sin A)(1+\sin A)}{\cos^{2}A}=\dfrac{1-\sin^{2}A}{\cos^{2}A}=\dfrac{\cos^{2}A}{\cos^{2}A}=1=$ RHS.

NCERT Example 11 — prove $\dfrac{\cot A-\cos A}{\cot A+\cos A}=\dfrac{\csc A-1}{\csc A+1}$

LHS $=\dfrac{\frac{\cos A}{\sin A}-\cos A}{\frac{\cos A}{\sin A}+\cos A}=\dfrac{\cos A\left(\frac{1}{\sin A}-1\right)}{\cos A\left(\frac{1}{\sin A}+1\right)}=\dfrac{\frac{1}{\sin A}-1}{\frac{1}{\sin A}+1}=\dfrac{\csc A-1}{\csc A+1}=$ RHS.

NCERT Example 12 — prove $\dfrac{\sin\theta-\cos\theta+1}{\sin\theta+\cos\theta-1}=\dfrac{1}{\sec\theta-\tan\theta}$ (use $\sec^{2}\theta=1+\tan^{2}\theta$)

Divide numerator and denominator of LHS by $\cos\theta$: $=\dfrac{\tan\theta-1+\sec\theta}{\tan\theta+1-\sec\theta}$. Multiply top and bottom by $(\tan\theta-\sec\theta)$ and use $\tan^{2}\theta-\sec^{2}\theta=-1$. After simplification the numerator becomes $-1-\tan\theta+\sec\theta$ and the whole expression reduces to $\dfrac{-1}{\tan\theta-\sec\theta}=\dfrac{1}{\sec\theta-\tan\theta}=$ RHS.

11. NCERT Exercise 8.3 — fully solved

Q1. Express $\sin A,\sec A,\tan A$ in terms of $\cot A$. Using $\csc^{2}A=1+\cot^{2}A$: $\sin A=\dfrac{1}{\csc A}=\dfrac{1}{\sqrt{1+\cot^{2}A}}$. Then $\tan A=\dfrac{1}{\cot A}$, and $\sec A=\dfrac{1}{\cos A}=\dfrac{\sqrt{1+\cot^{2}A}}{\cot A}$ (since $\cos A=\cot A\sin A=\dfrac{\cot A}{\sqrt{1+\cot^{2}A}}$).

Q2. Write all other ratios of $\angle A$ in terms of $\sec A$. $\cos A=\dfrac{1}{\sec A}$; $\sin A=\sqrt{1-\cos^{2}A}=\dfrac{\sqrt{\sec^{2}A-1}}{\sec A}$; $\tan A=\sqrt{\sec^{2}A-1}$; $\csc A=\dfrac{\sec A}{\sqrt{\sec^{2}A-1}}$; $\cot A=\dfrac{1}{\sqrt{\sec^{2}A-1}}.$

Q3. Choose the correct option:

  • (i) $9\sec^{2}A-9\tan^{2}A=9(\sec^{2}A-\tan^{2}A)=9(1)=9.$ (B).
  • (ii) $(1+\tan\theta+\sec\theta)(1+\cot\theta-\csc\theta)=2.$ (C) (expand and use identities).
  • (iii) $(\sec A+\tan A)(1-\sin A)=\left(\dfrac{1+\sin A}{\cos A}\right)(1-\sin A)=\dfrac{1-\sin^{2}A}{\cos A}=\dfrac{\cos^{2}A}{\cos A}=\cos A.$ (D).
  • (iv) $\dfrac{1+\tan^{2}A}{1+\cot^{2}A}=\dfrac{\sec^{2}A}{\csc^{2}A}=\dfrac{1/\cos^{2}A}{1/\sin^{2}A}=\tan^{2}A.$ (D).

Q4. Prove the following identities:

  • (i) $(\csc\theta-\cot\theta)^{2}=\dfrac{1-\cos\theta}{1+\cos\theta}$. LHS $=\left(\dfrac{1-\cos\theta}{\sin\theta}\right)^{2}=\dfrac{(1-\cos\theta)^{2}}{\sin^{2}\theta}=\dfrac{(1-\cos\theta)^{2}}{1-\cos^{2}\theta}=\dfrac{(1-\cos\theta)^{2}}{(1-\cos\theta)(1+\cos\theta)}=\dfrac{1-\cos\theta}{1+\cos\theta}.$
  • (ii) $\dfrac{\cos A}{1+\sin A}+\dfrac{1+\sin A}{\cos A}=2\sec A$. LHS $=\dfrac{\cos^{2}A+(1+\sin A)^{2}}{(1+\sin A)\cos A}=\dfrac{\cos^{2}A+1+2\sin A+\sin^{2}A}{(1+\sin A)\cos A}=\dfrac{2+2\sin A}{(1+\sin A)\cos A}=\dfrac{2(1+\sin A)}{(1+\sin A)\cos A}=\dfrac{2}{\cos A}=2\sec A.$
  • (iii) $\dfrac{\tan\theta}{1-\cot\theta}+\dfrac{\cot\theta}{1-\tan\theta}=1+\sec\theta\csc\theta$. Writing in $\sin,\cos$: $=\dfrac{\sin^{2}\theta}{\cos\theta(\sin\theta-\cos\theta)}+\dfrac{\cos^{2}\theta}{\sin\theta(\cos\theta-\sin\theta)}=\dfrac{\sin^{3}\theta-\cos^{3}\theta}{\sin\theta\cos\theta(\sin\theta-\cos\theta)}=\dfrac{\sin^{2}\theta+\sin\theta\cos\theta+\cos^{2}\theta}{\sin\theta\cos\theta}=\dfrac{1+\sin\theta\cos\theta}{\sin\theta\cos\theta}=1+\sec\theta\csc\theta.$
  • (iv) $\dfrac{1+\sec A}{\sec A}=\dfrac{\sin^{2}A}{1-\cos A}$. RHS $=\dfrac{1-\cos^{2}A}{1-\cos A}=\dfrac{(1-\cos A)(1+\cos A)}{1-\cos A}=1+\cos A$. LHS $=\dfrac{1}{\sec A}+1=\cos A+1$. Both equal $1+\cos A.$
  • (v) $\dfrac{\cos A-\sin A+1}{\cos A+\sin A-1}=\csc A+\cot A$ (use $\csc^{2}A=1+\cot^{2}A$). Divide numerator and denominator by $\sin A$ and simplify; the expression reduces to $\csc A+\cot A.$
  • (vi) $\sqrt{\dfrac{1+\sin A}{1-\sin A}}=\sec A+\tan A$. Multiply inside by $\dfrac{1+\sin A}{1+\sin A}$: $\sqrt{\dfrac{(1+\sin A)^{2}}{1-\sin^{2}A}}=\sqrt{\dfrac{(1+\sin A)^{2}}{\cos^{2}A}}=\dfrac{1+\sin A}{\cos A}=\sec A+\tan A.$
  • (vii) $\dfrac{\sin\theta-2\sin^{3}\theta}{2\cos^{3}\theta-\cos\theta}=\tan\theta$. LHS $=\dfrac{\sin\theta(1-2\sin^{2}\theta)}{\cos\theta(2\cos^{2}\theta-1)}$. Since $1-2\sin^{2}\theta=2\cos^{2}\theta-1$, the brackets cancel: $=\dfrac{\sin\theta}{\cos\theta}=\tan\theta.$
  • (viii) $(\sin A+\csc A)^{2}+(\cos A+\sec A)^{2}=7+\tan^{2}A+\cot^{2}A$. Expand: $\sin^{2}A+2+\csc^{2}A+\cos^{2}A+2+\sec^{2}A=1+4+\csc^{2}A+\sec^{2}A=5+(1+\cot^{2}A)+(1+\tan^{2}A)=7+\tan^{2}A+\cot^{2}A.$
  • (ix) $(\csc A-\sin A)(\sec A-\cos A)=\dfrac{1}{\tan A+\cot A}$. LHS $=\dfrac{1-\sin^{2}A}{\sin A}\cdot\dfrac{1-\cos^{2}A}{\cos A}=\dfrac{\cos^{2}A}{\sin A}\cdot\dfrac{\sin^{2}A}{\cos A}=\sin A\cos A$. RHS $=\dfrac{1}{\frac{\sin A}{\cos A}+\frac{\cos A}{\sin A}}=\dfrac{\sin A\cos A}{\sin^{2}A+\cos^{2}A}=\sin A\cos A.$ Equal.
  • (x) $\left(\dfrac{1+\tan^{2}A}{1+\cot^{2}A}\right)=\left(\dfrac{1-\tan A}{1-\cot A}\right)^{2}=\tan^{2}A$. First part $=\dfrac{\sec^{2}A}{\csc^{2}A}=\tan^{2}A$ (Q3(iv)). Second: $\dfrac{1-\tan A}{1-\cot A}=\dfrac{1-\tan A}{1-\frac{1}{\tan A}}=\dfrac{(1-\tan A)\tan A}{\tan A-1}=-\tan A$, whose square is $\tan^{2}A.$

12. Common mistakes to avoid

  • Mixing up opposite and adjacent after switching the angle — always fix the angle first.
  • Treating $\sin A$ as "$\sin\times A$" — it is one inseparable symbol; you cannot cancel "$\sin$".
  • Confusing $\csc A=(\sin A)^{-1}$ with $\sin^{-1}A$ — they are completely different.
  • Writing $\sin\theta=\dfrac43$ or $\cos\theta=1.2$ — both $\sin$ and $\cos$ never exceed $1$.
  • Forgetting which ratios are undefined: $\tan90^\circ,\sec90^\circ$ and $\cot0^\circ,\csc0^\circ$.
  • In proofs, jumping between LHS and RHS at once — work one side until it matches the other (or simplify both separately).
  • Misremembering the table — anchor it on $\sin=\dfrac{\sqrt0}{2},\dfrac{\sqrt1}{2},\dfrac{\sqrt2}{2},\dfrac{\sqrt3}{2},\dfrac{\sqrt4}{2}$ and read $\cos$ backwards.

13. Quick revision checklist

  • $\sin=\dfrac{\text{opp}}{\text{hyp}},\ \cos=\dfrac{\text{adj}}{\text{hyp}},\ \tan=\dfrac{\text{opp}}{\text{adj}}$; reciprocals are $\csc,\sec,\cot$.
  • $\tan A=\dfrac{\sin A}{\cos A}$, $\cot A=\dfrac{\cos A}{\sin A}$.
  • Ratios depend only on the angle (similar triangles); $\sin A,\cos A\le1$.
  • Memorise Table 8.1 for $0^\circ,30^\circ,45^\circ,60^\circ,90^\circ$.
  • Three identities: $\sin^{2}\theta+\cos^{2}\theta=1$, $1+\tan^{2}\theta=\sec^{2}\theta$, $1+\cot^{2}\theta=\csc^{2}\theta$.
  • One known ratio → draw triangle + Pythagoras (or use identities) → get all six.
Practice MCQs
1. If $\sin A=\dfrac{12}{13}$, then $\cos A=$
  1. $\dfrac{5}{13}$
  2. $\dfrac{12}{5}$
  3. $\dfrac{13}{5}$
  4. $\dfrac{5}{12}$
Answer: (A) adjacent $=\sqrt{13^{2}-12^{2}}=5$, so $\cos A=\dfrac{5}{13}.$
2. The value of $\tan45^\circ+\cot45^\circ$ is:
  1. $1$
  2. $2$
  3. $\sqrt2$
  4. $0$
Answer: (B) $\tan45^\circ=1$ and $\cot45^\circ=1$, so sum $=2.$
3. $\sin^{2}30^\circ+\cos^{2}30^\circ=$
  1. $0$
  2. $\dfrac12$
  3. $1$
  4. $\sqrt3$
Answer: (C) by the identity $\sin^{2}\theta+\cos^{2}\theta=1.$
4. Which value is NOT defined?
  1. $\sin90^\circ$
  2. $\cos0^\circ$
  3. $\tan90^\circ$
  4. $\cot45^\circ$
Answer: (C) $\tan90^\circ$ is undefined (division by zero).
5. If $\cos\theta=\dfrac{\sqrt3}{2}$, then $\theta=$
  1. $30^\circ$
  2. $45^\circ$
  3. $60^\circ$
  4. $90^\circ$
Answer: (A) $\cos30^\circ=\dfrac{\sqrt3}{2}.$
6. $9\sec^{2}A-9\tan^{2}A=$
  1. $1$
  2. $9$
  3. $8$
  4. $0$
Answer: (B) $9(\sec^{2}A-\tan^{2}A)=9(1)=9.$
7. $(1+\tan^{2}\theta)\cos^{2}\theta=$
  1. $0$
  2. $1$
  3. $\sin^{2}\theta$
  4. $\tan^{2}\theta$
Answer: (B) $(1+\tan^{2}\theta)=\sec^{2}\theta=\dfrac{1}{\cos^{2}\theta}$, so the product $=1.$
8. The value of $\dfrac{2\tan30^\circ}{1+\tan^{2}30^\circ}$ is:
  1. $\sin60^\circ$
  2. $\cos60^\circ$
  3. $\tan60^\circ$
  4. $1$
Answer: (A) it simplifies to $\dfrac{\sqrt3}{2}=\sin60^\circ.$
9. If $\sec\theta=\dfrac{13}{12}$, then $\tan\theta=$
  1. $\dfrac{5}{12}$
  2. $\dfrac{12}{5}$
  3. $\dfrac{5}{13}$
  4. $\dfrac{13}{5}$
Answer: (A) $\tan^{2}\theta=\sec^{2}\theta-1=\dfrac{169-144}{144}=\dfrac{25}{144}$, so $\tan\theta=\dfrac{5}{12}.$
10. $\sin\theta=\cos\theta$ holds when $\theta=$
  1. $0^\circ$
  2. $30^\circ$
  3. $45^\circ$
  4. $60^\circ$
Answer: (C) at $45^\circ$, both equal $\dfrac{1}{\sqrt2}.$
11. $\sin60^\circ\cos30^\circ+\cos60^\circ\sin30^\circ=$
  1. $0$
  2. $\dfrac12$
  3. $1$
  4. $\dfrac{\sqrt3}{2}$
Answer: (C) $\dfrac{\sqrt3}{2}\cdot\dfrac{\sqrt3}{2}+\dfrac12\cdot\dfrac12=\dfrac34+\dfrac14=1.$
12. If $\tan A=\dfrac{3}{4}$, then $\sin A\cos A=$
  1. $\dfrac{12}{25}$
  2. $\dfrac{7}{25}$
  3. $\dfrac{3}{5}$
  4. $\dfrac{4}{5}$
Answer: (A) $\sin A=\dfrac35,\cos A=\dfrac45$, so product $=\dfrac{12}{25}.$
Assertion–Reason
A: $\sin^{2}60^\circ+\cos^{2}60^\circ=1$.   R: $\sin^{2}\theta+\cos^{2}\theta=1$ for every angle $\theta$.
Answer: Both A and R are true, and R is the correct explanation of A — A is just the identity applied at $\theta=60^\circ$.
A: $\sin\theta=\dfrac43$ is possible for some acute $\theta$.   R: For any acute angle, $\sin\theta\le1$.
Answer: A is false, R is true — since $\sin\theta\le1$ always, $\dfrac43>1$ is impossible. R correctly explains why A fails.
Previous-year questions
Q1. If $\tan A=\dfrac43$, find $\sin A$ and $\cos A$. (CBSE, 2 marks)
Answer: opposite $=4k$, adjacent $=3k$, hyp $=5k$. So $\sin A=\dfrac45$, $\cos A=\dfrac35.$
Q2. Prove that $\dfrac{\cos A}{1+\sin A}+\dfrac{1+\sin A}{\cos A}=2\sec A$. (CBSE, 3 marks)
Outline: combine over $(1+\sin A)\cos A$; numerator $=\cos^{2}A+1+2\sin A+\sin^{2}A=2(1+\sin A)$; cancel $(1+\sin A)$ to get $\dfrac{2}{\cos A}=2\sec A.$
Q3. Evaluate $\dfrac{5\cos^{2}60^\circ+4\sec^{2}30^\circ-\tan^{2}45^\circ}{\sin^{2}30^\circ+\cos^{2}30^\circ}$. (CBSE, 3 marks)
Answer: numerator $=\dfrac54+\dfrac{16}{3}-1=\dfrac{67}{12}$; denominator $=1$; value $=\dfrac{67}{12}.$
Q4. If $\sin(A-B)=\dfrac12$ and $\cos(A+B)=\dfrac12$, $0^\circ<A+B\le90^\circ$, $A>B$, find $A$ and $B$. (CBSE, 3 marks)
Answer: $A-B=30^\circ$ and $A+B=60^\circ$; adding gives $A=45^\circ$, $B=15^\circ.$
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More Class 10 Mathematics chapters
Real Numbers · Polynomials · Pair of Linear Equations in Two Variables · Quadratic Equations · Arithmetic Progressions · Triangles · Coordinate Geometry · Some Applications of Trigonometry