Introduction to Trigonometry

The word trigonometry is derived from the Greek words 'tri' (Meaning three), 'gon' (Meaning sides) and 'metron' (meaning measure). Infact trigonometry is the study of relationships between the sides and angles of a triangle. Let us take some examples from our surroundings where right triangles can be imagined to be formed.

Suppose the students of a school are visiting Statue of Unity. Now if a student is looking at the top of the tower, a right-angle triangle can be imagined to be made, as shown in fig.

Finding ratios of side lengths in similar right triangles. Work with a partner. You will need.

• Graph paper
• A protractor

Exploring the concept

• Draw a segment along a horizontal grid line on your graph paper. Label one end point $A$.
  
• Use a protractor to draw a $60^{\circ }$ angle at A.
• Draw segments along vertical grid lines to form right triangles, such as $△ABC,△ADE$ and $△AFG$ shown in figure.
  

Drawing conclusions

• Why are $△\mathrm{ABC},△\mathrm{ADE}$ and $△\mathrm{AFG}$ similar triangles?
• Cut a strip of graph paper to use as a ruler. Use your ruler to find the value of following ratio for $△ABC$ and for $△ADE$ : $\frac{\text{ Length of leg opposite }\angle \mathrm{A}}{\text{ Length of hypotenuse }}=\frac{\text{ Perpendicular }}{\text{ hypotenuse }}$

What do you notice? 3. Make a conjecture about the value of the ratio for any triangle similar to $△\mathrm{ABC}$ to test your conjecture, find the value of this ratio for $△AFG$ [i.e. perpendicular (p), Base(b) and hypotenuse(h)]. In the Exploration you found that corresponding ratios of side lengths in similar right triangles do not depend on the lengths of the sides. These ratios depend only on the shape of the triangles as determined by the measures of the acute angles. These constant ratios are so important that they are given names as sine, cosine, tangent, secant, cosecant, cotangent. Here side opposite to angle A is perpendicular; side adjacent to angle A is base & side opposite to right angle is hypotenuse.

1.0Trigonometric ratios

Sine and Cosine of an angle

In right $△ABC$, the sine of $\angle A$, which is written " $\sin A$ ", is given by

Numerical Ability 1 In fig., for $△ABC$, find $\sin A$ and $\cos A$.

Building Concepts 1 Between A.D. 1000 and 1300, the Anasazi people lived in cliff dwellings in the southwestern part of the United States. The doors to the cliff dwellings opened onto balconies that were reached by climbing ladders. Suppose the ladder shown rests on the ground and extends $3\mathbf{ft}$ above the balcony. How long is the ladder? (use $\sin 60^{\circ }=\sqrt{3}/2$ ) Explanation: Use the sine ratio to find the unknown side length. $\sin 60^{\circ }=\frac{\text{ opposite }}{\text{ hypotenuse }}=\frac{10}{\mathrm{x}}$ $x=\frac{10}{\sin 60^{\circ }}=\frac{10}{\sqrt{3}/2}=\frac{20}{\sqrt{3}}\times \frac{\sqrt{3}}{\sqrt{3}}=\frac{20\sqrt{3}}{3}=11.54\approx 12$

Building Concepts 2 Indicate the perpendicular, the hypotenuse and the base (in that order) with respect to the angle marked $x$.

2.0Tangent of an angle

In addition to the sine and cosine ratios, you can use the tangent ratio to find the measures of the sides and angles of a right triangle.

In right $△ABC$, the tangent of $\angle A$, which is written "tan $A$ ", is given by $\frac{\text{ length of leg opposite to }\angle A}{\text{ length of legadjacent to }\angle A}=\frac{a}{b}$

Building Concepts 3 Surveying surveyors use trigonometry to calculate distances that would otherwise be difficult to find, such as the distance between two houses located across a lake from each other, as shown in the diagram. What is this distance? (Use $\tan 45^{\circ }=1$ ) Explanation Let $\mathrm{x}=$ the distance in meters across the lake. Write an equation involving x and a trigonometric ratio. $\Rightarrow \frac{x}{48}=\tan 45^{\circ }$ $\Rightarrow \mathrm{x}=48\tan 45^{\circ }$ $\Rightarrow \mathrm{x}=48(1)=48$ The distance between the two houses is about 48 m .

Cosecant, secant and cotangent of an Angle

• The sine, cosine, and tangent ratios in a right triangle can be remembered by representing them as strings of letters, as in SOH-CAH-TOA. Sine $=$ Opposite $÷$ Hypotenuse Cosine $=$ Adjacent $÷$ Hypotenuse Tangent $=$ Opposite $÷$ Adjacent The memorization of this mnemonic can be aided by expanding it into a phrase, such as "Some Officers Have Curly Auburn Hair Till Old Age".

Numerical Ability 2 Using the information given in figure. write the values of all trigonometric ratios of angle $C$.

• Pythagoras theorem : In a right angle triangle, Hypotenuse ${}^2=$ Perpendicular ${}^2+$ Base ${}^2$

3.0Reciprocal Relations

Clearly, we have : (i) $\mathrm{cosec}\theta =\frac{1}{\sin \theta }$ (ii) $\sec \theta =\frac{1}{\cos \theta }$ (iii) $\cot \theta =\frac{1}{\tan \theta }$

Thus, we have : (i) $\sin \theta \cdot \mathrm{cosec}\theta =1$ (ii) $\cos \theta \cdot \sec \theta =1$ (iii) $\tan \theta \cdot \cot \theta =1$

4.0Quotient Relations

Consider a right angled triangle in which for an acute angle $\theta$, we have : $\sin \theta =\frac{\text{ Perpendicular }}{\text{ Hypotenuse }}=\frac{\mathrm{P}}{\mathrm{H}}$ $\cos \theta =\frac{\text{ Base }}{\text{ Hypotenuse }}=\frac{B}{\mathrm{H}}$ Now, $\frac{\sin \theta }{\cos \theta }=\frac{\frac{P}{\frac{B}{B}}}{\frac{H}{H}}=\frac{P}{H}\times \frac{H}{B}=\frac{P}{B}=\tan \theta$ (by def.) and $\frac{\cos \theta }{\sin \theta }=\frac{\frac{B}{H}}{\frac{P}{H}}=\frac{B}{H}\times \frac{H}{P}=\frac{B}{P}=\cot \theta$ (by def.) Thus, $\tan \theta =\frac{\sin \theta }{\cos \theta }$ and $\cot \theta =\frac{\cos \theta }{\sin \theta }$

Numerical Ability 3 In a right $△ABC$, if $\angle A$ is acute and $\tan A=\frac{3}{4}$, find the remaining trigonometric ratios of $\angle A$. Solution: Consider a $△\mathrm{ABC}$ in which $\angle \mathrm{B}=90^{\circ }$. For $\angle \mathrm{A}$, we have : Base $=\mathrm{AB}$, perpendicular $=\mathrm{BC}$ and Hypotenuse $=\mathrm{AC}$. $\therefore \tan \mathrm{A}=\frac{\text{ Perpendicular }}{\text{ Base }}=\frac{3}{4}$

5.0Power of T-ratios

We denote: (i) $(\sin \theta {)}^2$ by ${\sin }^2\theta$ (ii) $(\cos \theta {)}^2$ by ${\cos }^2\theta$ (iii) $(\sin \theta {)}^3$ by ${\sin }^3\theta$ (iv) $(\cos \theta {)}^3$ by ${\cos }^3\theta$ and so on.

• The symbol $\sin A$ is used as an abbreviation for 'the sine of the angle $A$ '. $\sin A$ is not the product of 'sin' and A. 'sin' separated from A has no meaning. Similarly, $\cos$ A is not the product of 'cos' and A. Similar interpretations follow for other trigonometric ratios also.
• We may write ${\sin }^{2}A,{\cos }^{2}A$, etc., in place of $(\sin A{)}^2$, $(\cos A{)}^2$, etc., respectively. But $\mathrm{cosec}A=(\sin A{)}^{-1}\ne {\sin }^{-1}A$ (it is called sine inverse $A$ ). ${\sin }^{-1}A$ has a different meaning, which will be discussed in higher classes. Similar conventions hold for the other trigonometric ratios as well.
• Since the hypotenuse is the longest side in a right triangle, the value of $\sin \mathrm{A}$ or $\cos \mathrm{A}$ is always less than 1 (or, in particular, equal to 1 ).

Numerical Ability 4 If $\sin A=\frac{1}{2}$, verify that $2\sin A\cos A=\frac{2\tan A}{1+{\tan }^{2}A}$ Solution:

6.0Trigonometric ratio of standard angles

T-ratios of 45^{\circ}

Consider a $△\mathrm{ABC}$ in which $\angle \mathrm{B}=90^{\circ }$ and $\angle \mathrm{A}=45^{\circ }$. Then, clearly, $\angle \mathrm{C}=45^{\circ }$. $$\begin{gathered} \therefore AB=BC=a(\text{ say }). \\ \therefore AC=\sqrt{A{B}^2+B{C}^2}=\sqrt{{\mathrm{a}}^2+{\mathrm{a}}^2}=\sqrt{2{\mathrm{a}}^2}=\sqrt{2\mathrm{a}}. \\ \therefore \sin 45^{\circ }=\frac{\mathrm{BC}}{\mathrm{AC}}=\frac{\mathrm{a}}{\sqrt{2}\mathrm{a}}=\frac{1}{\sqrt{2}};\cos 45^{\circ }=\frac{\mathrm{AB}}{\mathrm{AC}}=\frac{\mathrm{a}}{\sqrt{2}\mathrm{a}}=\frac{1}{\sqrt{2}} \\ \tan 45^{\circ }=\frac{\mathrm{BC}}{\mathrm{AB}}=\frac{\mathrm{a}}{\mathrm{a}}=1 \\ \mathrm{cosec}45^{\circ }=\frac{1}{\sin 45^{\circ }}=\sqrt{2};\sec 45=\frac{1}{\cos 45^{\circ }}=\sqrt{2} \\ \cot 45^{\circ }=\frac{\mathrm{AB}}{\mathrm{BC}}=\frac{1}{\tan 45^{\circ }}=1 \end{gathered}$$

T-ratios of \mathbf{6 0} and \mathbf{3 0 ^ { \circ }}

Draw an equilateral $△\mathrm{ABC}$ with each side $=2\mathrm{a}$. Then, $\angle \mathrm{A}=\angle \mathrm{B}=\angle \mathrm{C}=60^{\circ }$. From A, draw $AD\perp BC$. Then, $\mathrm{BD}=\mathrm{DC}=\mathrm{a},\angle \mathrm{BAD}=30^{\circ }$ and $\angle \mathrm{ADB}=90^{\circ }$.

T-ratios of 60^{\circ}

In $△\mathrm{ADB}$ we have: $\angle \mathrm{ADB}=90^{\circ }$ and $\angle \mathrm{ABD}=60^{\circ }$. Base $=\mathrm{BD}=\mathrm{a}$, Perpendicular $=\mathrm{AD}=\sqrt{3}\mathrm{a}$ and Hypotenuse $\mathrm{AB}=2\mathrm{a}$.

$$\begin{gathered} \therefore \sin 60^{\circ }=\frac{\mathrm{AD}}{\mathrm{AB}}=\frac{\sqrt{3a}}{2\mathrm{a}}=\frac{\sqrt{3}}{2};\cos 60^{\circ }=\frac{\mathrm{BD}}{\mathrm{AB}}=\frac{\mathrm{a}}{2\mathrm{a}}=\frac{1}{2} \\ \tan 60^{\circ }=\frac{\mathrm{AD}}{\mathrm{BD}}=\frac{\sqrt{3}\mathrm{a}}{\mathrm{a}}=\sqrt{3} \\ \therefore \mathrm{cosec}60^{\circ }=\frac{1}{\sin 60^{\circ }}=\frac{2}{\sqrt{3}};\sec 60^{\circ }=\frac{1}{\cos 60^{\circ }}=2 \\ \cot 60^{\circ }=\frac{1}{\tan 60^{\circ }}=\frac{1}{\sqrt{3}} \end{gathered}$$

T-ratios of \mathbf{3 0}{ }^{\circ}

In $△\mathrm{ADB}$ we have: $\angle \mathrm{ADB}=90^{\circ }$ and $\angle \mathrm{BAD}=30^{\circ }$. $\therefore$ Base $=AD=\sqrt{3}a$, Perpendicular $=BD=a$ and Hypotenuse $=AB=2a$. $\therefore \sin 30^{\circ }=\frac{\mathrm{BD}}{\mathrm{AB}}=\frac{\mathrm{a}}{2\mathrm{a}}=\frac{1}{2};\cos 30^{\circ }=\frac{\mathrm{AD}}{\mathrm{AB}}=\frac{\sqrt{3}\mathrm{a}}{2\mathrm{a}}=\frac{\sqrt{3}}{2}$ $\tan 30^{\circ }=\frac{\mathrm{BD}}{\mathrm{AD}}=\frac{\mathrm{a}}{\sqrt{3}\mathrm{a}}=\frac{1}{\sqrt{3}}$ $\mathrm{cosec}30^{\circ }=\frac{1}{\sin 30^{\circ }}=2;\sec 30^{\circ }=\frac{1}{\cos 30^{\circ }}=\frac{2}{\sqrt{3}}$; $\cot 30^{\circ }=\frac{1}{\tan 30^{\circ }}=\sqrt{3}$ Table for T-ratios of standard angles

Quick Tips (i) As $\theta$ increases from $0^{\circ }$ to $90^{\circ },\sin \theta$ increases from 0 to 1 . (ii) As $\theta$ increases from $0^{\circ }$ to $90^{\circ },\cos \theta$ decreases from 1 to 0 . (iii) As $\theta$ increases from $0^{\circ }$ to $90^{\circ },\tan \theta$ increases from 0 to $\infty$. (iv) The maximum value of $\frac{1}{\sec \theta },0^{\circ }\le \theta \le 90^{\circ }$ is one. (v) As $\cos \theta$ decreases from 1 to $0,\theta$ increases from $0^{\circ }$ to $90^{\circ }$. (vi) $\sin \theta$ and $\cos \theta$ can not be greater than one numerically. (vii) $\sec \theta$ and $\mathrm{cosec}\theta$ can not be less than one numerically. (viii) $\tan \theta$ and $\cot \theta$ can have any value.

Numerical Ability 5 In $△ABC$, right angled at $B,BC=5\mathrm{cm},\angle BAC=30^{\circ }$, find the length of the sides $AB$ and AC. Solution: We are given $\angle \mathrm{BAC}=30^{\circ }$, i.e., $\angle \mathrm{A}=30^{\circ }$ and $\mathrm{BC}=5\mathrm{cm}$ Now, $\sin \mathrm{A}=\frac{\mathrm{BC}}{\mathrm{AC}}$ or $\sin 30^{\circ }=\frac{5}{\mathrm{AC}}$ or $\frac{5}{\mathrm{AC}}=\frac{1}{2}$ $\left[\sin 30^{\circ }=\frac{1}{2}\right]$ or $\mathrm{AC}=2\times 5$ or 10 cm

Numerical Ability 6 In $△ABC$, right angled at $C$, if $AC=4\mathrm{cm}$ and $AB=8\mathrm{cm}$. Find $\angle A$ and $\angle B$. Solution

Numerical Ability 7 Find the value of $\theta$ in each of the following: (i) $2\sin 2\theta =\sqrt{3}$ (ii) $2\cos 3\theta =1$ (iii) $\sqrt{3}\tan 2\theta -3=0$ Solution: (i) we have, $2\sin 2\theta =\sqrt{3}\Rightarrow \sin 2\theta =\frac{\sqrt{3}}{2}$ $\Rightarrow \sin 2\theta =\sin 60^{\circ }$ $\Rightarrow 2\theta =60^{\circ }$ $\Rightarrow \theta =30^{\circ }$ (ii) we have, $2\cos 3\theta =1$ $\Rightarrow \cos 3\theta =\frac{1}{2}$ $\Rightarrow \cos 3\theta =\cos 60^{\circ }$ $\Rightarrow 3\theta =60^{\circ }$ $\Rightarrow \theta =20^{\circ }$ (iii) we have, $$\begin{gathered} \sqrt{3}\tan 2\theta -3=0 \\ \Rightarrow \sqrt{3}\tan 2\theta =3 \\ \Rightarrow \tan 2\theta =\frac{3}{\sqrt{3}}=\sqrt{3} \\ \Rightarrow \tan 2\theta =\tan 60^{\circ } \\ \Rightarrow 2\theta =60^{\circ } \\ \Rightarrow \theta =30^{\circ } \end{gathered}$$

7.0T-ratios of complementary angles

Complementary angles

Two angles are said to be complementary, if their sum is $90^{\circ }$. Thus, ${\theta }^{\circ }$ and $\left(90^{\circ }-\theta \right)$ are complementary angles.

T-ratios of complementary angles Consider $△\mathrm{ABC}$ in which $\angle \mathrm{B}=90^{\circ }$ and $\angle \mathrm{A}={\theta }^{\circ }$.

$\therefore \angle \mathrm{C}=\left(90^{\circ }-\theta \right)$ Let $\mathrm{AB}=\mathrm{x}.\mathrm{BC}=\mathrm{y}$ and $\mathrm{AC}=\mathrm{r}$. Then, $\sin \theta =\frac{\mathrm{y}}{\mathrm{r}},\cos \theta =\frac{\mathrm{x}}{\mathrm{r}}$ and $\tan \theta =\frac{\mathrm{y}}{\mathrm{x}}$. When we consider the T-ratios of $\left(90^{\circ }-\theta \right)$, then Base $=BC$, Perpendicular $=AB$ and Hypotenuse $=AC$.

$\therefore \quad \sin \left(90^{\circ}-\theta\right)=\frac{A B}{A C}=\frac{x}{r}=\cos \theta \ \cos \left(90^{\circ}-\theta\right)=\frac{\mathrm{BC}}{\mathrm{AC}}=\frac{\mathrm{y}}{\mathrm{r}}=\sin \theta . \ \tan \left(90^{\circ}-\theta\right)=\frac{\mathrm{AB}}{\mathrm{BC}}=\frac{x}{y}=\cot \theta \

\therefore \quad \operatorname{cosec}\left(90^{\circ}-\theta\right)=\frac{1}{\sin \left(90^{\circ}-\theta\right)}=\frac{1}{\cos \theta}=\sec \theta \ \sec \left(90^{\circ}-\theta\right)=\frac{1}{\cos \left(90^{\circ}-\theta\right)}=\frac{1}{\sin \theta}=\operatorname{cosec} \theta \ \cot \left(90^{\circ}-\theta\right)=\frac{1}{\tan \left(90^{\circ}-\theta\right)}=\frac{1}{\cot \theta}=\tan \theta$

(i) $\sin \left(90^{\circ }-\theta \right)=\cos \theta$ (ii) $\cos \left(90^{\circ }-\theta \right)=\sin \theta$ (iii) $\tan \left(90^{\circ }-\theta \right)=\cot \theta$ (iv) $\mathrm{cosec}\left(90^{\circ }-\theta \right)=\sec \theta$ (v) $\sec \left(90^{\circ }-\theta \right)=\mathrm{cosec}\theta$ (vi) $\cot \left(90^{\circ }-\theta \right)=\tan \theta$

In other words: $\sin ($ angle $)=\cos ($ complement $);\cos$ (angle) $=\sin ($ complement $)$ $\tan$ (angle) $=\cot$ (complement); $\cot$ (angle) $=\tan$ (complement) $\sec ($ angle $)=\mathrm{cosec}($ complement $);\mathrm{cosec}$ (angle) $=\sec$ (complement) $(\because$ where complement $=90^{\circ }-$ angle $)$

Numerical Ability 8 Without using tables, evaluate: (i) $\frac{\sin 53^{\circ }}{\cos 37^{\circ }}$ (ii) $\frac{\cos 49^{\circ }}{\sin 41^{\circ }}$ (iii) $\frac{\tan 66^{\circ }}{\cot 24^{\circ }}$ Solution (i) $\frac{\sin 53^{\circ }}{\cos 37^{\circ }}=\frac{\sin \left(90^{\circ }-37^{\circ }\right)}{\cos 37^{\circ }}=\frac{\cos 37^{\circ }}{\cos 37^{\circ }}=1\left[\therefore \sin \left(90^{\circ }-\theta \right)=\cos \theta \right]$ (ii) $\frac{\cos 49^{\circ }}{\sin 41^{\circ }}=\frac{\cos \left(90^{\circ }-41^{\circ }\right)}{\sin 41^{\circ }}=\frac{\sin 41^{\circ }}{\sin 41^{\circ }}=1\left[\therefore \cos \left(90^{\circ }-\theta \right)=\sin \theta \right]$ (iii) $\frac{\tan 66^{\circ }}{\cot 24^{\circ }}=\frac{\tan \left(90^{\circ }-24^{\circ }\right)}{\cot 24^{\circ }}=\frac{\cot 24^{\circ }}{\cot 24^{\circ }}=1\left[\therefore \tan \left(90^{\circ }-\theta \right)=\cot \theta \right]$

• The above example suggests that out of the two t-ratios, we convert one in term of the t-ratio of the complement.
• For uniformity, we usually convert the angle greater than $45^{\circ }$ in terms of its complement.

Numerical Ability 9 Without using tables, show that $\left(\cos 35^{\circ }\cos 55^{\circ }-\sin 35^{\circ }\sin 55^{\circ }\right)=0$. Solution: $$\begin{gathered} \text{ LHS }=\left(\cos 35^{\circ }\cos 55^{\circ }-\sin 35^{\circ }\sin 55^{\circ }\right) \\ =[\left(\cos 35^{\circ }\cos 55^{\circ }-\sin \left(90^{\circ }-55^{\circ }\right)\sin \left(90^{\circ }-35^{\circ }\right)\right] \\ =\left(\cos 35^{\circ }\cos 55^{\circ }-\cos 55^{\circ }\cos 35^{\circ }\right)=0=\mathrm{RHS}. \\ \left[\therefore \sin \left(90^{\circ }-\theta \right)=\cos \theta \text{ and }\cos \left(90^{\circ }-\theta \right)=\sin \theta \right] \end{gathered}$$

Numerical Ability 10 Express $\left(\sin 85^{\circ }+\mathrm{cosec}85^{\circ }\right)$ in terms of trigonometric ratios of angles between $0^{\circ }$ and $45^{\circ }$. Solution: $\left(\sin 85^{\circ }+\mathrm{cosec}85^{\circ }\right)=\sin \left(90^{\circ }-5^{\circ }\right)+\mathrm{cosec}\left(90^{\circ }-5^{\circ }\right)$ $=\left(\cos 5^{\circ }+\sec 5^{\circ }\right)$.

Numerical Ability 11 Evaluate : $\frac{\sec 29^{\circ }}{\mathrm{cosec}61^{\circ }}+2\cot 8^{\circ }\cot 17^{\circ }\cot 45^{\circ }\cot 73^{\circ }\cot 82^{\circ }$ Solution: $\frac{\sec 29^{\circ }}{\mathrm{cosec}61^{\circ }}+2\cot 8^{\circ }\cot 17^{\circ }\cot 45^{\circ }\cot 73^{\circ }\cot 82^{\circ }$. $=\frac{\sec 29^{\circ }}{\mathrm{cosec}\left(90^{\circ }-29^{\circ }\right)}+2\cot 8^{\circ }\cot 17^{\circ }(1)\cot \left(90^{\circ }-17^{\circ }\right)\cot \left(90^{\circ }-8^{\circ }\right)$ $=\frac{\sec 29^{\circ }}{\sec 29^{\circ }}+2\cot 8^{\circ }\cot 17^{\circ }\tan 17^{\circ }\tan 8^{\circ }.\left[\cot \left(90^{\circ }-\theta \right)=\tan \theta \right]$ $=1+2\cot 8^{\circ }\cot 17^{\circ }\frac{1}{\cot 17^{\circ }}\frac{1}{\cot 8^{\circ }}\left(\because \tan \theta =\frac{1}{\cot \theta }\right)$ $=1+2=3$.

8.0Trigonometric identities

We know that an equation is called an identity when it is true for all values of the variables involved. Similarly, "an equation involving trigonometric ratios of an angle is called a trigonometric identity, if it is true for all values of the angle(s) involved."

The three Fundamental Trigonometric Identities are - (i) ${\cos }^2\mathrm{A}+{\sin }^2\mathrm{A}=1;0^{\circ }\le \mathrm{A}\le 90^{\circ }$ (ii) $1+{\tan }^2\mathrm{A}={\sec }^2\mathrm{A};0^{\circ }\le \mathrm{A}<90^{\circ }$ (iii) $1+{\cot }^2\mathrm{A}={\mathrm{cosec}}^2\mathrm{A};0^{\circ }<\mathrm{A}\le 90^{\circ }$

Geometrical proof

Consider a $△ABC$, right angled at $B$. Then we have : ${\mathrm{AB}}^2+{\mathrm{BC}}^2={\mathrm{AC}}^2$ By Pythagoras theorem (i) ${\cos }^2\mathrm{A}+{\sin }^2\mathrm{A}=1;0^{\circ }\le \mathrm{A}\le 90^{\circ }$ Dividing each term of (i) by ${\mathrm{AC}}^2$, we get $\frac{{\mathrm{AB}}^2}{{\mathrm{AC}}^2}+\frac{{\mathrm{BC}}^2}{{\mathrm{AC}}^2}=\frac{{\mathrm{AC}}^2}{{\mathrm{AC}}^2}\text{ i.e. }{\left(\frac{\mathrm{AB}}{\mathrm{AC}}\right)}^2+{\left(\frac{\mathrm{BC}}{\mathrm{AC}}\right)}^2={\left(\frac{\mathrm{AC}}{\mathrm{AC}}\right)}^2$ i.e., $(\cos A{)}^2+(\sin A{)}^2=1$ i.e., ${\cos }^2\mathrm{A}+{\sin }^2\mathrm{A}=1$ This is true for all A such that $0^{\circ }\le \mathrm{A}\le 90^{\circ }$ So, this is a trigonometric identity. (ii) $1+{\tan }^2\mathrm{A}={\sec }^2\mathrm{A};0^{\circ }\le \mathrm{A}<90^{\circ }$ Let us now divide (i) by ${\mathrm{AB}}^2$. We get $\frac{{\mathrm{AB}}^2}{{\mathrm{AB}}^2}+\frac{{\mathrm{BC}}^2}{{\mathrm{AB}}^2}=\frac{{\mathrm{AC}}^2}{{\mathrm{AB}}^2}\text{ or, }{\left(\frac{\mathrm{AB}}{\mathrm{AB}}\right)}^2+{\left(\frac{\mathrm{BC}}{\mathrm{AB}}\right)}^2={\left(\frac{\mathrm{AC}}{\mathrm{AB}}\right)}^2$ i.e., $1+{\tan }^2\mathrm{A}={\sec }^2\mathrm{A}$ This equation is true for $A=0^{\circ }$. Since $\tan A$ and sec $A$ are not defined for $A=90^{\circ }$, so (iii) is true for all A such that $0^{\circ }\le \mathrm{A}<90^{\circ }$ (iii) $1+{\cot }^2\mathrm{A}={\mathrm{cosec}}^2\mathrm{A};0^{\circ }<\mathrm{A}\le 90^{\circ }$ Again, let us divide (i) by ${\mathrm{BC}}^2$, we get $$\begin{gathered} \frac{{\mathrm{AB}}^2}{{\mathrm{BC}}^2}+\frac{{\mathrm{BC}}^2}{{\mathrm{BC}}^2}=\frac{{\mathrm{AC}}^2}{{\mathrm{BC}}^2} \\ \Rightarrow {\left(\frac{\mathrm{AB}}{\mathrm{BC}}\right)}^2+{\left(\frac{\mathrm{BC}}{\mathrm{BC}}\right)}^2={\left(\frac{\mathrm{AC}}{\mathrm{BC}}\right)}^2 \\ \Rightarrow 1+{\cot }^2\mathrm{A}={\mathrm{cosec}}^2\mathrm{A} \end{gathered}$$ Since $\mathrm{cosec}\mathrm{A}$ and $\cot \mathrm{A}$ are not defined for $\mathrm{A}=0^{\circ }$, therefore (iv) is true for all A such that $0^{\circ }<\mathrm{A}\le 90^{\circ }$

Using the above trigonometric identities, we can express each trigonometric ratio in terms of the other trigonometric ratios, i.e., if any one of the ratios is known, we can also determine the values of other trigonometric ratios.

Fundamental identities (results)

To prove trigonometrical identities

The following methods are to be followed: Method-I: Take the more complicated side of the identity (L.H.S. or R.H.S. as the case may be) and by using suitable trigonometric and algebraic formulae prove it equal to the other side. Method-II: When neither side of the identity is in a simple form, simplify the L.H.S. and R.H.S. separately by using suitable formulae (by expressing all the T-ratios occurring in the identity in terms of the sine and cosine and show that the results are equal).

Numerical Ability 12 Prove $\sqrt{{\sec }^2\theta +{\mathrm{cosec}}^2\theta }=\tan \theta +\cot \theta$ Solution: $$\begin{gathered} \text{ LHS }=\sqrt{{\sec }^2\theta +{\mathrm{cosec}}^2\theta }=\sqrt{\left(1+{\tan }^2\theta \right)+\left(1+{\cot }^2\theta \right)} \\ =\sqrt{{\tan }^2\theta +{\cot }^2\theta +2\tan \theta \cot \theta }(\tan \theta \cdot \cot \theta =1) \\ =\sqrt{(\tan \theta +\cot \theta {)}^2} \\ =\tan \theta +\cot \theta =\text{ RHS } \\ \text{ Hence proved. } \end{gathered}$$

Numerical Ability 13 Prove that : $\frac{{\sec }^2\theta {\sin }^2\theta -{\mathrm{cosec}}^2\theta +{\mathrm{cosec}}^2\theta {\cos }^2\theta }{{\sec }^2\theta {\sin }^2\theta -{\mathrm{cosec}}^2\theta {\cos }^2\theta }={\sin }^2\theta$ Solution: $$\begin{gathered} \text{ LHS }=\frac{{\sec }^2\theta {\sin }^2\theta -{\mathrm{cosec}}^2\theta +{\mathrm{cosec}}^2\theta {\cos }^2\theta }{{\sec }^2\theta {\sin }^2\theta -{\mathrm{cosec}}^2\theta {\cos }^2\theta } \\ =\frac{\frac{{\sin }^2\theta }{{\cos }^2\theta }-\frac{1}{{\sin }^2\theta }+\frac{{\cos }^2\theta }{{\sin }^2\theta }}{\frac{{\sin }^2\theta }{{\cos }^2\theta }-\frac{{\cos }^2\theta }{{\sin }^2\theta }} \end{gathered}$$ $$\begin{gathered} =\frac{\frac{{\sin }^2\theta }{{\cos }^2\theta }-\left(\frac{1-{\cos }^2\theta }{{\sin }^2\theta }\right)}{\frac{{\sin }^4\theta -{\cos }^4\theta }{{\sin }^2\theta {\cos }^2\theta }} \\ =\frac{\frac{{\sin }^2\theta }{{\cos }^2\theta }-1}{\frac{\left({\sin }^2\theta -{\cos }^2\theta \right)\left({\sin }^2\theta +{\cos }^2\theta \right)}{{\sin }^2\theta {\cos }^2\theta }} \\ =\frac{{\sin }^2\theta -{\cos }^2\theta }{{\cos }^2\theta }\times \frac{{\sin }^2\theta {\cos }^2\theta }{{\sin }^2\theta -{\cos }^2\theta } \\ =\sin 2=\text{ RHS } \\ \text{ Hence Proved. } \end{gathered}$$

Numerical Ability 14 Prove that : $2\left({\sin }^6\theta +{\cos }^6\theta \right)-3\left({\sin }^4\theta +{\cos }^4\theta \right)+1=0$ Solution: $$\begin{gathered} \text{ LHS }=2\left({\sin }^6\theta +{\cos }^6\theta \right)-3\left({\sin }^4\theta +{\cos }^4\theta \right)+1 \\ =2\left[{\left({\sin }^2\theta \right)}^3+{\left({\cos }^2\theta \right)}^3\right]-3\left[{\left({\sin }^2\theta \right)}^2+{\left({\cos }^2\theta \right)}^2\right]+1 \\ =2\left[{\left({\sin }^2\theta +{\cos }^2\theta \right)}^3\right]-3{\sin }^2\theta {\cos }^2\theta \left({\sin }^2\theta +{\cos }^2\theta \right)]-3[{\left({\sin }^2\theta +{\cos }^2\theta \right)}^2-2 \\ {\sin }^2\theta {\cos }^2\theta ]+1 \\ =2\left\{1-3{\sin }^2\theta {\cos }^2\theta \right\}-3\left\{1-2{\sin }^2\theta {\cos }^2\theta \right\}+1 \\ =2-6{\sin }^2\theta {\cos }^2\theta -3+6{\sin }^2\theta {\cos }^2\theta +1=3-3=0=\text{ RHS. } \end{gathered}$$ Hence Proved.

Numerical Ability 15 Prove the following by using suitable identities: $\frac{1-\sin x}{1+\sec x}-\frac{1+\sin x}{1-\sec x}=2\cos x\left(\cot x+{\mathrm{cosec}}^{2}x\right)$ Solution: $$\begin{gathered} \text{ LHS }=\frac{1-\sin x}{1+\sec x}-\frac{1+\sin x}{1-\sec x}=\frac{1-\sin x}{1+\frac{1}{\cos x}}-\frac{1+\sin x}{1-\frac{1}{\cos x}} \\ =\frac{\cos x(1-\sin x)}{\cos x+1}-\frac{\cos x(1+\sin x)}{\cos x-1}=\cos x\left[\frac{(1-\sin x)}{\cos x+1}-\frac{1+\sin x}{\cos x-1}\right] \end{gathered}$$

$$\begin{gathered} =\cos x\left\{\frac{(1-\sin x)(\cos x-1)-(\cos x+1)(1+\sin x)}{{\cos }^{2}x-1}\right\} \\ =\cos x\left\{\frac{(1-\sin x)(1-\cos x)+(1+\cos x)(1+\sin x)}{1-{\cos }^{2}x}\right\} \\ =\cos x\left\{\frac{1-\cos x-\sin x+\sin x\cos x+1+\sin x+\cos x+\sin x\cos x}{{\sin }^{2}x}\right\} \\ =\cos x\left\{\frac{2+2\sin x\mathrm{cox}}{{\sin }^{2}x}\right\}=2\cos x\left(\frac{1+\sin x\cos x}{{\sin }^{2}x}\right) \\ \text{ LHS }=2\cos x\left\{\frac{1}{{\sin }^{2}x}+\frac{\cos x}{\sin x}\right\}=2\cos x\left({\mathrm{cosec}}^{2}x+\cot x\right)=\text{ RHS. } \end{gathered}$$ Hence Proved.

• In proving question, generally we convert all the trigonometric ratios in terms of $\sin \theta$ and $\cos \theta$ to make it easy and convenient.

Numerical Ability 16 Prove that $:\frac{\tan \theta +\sec \theta -1}{\tan \theta -\sec \theta +1}=\sec \theta +\tan \theta =\frac{1+\sin \theta }{\cos \theta }=\frac{\cos \theta }{1-\sin \theta }=\sqrt{\frac{1+\sin \theta }{1-\sin \theta }}$ Solution: Consider the numerator of the LHS of the given expression. i.e. $\tan \theta +\sec \theta -1$ [Use $1={\sec }^2\theta -{\tan }^2\theta$ ] $=(\tan \theta +\sec \theta )-\left({\sec }^2\theta -{\tan }^2\theta \right)$ [Use ${\mathrm{a}}^2-{\mathrm{b}}^2=(\mathrm{a}+\mathrm{b})(\mathrm{a}-\mathrm{b})$ ] $=(\tan \theta +\sec \theta )-(\sec \theta +\tan \theta )(\sec \theta -\tan \theta )[$ Take $(\sec \theta +\tan \theta )$ as common] $=(\sec \theta +\tan \theta )(1-\sec \theta +\tan \theta )$ [Note this step very carefully] $=(\sec \theta +\tan \theta )(\tan \theta -\sec \theta +1)$ ... (1) [Rearranging] LHS $=\frac{\tan \theta +\sec \theta -1}{\tan \theta -\sec \theta +1}$ $=\frac{(\sec \theta +\tan \theta )(\tan \theta -\sec \theta +1)}{(\tan \theta -\sec \theta +1)}$ $$\begin{gathered} =\sec \theta +\tan \theta =\text{ RHS }\text{ [First form] } \\ =\frac{1}{\cos \theta }+\frac{\sin \theta }{\cos \theta }=\frac{1+\sin \theta }{\cos \theta }=\text{ RHS }\text{... (2) } \\ =\frac{(1+\sin \theta )(1-\sin \theta )}{\cos \theta (1-\sin \theta )}=\frac{1-{\sin }^2\theta }{\cos \theta (1-\sin \theta )}=\frac{{\cos }^2\theta }{\cos \theta (1-\sin \theta )} \\ =\frac{\cos \theta }{(1-\sin \theta )}\text{ [Third form] } \\ =\frac{\sqrt{1+\sin \theta }\cdot \sqrt{1-\sin \theta }}{\sqrt{1-\sin \theta }\sqrt{1-\sin \theta }}\text{ [use }{\cos }^2\theta =1-{\sin }^2\theta ] \\ =\frac{\sqrt{1+\sin \theta }}{\sqrt{1-\sin \theta }}=\text{ RHS }\text{ [Fourth form] } \end{gathered}$$