NCERT Solutions For Class 10 Maths: Introduction to Trigonometry

EXERCISE 8.1

Question 1: In $\Delta$ ABC, right-angled at B, AB = 24 cm, BC = 7 cm. Determine :
(i) sin A, cos A
(ii) sin C, cos C

Solution-
1.First, find the hypotenuse AC using the Pythagoras theorem.
$AC = \sqrt{AB^2 + BC^2} = \sqrt{24^2 + 7^2} = \sqrt{576 + 49} = \sqrt{625} = 25$ cm.

(i) Ratios for $\angle A$:
1.sin A = $\frac{\text{side opposite to angle A}}{\text{hypotenuse}} = \frac{BC}{AC} = \frac{7}{25}$.

2.cos A = $\frac{\text{side adjacent to angle A}}{\text{hypotenuse}} = \frac{AB}{AC} = \frac{24}{25}$.

(ii) Ratios for $\angle C$:
1.sin C = $\frac{\text{side opposite to angle C}}{\text{hypotenuse}} = \frac{AB}{AC} = \frac{24}{25}$.

2.cos C = $\frac{\text{side adjacent to angle C}}{\text{hypotenuse}} = \frac{BC}{AC} = \frac{7}{25}$.

Question 2: In Fig. 8.13, find tan P – cot R.

Solution-
1.In $\Delta PQR$, right-angled at Q, PQ = 12 cm and PR = 13 cm.
2.Find the side QR using the Pythagoras theorem:
$QR = \sqrt{PR^2 - PQ^2} = \sqrt{13^2 - 12^2} = \sqrt{169 - 144} = \sqrt{25} = 5$ cm.

3.Calculate tan P:
$\tan P = \frac{\text{side opposite to angle P}}{\text{side adjacent to angle P}} = \frac{QR}{PQ} = \frac{5}{12}$.

4.Calculate cot R:
$\cot R = \frac{\text{side adjacent to angle R}}{\text{side opposite to angle R}} = \frac{QR}{PQ} = \frac{5}{12}$.

5.Determine tan P – cot R:
$\tan P - \cot R = \frac{5}{12} - \frac{5}{12} = 0$.

Question 3: If $\sin A = \frac{3}{4}$, calculate $\cos A$ and $\tan A$.

Solution-
1.Given $\sin A = \frac{3}{4} = \frac{\text{Opposite Side}}{\text{Hypotenuse}}$. Let BC = 3k and AC = 4k.
2.Find the adjacent side AB using the Pythagoras theorem:
$AB = \sqrt{AC^2 - BC^2} = \sqrt{(4k)^2 - (3k)^2} = \sqrt{16k^2 - 9k^2} = \sqrt{7k^2} = k\sqrt{7}$.

3.Calculate $\cos A$:
$\cos A = \frac{\text{Adjacent Side}}{\text{Hypotenuse}} = \frac{AB}{AC} = \frac{k\sqrt{7}}{4k} = \frac{\sqrt{7}}{4}$.

4.Calculate $\tan A$:
$\tan A = \frac{\text{Opposite Side}}{\text{Adjacent Side}} = \frac{BC}{AB} = \frac{3k}{k\sqrt{7}} = \frac{3}{\sqrt{7}}$. (Alternatively, $\tan A = \frac{\sin A}{\cos A}$).

Question 4: Given 15 cot A = 8, find $\sin A$ and $\sec A$.

Solution-
1.Given $\cot A = \frac{8}{15} = \frac{\text{Adjacent Side (AB)}}{\text{Opposite Side (BC)}}$. Let AB = 8k and BC = 15k.
2.Find the hypotenuse AC using the Pythagoras theorem:
$AC = \sqrt{AB^2 + BC^2} = \sqrt{(8k)^2 + (15k)^2} = \sqrt{64k^2 + 225k^2} = \sqrt{289k^2} = 17k$.

3.Calculate $\sin A$:
$\sin A = \frac{\text{Opposite Side}}{\text{Hypotenuse}} = \frac{BC}{AC} = \frac{15k}{17k} = \frac{15}{17}$.

4.Calculate $\sec A$:
$\sec A = \frac{\text{Hypotenuse}}{\text{Adjacent Side}} = \frac{AC}{AB} = \frac{17k}{8k} = \frac{17}{8}$.

Question 5: Given $\sec \theta = \frac{13}{12}$, calculate all other trigonometric ratios.

Solution-
1.Given $\sec \theta = \frac{13}{12} = \frac{\text{Hypotenuse}}{\text{Adjacent Side}}$. Let Hypotenuse = 13k and Adjacent Side = 12k.
2.Find the Opposite Side using the Pythagoras theorem:
Opposite Side $= \sqrt{(13k)^2 - (12k)^2} = \sqrt{169k^2 - 144k^2} = \sqrt{25k^2} = 5k$.

3.Calculate the other ratios:
1.$\cos \theta = \frac{1}{\sec \theta} = \frac{12}{13}$.

2.$\sin \theta = \frac{\text{Opposite Side}}{\text{Hypotenuse}} = \frac{5k}{13k} = \frac{5}{13}$.

3.$\tan \theta = \frac{\text{Opposite Side}}{\text{Adjacent Side}} = \frac{5k}{12k} = \frac{5}{12}$.

4.$\operatorname{cosec} \theta = \frac{1}{\sin \theta} = \frac{13}{5}$.

5.$\cot \theta = \frac{1}{\tan \theta} = \frac{12}{5}$.

Question 6: If $\angle A$ and $\angle B$ are acute angles such that $\cos A = \cos B$, then show that $\angle A = \angle B$.

Solution-
1.Consider two right triangles, $\triangle ACB$ (right-angled at C) and $\triangle PRQ$ (right-angled at R), or consider a single triangle $\triangle ABC$ right-angled at C (since A and B are acute angles).
2.If $\cos A = \cos B$ in $\triangle ABC$ (right-angled at C), then $\frac{AC}{AB} = \frac{BC}{AB}$.
3.This implies $AC = BC$.
4.In a triangle, the sides opposite to equal angles are equal. Since side AC is opposite to $\angle B$ and side BC is opposite to $\angle A$, having $AC=BC$ implies $\angle A = \angle B$.
(Alternatively, using the similar triangle approach demonstrated in Example 2, if $\cos A = \cos B$, we can show that $\triangle ACB \sim \triangle BDA$ or use similar construction to prove the angles are equal).

Question 7: If $\cot \theta = \frac{7}{8}$, evaluate : (i) $\frac{(1 + \sin \theta) (1 - \sin \theta)}{(1 + \cos \theta) (1 - \cos \theta)}$ (ii) $\cot^2 \theta$

Solution-
(i) Evaluate $\frac{(1 + \sin \theta) (1 - \sin \theta)}{(1 + \cos \theta) (1 - \cos \theta)}$:
1.Using the algebraic identity $(a+b)(a-b) = a^2 - b^2$, the expression simplifies to $\frac{1 - \sin^2 \theta}{1 - \cos^2 \theta}$.

2.Using the trigonometric identity $\sin^2 \theta + \cos^2 \theta = 1$, we know $1 - \sin^2 \theta = \cos^2 \theta$ and $1 - \cos^2 \theta = \sin^2 \theta$.

3.The expression becomes $\frac{\cos^2 \theta}{\sin^2 \theta}$. Since $\frac{\cos \theta}{\sin \theta} = \cot \theta$, this is equal to $\cot^2 \theta$.

4.Substitute the given value: $\cot^2 \theta = \left(\frac{7}{8}\right)^2 = \frac{49}{64}$.

(ii) Evaluate $\cot^2 \theta$:
1.Given $\cot \theta = \frac{7}{8}$.

2.$\cot^2 \theta = \left(\frac{7}{8}\right)^2 = \frac{49}{64}$.

Question 8: If 3 cot A = 4, check whether $\frac{1 - \tan^2 A}{1 + \tan^2 A} = \cos^2 A - \sin^2 A$ or not.

Solution-
1.Given $3 \cot A = 4$, so $\cot A = \frac{4}{3}$. Thus $\tan A = \frac{1}{\cot A} = \frac{3}{4}$.

2.Assume a right triangle where Adjacent $= 4k$ and Opposite $= 3k$. Hypotenuse $= \sqrt{4^2 + 3^2} = 5k$.

3.Calculate $\sin A = \frac{3}{5}$ and $\cos A = \frac{4}{5}$.

4.Evaluate the Left Hand Side (LHS):
$LHS = \frac{1 - \tan^2 A}{1 + \tan^2 A} = \frac{1 - (3/4)^2}{1 + (3/4)^2} = \frac{1 - 9/16}{1 + 9/16} = \frac{7/16}{25/16} = \frac{7}{25}$.

5.Evaluate the Right Hand Side (RHS):
$RHS = \cos^2 A - \sin^2 A = \left(\frac{4}{5}\right)^2 - \left(\frac{3}{5}\right)^2 = \frac{16}{25} - \frac{9}{25} = \frac{7}{25}$.

6.Since LHS = RHS ($\frac{7}{25} = \frac{7}{25}$), the given statement is true.

Question 9: In triangle ABC, right-angled at B, if $\tan A = \frac{1}{\sqrt{3}}$, find the value of:
(i) $\sin A \cos C + \cos A \sin C$
(ii) $\cos A \cos C – \sin A \sin C$

Solution-
1.Given $\tan A = \frac{1}{\sqrt{3}}$. We recognize this standard ratio corresponds to $A = 30^\circ$.
2.Since $\Delta ABC$ is right-angled at B ($90^\circ$), $\angle C = 180^\circ - 90^\circ - 30^\circ = 60^\circ$.
3.Using trigonometric ratio values for $30^\circ$ and $60^\circ$:
$\sin A = \sin 30^\circ = 1/2$. $\cos A = \cos 30^\circ = \sqrt{3}/2$.
$\sin C = \sin 60^\circ = \sqrt{3}/2$. $\cos C = \cos 60^\circ = 1/2$.

(i) Calculate $\sin A \cos C + \cos A \sin C$:
1.$\left(\frac{1}{2}\right)\left(\frac{1}{2}\right) + \left(\frac{\sqrt{3}}{2}\right)\left(\frac{\sqrt{3}}{2}\right) = \frac{1}{4} + \frac{3}{4} = \frac{4}{4} = 1$.

(ii) Calculate $\cos A \cos C – \sin A \sin C$:
1.$\left(\frac{\sqrt{3}}{2}\right)\left(\frac{1}{2}\right) - \left(\frac{1}{2}\right)\left(\frac{\sqrt{3}}{2}\right) = \frac{\sqrt{3}}{4} - \frac{\sqrt{3}}{4} = 0$.

Question 10: In $\Delta$ PQR, right-angled at Q, PR + QR = 25 cm and PQ = 5 cm. Determine the values of $\sin P$, $\cos P$ and $\tan P$.

Solution-
1.In $\Delta PQR$, by Pythagoras theorem, $PQ^2 + QR^2 = PR^2$.
2.We are given $PQ = 5$ cm and $PR + QR = 25$ cm. Let $PR = 25 - QR$.
3.Substitute into the Pythagoras equation:
$5^2 + QR^2 = (25 - QR)^2$.
$25 + QR^2 = 625 - 50QR + QR^2$.
$25 = 625 - 50QR$.
$50QR = 600$.
$QR = 12$ cm.

4.Find PR:
$PR = 25 - QR = 25 - 12 = 13$ cm.

5.Determine the trigonometric ratios for $\angle P$:
1.sin P = $\frac{\text{Opposite Side (QR)}}{\text{Hypotenuse (PR)}} = \frac{12}{13}$.

2.cos P = $\frac{\text{Adjacent Side (PQ)}}{\text{Hypotenuse (PR)}} = \frac{5}{13}$.

3.tan P = $\frac{\text{Opposite Side (QR)}}{\text{Adjacent Side (PQ)}} = \frac{12}{5}$.

Question 11: State whether the following are true or false. Justify your answer.
(i) The value of $\tan A$ is always less than 1.
(ii) $\sec A = \frac{12}{5}$ for some value of angle A.
(iii) $\cos A$ is the abbreviation used for the cosecant of angle A.
(iv) $\cot A$ is the product of $\cot$ and $A$.
(v) $\sin \theta = \frac{4}{3}$ for some angle $\theta$.

Solution-
(i) The value of $\tan A$ is always less than 1.
1.False. $\tan A$ is defined as the ratio of the side opposite to $\angle A$ to the side adjacent to $\angle A$. This ratio can be greater than 1, equal to 1 (when $A=45^\circ$), or less than 1. For example, $\tan 60^\circ = \sqrt{3} \approx 1.732$, which is greater than 1.

(ii) $\sec A = \frac{12}{5}$ for some value of angle A.
1.True. $\sec A$ is the ratio of the hypotenuse to the adjacent side. Since the hypotenuse is the longest side in a right triangle, $\sec A$ must always be greater than or equal to 1. Since $\frac{12}{5} = 2.4$, which is greater than 1, this value is possible.

(iii) $\cos A$ is the abbreviation used for the cosecant of angle A.
1.False. $\cos A$ is the abbreviation for the cosine of angle A. The abbreviation for cosecant of angle A is $\operatorname{cosec} A$.

(iv) $\cot A$ is the product of $\cot$ and $A$.
1.False. $\cot A$ is used as an abbreviation for 'the cotangent of angle A'. 'cot' separated from A has no meaning, and it is not a product.

(v) $\sin \theta = \frac{4}{3}$ for some angle $\theta$.
1.False. $\sin \theta$ is the ratio of the side opposite to $\theta$ to the hypotenuse. Since the hypotenuse is the longest side, the side opposite can never be greater than the hypotenuse, meaning the value of $\sin \theta$ can never exceed 1. Since $\frac{4}{3} \approx 1.33$, this value is impossible.

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EXERCISE 8.2

Question 1: Evaluate the following :
(i) $\sin 60^\circ \cos 30^\circ + \sin 30^\circ \cos 60^\circ$
(ii) $2 \tan^2 45^\circ + \cos^2 30^\circ – \sin^2 60^\circ$
(iii) $\frac{\cos 45^\circ}{\sec 30^\circ + \operatorname{cosec} 30^\circ}$
(iv) $\frac{\sin 30^\circ + \tan 45^\circ – \operatorname{cosec} 60^\circ}{\sec 30^\circ + \cos 60^\circ + \cot 45^\circ}$
(v) $\frac{5 \cos^2 60^\circ + 4 \sec^2 30^\circ – \tan^2 45^\circ}{\sin^2 30^\circ + \cos^2 30^\circ}$

Solution-
(i) $\sin 60^\circ \cos 30^\circ + \sin 30^\circ \cos 60^\circ$:
1.Substitute values from Table 8.1: $\left(\frac{\sqrt{3}}{2}\right)\left(\frac{\sqrt{3}}{2}\right) + \left(\frac{1}{2}\right)\left(\frac{1}{2}\right)$.
2.$=\frac{3}{4} + \frac{1}{4} = \frac{4}{4} = 1$.

(ii) $2 \tan^2 45^\circ + \cos^2 30^\circ – \sin^2 60^\circ$:
1.Substitute values: $2(1)^2 + \left(\frac{\sqrt{3}}{2}\right)^2 - \left(\frac{\sqrt{3}}{2}\right)^2$.
2.$= 2 + \frac{3}{4} - \frac{3}{4} = 2$.

(iii) $\frac{\cos 45^\circ}{\sec 30^\circ + \operatorname{cosec} 30^\circ}$:
1.Substitute values: $\frac{1/\sqrt{2}}{2/\sqrt{3} + 2}$.
2.$=\frac{1/\sqrt{2}}{\frac{2 + 2\sqrt{3}}{\sqrt{3}}} = \frac{1}{\sqrt{2}} \times \frac{\sqrt{3}}{2(1 + \sqrt{3})} = \frac{\sqrt{3}}{2\sqrt{2}(1 + \sqrt{3})}$.
3.Rationalize the denominator by multiplying by $\frac{(\sqrt{3}-1)\sqrt{2}}{(\sqrt{3}-1)\sqrt{2}}$:
$=\frac{\sqrt{6}(\sqrt{3}-1)}{2\sqrt{2}\sqrt{2}(3-1)} = \frac{\sqrt{18} - \sqrt{6}}{2(2)(2)} = \frac{3\sqrt{2} - \sqrt{6}}{8}$.

(iv) $\frac{\sin 30^\circ + \tan 45^\circ – \operatorname{cosec} 60^\circ}{\sec 30^\circ + \cos 60^\circ + \cot 45^\circ}$:
1.Substitute values: $\frac{\frac{1}{2} + 1 - \frac{2}{\sqrt{3}}}{\frac{2}{\sqrt{3}} + \frac{1}{2} + 1}$.
2.Numerator $= \frac{3}{2} - \frac{2}{\sqrt{3}} = \frac{3\sqrt{3} - 4}{2\sqrt{3}}$.
3.Denominator $= \frac{2}{\sqrt{3}} + \frac{3}{2} = \frac{4 + 3\sqrt{3}}{2\sqrt{3}}$.
4.Expression $= \frac{3\sqrt{3} - 4}{2\sqrt{3}} \div \frac{3\sqrt{3} + 4}{2\sqrt{3}} = \frac{3\sqrt{3} - 4}{3\sqrt{3} + 4}$.
5.Rationalize: $\frac{(3\sqrt{3} - 4)^2}{(3\sqrt{3})^2 - 4^2} = \frac{27 - 24\sqrt{3} + 16}{27 - 16} = \frac{43 - 24\sqrt{3}}{11}$.

(v) $\frac{5 \cos^2 60^\circ + 4 \sec^2 30^\circ – \tan^2 45^\circ}{\sin^2 30^\circ + \cos^2 30^\circ}$:
1.The denominator is $\sin^2 30^\circ + \cos^2 30^\circ$. Using the identity $\sin^2 A + \cos^2 A = 1$, the denominator equals 1.
2.Evaluate the numerator using values: $5\left(\frac{1}{2}\right)^2 + 4\left(\frac{2}{\sqrt{3}}\right)^2 - (1)^2$.
3.$= 5\left(\frac{1}{4}\right) + 4\left(\frac{4}{3}\right) - 1 = \frac{5}{4} + \frac{16}{3} - 1$.
4.Find the common denominator (12): $\frac{15}{12} + \frac{64}{12} - \frac{12}{12} = \frac{15 + 64 - 12}{12} = \frac{67}{12}$.

Question 2: Choose the correct option and justify your choice :
(i) $\frac{2 \tan 30^\circ}{1 + \tan^2 30^\circ}$
(A) $\sin 60^\circ$ (B) $\cos 60^\circ$ (C) $\tan 60^\circ$ (D) $\sin 30^\circ$

(ii) $\frac{1 - \tan^2 45^\circ}{1 + \tan^2 45^\circ}$
(A) $\tan 90^\circ$ (B) 1 (C) $\sin 45^\circ$ (D) 0

(iii) $\sin 2A = 2 \sin A$ is true when $A =$
(A) $0^\circ$ (B) $30^\circ$ (C) $45^\circ$ (D) $60^\circ$

(iv) $\frac{2 \tan 30^\circ}{1 - \tan^2 30^\circ}$
(A) $\cos 60^\circ$ (B) $\sin 60^\circ$ (C) $\tan 60^\circ$ (D) $\sin 30^\circ$

Solution-
(i) $\frac{2 \tan 30^\circ}{1 + \tan^2 30^\circ}$:
1.Substitute $\tan 30^\circ = 1/\sqrt{3}$: $\frac{2(1/\sqrt{3})}{1 + (1/\sqrt{3})^2} = \frac{2/\sqrt{3}}{1 + 1/3} = \frac{2/\sqrt{3}}{4/3}$.
2.$= \frac{2}{\sqrt{3}} \times \frac{3}{4} = \frac{6}{4\sqrt{3}} = \frac{3}{2\sqrt{3}} = \frac{\sqrt{3}}{2}$.
3.Since $\sin 60^\circ = \frac{\sqrt{3}}{2}$, the correct option is (A) $\sin 60^\circ$.

(ii) $\frac{1 - \tan^2 45^\circ}{1 + \tan^2 45^\circ}$:
1.Substitute $\tan 45^\circ = 1$: $\frac{1 - (1)^2}{1 + (1)^2} = \frac{1 - 1}{1 + 1} = \frac{0}{2} = 0$.
2.The correct option is (D) 0.

(iii) $\sin 2A = 2 \sin A$ is true when $A =$:
1.We check $A=0^\circ$. $\sin (2 \times 0^\circ) = \sin 0^\circ = 0$.
2.$2 \sin 0^\circ = 2(0) = 0$. Since $0=0$, the condition is true for $A=0^\circ$.
3.The correct option is (A) $0^\circ$.

(iv) $\frac{2 \tan 30^\circ}{1 - \tan^2 30^\circ}$:
1.Substitute $\tan 30^\circ = 1/\sqrt{3}$: $\frac{2(1/\sqrt{3})}{1 - (1/\sqrt{3})^2} = \frac{2/\sqrt{3}}{1 - 1/3} = \frac{2/\sqrt{3}}{2/3}$.
2.$= \frac{2}{\sqrt{3}} \times \frac{3}{2} = \frac{3}{\sqrt{3}} = \sqrt{3}$.
3.Since $\tan 60^\circ = \sqrt{3}$, the correct option is (C) $\tan 60^\circ$.

Question 3: If $\tan (A + B) = \sqrt{3}$ and $\tan (A – B) = \frac{1}{\sqrt{3}}$; $0^\circ < A + B \leq 90^\circ$; $A > B$, find A and B.

Solution-
1.Given $\tan (A + B) = \sqrt{3}$. Since $\tan 60^\circ = \sqrt{3}$, we have $A + B = 60^\circ$ (1).

2.Given $\tan (A - B) = \frac{1}{\sqrt{3}}$. Since $\tan 30^\circ = \frac{1}{\sqrt{3}}$, we have $A - B = 30^\circ$ (2).

3.Add equations (1) and (2):
$(A + B) + (A - B) = 60^\circ + 30^\circ$.
$2A = 90^\circ$.
$A = 45^\circ$.

4.Substitute $A = 45^\circ$ into equation (1):
$45^\circ + B = 60^\circ$.
$B = 15^\circ$.

5.The values are $\mathbf{A = 45^\circ}$ and $\mathbf{B = 15^\circ}$.

Question 4: State whether the following are true or false. Justify your answer.
(i) $\sin (A + B) = \sin A + \sin B$.
(ii) The value of $\sin \theta$ increases as $\theta$ increases.
(iii) The value of $\cos \theta$ increases as $\theta$ increases.
(iv) $\sin \theta = \cos \theta$ for all values of $\theta$.
(v) $\cot A$ is not defined for $A = 0^\circ$.

Solution-
(i) $\sin (A + B) = \sin A + \sin B$.
1.False. Let $A = 60^\circ$ and $B = 30^\circ$.
$LHS = \sin (60^\circ + 30^\circ) = \sin 90^\circ = 1$.
$RHS = \sin 60^\circ + \sin 30^\circ = \frac{\sqrt{3}}{2} + \frac{1}{2} \neq 1$.

(ii) The value of $\sin \theta$ increases as $\theta$ increases.
1.True. As $\angle A$ increases from $0^\circ$ to $90^\circ$, the value of $\sin A$ increases monotonically from 0 to 1.

(iii) The value of $\cos \theta$ increases as $\theta$ increases.
1.False. As $\angle A$ increases from $0^\circ$ to $90^\circ$, the value of $\cos A$ decreases monotonically from 1 to 0.

(iv) $\sin \theta = \cos \theta$ for all values of $\theta$.
1.False. This equality is true only when $\theta = 45^\circ$, where both $\sin 45^\circ$ and $\cos 45^\circ$ equal $\frac{1}{\sqrt{2}}$.

(v) $\cot A$ is not defined for $A = 0^\circ$.
1.True. $\cot 0^\circ$ is defined as $\frac{1}{\tan 0^\circ}$. Since $\tan 0^\circ = 0$, $\cot 0^\circ = \frac{1}{0}$, which is not defined.

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EXERCISE 8.3

Question 1: Express the trigonometric ratios $\sin A$, $\sec A$ and $\tan A$ in terms of $\cot A$.

Solution-
1.Express $\sin A$ in terms of $\cot A$:
We use the identity $\operatorname{cosec}^2 A = 1 + \cot^2 A$.
$\operatorname{cosec} A = \sqrt{1 + \cot^2 A}$.
Since $\sin A = \frac{1}{\operatorname{cosec} A}$, $\sin A = \frac{1}{\sqrt{1 + \cot^2 A}}$.

2.Express $\tan A$ in terms of $\cot A$:
By reciprocal relation, $\tan A = \frac{1}{\cot A}$.

3.Express $\sec A$ in terms of $\cot A$:
We use the identity $\sec^2 A = 1 + \tan^2 A$.
$\sec A = \sqrt{1 + \tan^2 A}$.
Substitute $\tan A = 1/\cot A$: $\sec A = \sqrt{1 + \left(\frac{1}{\cot A}\right)^2} = \sqrt{\frac{\cot^2 A + 1}{\cot^2 A}}$.
$\sec A = \frac{\sqrt{1 + \cot^2 A}}{\cot A}$.

Question 2: Write all the other trigonometric ratios of $\angle A$ in terms of $\sec A$.

Solution-
1.Express $\cos A$:
$\cos A = \frac{1}{\sec A}$.

2.Express $\sin A$:
Using the identity $\sin^2 A + \cos^2 A = 1$, we have $\sin A = \sqrt{1 - \cos^2 A}$.
Substituting $\cos A = 1/\sec A$: $\sin A = \sqrt{1 - \left(\frac{1}{\sec A}\right)^2} = \sqrt{\frac{\sec^2 A - 1}{\sec^2 A}}$.
$\sin A = \frac{\sqrt{\sec^2 A - 1}}{\sec A}$.

3.Express $\tan A$:
Using the identity $1 + \tan^2 A = \sec^2 A$, we have $\tan A = \sqrt{\sec^2 A - 1}$.

4.Express $\cot A$:
$\cot A = \frac{1}{\tan A}$.
$\cot A = \frac{1}{\sqrt{\sec^2 A - 1}}$.

5.Express $\operatorname{cosec} A$:
$\operatorname{cosec} A = \frac{1}{\sin A}$.
$\operatorname{cosec} A = \frac{\sec A}{\sqrt{\sec^2 A - 1}}$.

Question 3: Choose the correct option. Justify your choice.
(i) $9 \sec^2 A – 9 \tan^2 A =$
(A) 1 (B) 9 (C) 8 (D) 0

(ii) $(1 + \tan \theta + \sec \theta) (1 + \cot \theta – \operatorname{cosec} \theta) =$
(A) 0 (B) 1 (C) 2 (D) –1

(iii) $(\sec A + \tan A) (1 – \sin A) =$
(A) $\sec A$ (B) $\sin A$ (C) $\operatorname{cosec} A$ (D) $\cos A$

(iv) $\frac{1 + \tan^2 A}{1 + \cot^2 A} =$
(A) $\sec^2 A$ (B) –1 (C) $\cot^2 A$ (D) $\tan^2 A$

Solution-
(i) $9 \sec^2 A – 9 \tan^2 A$:
1.Factor out 9: $9(\sec^2 A - \tan^2 A)$.
2.Using the identity $\sec^2 A = 1 + \tan^2 A$, we have $\sec^2 A - \tan^2 A = 1$.
3.Result: $9(1) = 9$.
4.The correct option is (B) 9.

(ii) $(1 + \tan \theta + \sec \theta) (1 + \cot \theta – \operatorname{cosec} \theta)$:
1.Convert to $\sin \theta$ and $\cos \theta$:
LHS $= (1 + \frac{\sin \theta}{\cos \theta} + \frac{1}{\cos \theta})(1 + \frac{\cos \theta}{\sin \theta} - \frac{1}{\sin \theta})$.
LHS $= \left(\frac{\cos \theta + \sin \theta + 1}{\cos \theta}\right)\left(\frac{\sin \theta + \cos \theta - 1}{\sin \theta}\right)$.
2.Multiply numerators using $(X+1)(X-1) = X^2 - 1$, where $X = (\sin \theta + \cos \theta)$:
Numerator $= (\sin \theta + \cos \theta)^2 - 1 = (\sin^2 \theta + \cos^2 \theta + 2 \sin \theta \cos \theta) - 1$.
3.Using $\sin^2 \theta + \cos^2 \theta = 1$, Numerator $= (1 + 2 \sin \theta \cos \theta) - 1 = 2 \sin \theta \cos \theta$.
4.LHS $= \frac{2 \sin \theta \cos \theta}{\cos \theta \sin \theta} = 2$.
5.The correct option is (C) 2.

(iii) $(\sec A + \tan A) (1 – \sin A)$:
1.Convert to $\sin A$ and $\cos A$:
LHS $= \left(\frac{1}{\cos A} + \frac{\sin A}{\cos A}\right) (1 – \sin A)$.
LHS $= \frac{(1 + \sin A)}{\cos A} (1 – \sin A) = \frac{1 - \sin^2 A}{\cos A}$.
2.Using $1 - \sin^2 A = \cos^2 A$: LHS $= \frac{\cos^2 A}{\cos A} = \cos A$.
3.The correct option is (D) $\cos A$.

(iv) $\frac{1 + \tan^2 A}{1 + \cot^2 A}$:
1.Use identities: $1 + \tan^2 A = \sec^2 A$ and $1 + \cot^2 A = \operatorname{cosec}^2 A$.
LHS $= \frac{\sec^2 A}{\operatorname{cosec}^2 A} = \frac{1/\cos^2 A}{1/\sin^2 A} = \frac{\sin^2 A}{\cos^2 A}$.
2.LHS $= \tan^2 A$.
3.The correct option is (D) $\tan^2 A$.

Question 4: Prove the following identities, where the angles involved are acute angles for which the expressions are defined.
(i) $(\operatorname{cosec} \theta – \cot \theta)^2 = \frac{1 - \cos \theta}{1 + \cos \theta}$

Solution-
1.Start with LHS, converting to $\sin \theta$ and $\cos \theta$:
LHS $= (\operatorname{cosec} \theta – \cot \theta)^2 = \left(\frac{1}{\sin \theta} - \frac{\cos \theta}{\sin \theta}\right)^2$.

2.LHS $= \left(\frac{1 - \cos \theta}{\sin \theta}\right)^2 = \frac{(1 - \cos \theta)^2}{\sin^2 \theta}$.

3.Substitute $\sin^2 \theta = 1 - \cos^2 \theta$:
LHS $= \frac{(1 - \cos \theta)^2}{1 - \cos^2 \theta}$.

4.Use the difference of squares identity $1 - \cos^2 \theta = (1 - \cos \theta)(1 + \cos \theta)$:
LHS $= \frac{(1 - \cos \theta)(1 - \cos \theta)}{(1 - \cos \theta)(1 + \cos \theta)} = \frac{1 - \cos \theta}{1 + \cos \theta}$.

5.LHS = RHS. Hence proved.

Question 4: Prove the following identities, where the angles involved are acute angles for which the expressions are defined.
(ii) $\frac{\cos A}{1 + \sin A} + \frac{1 + \sin A}{\cos A} = 2 \sec A$

Solution-
1.Start with LHS, combining fractions by finding a common denominator:
LHS $= \frac{\cos^2 A + (1 + \sin A)^2}{(1 + \sin A) \cos A}$.

2.Expand the term in the numerator:
LHS $= \frac{\cos^2 A + (1 + 2 \sin A + \sin^2 A)}{(1 + \sin A) \cos A}$.

3.Group $\sin^2 A + \cos^2 A = 1$:
LHS $= \frac{(\sin^2 A + \cos^2 A) + 1 + 2 \sin A}{(1 + \sin A) \cos A} = \frac{1 + 1 + 2 \sin A}{(1 + \sin A) \cos A}$.

4.Simplify the numerator and factor:
LHS $= \frac{2 + 2 \sin A}{(1 + \sin A) \cos A} = \frac{2(1 + \sin A)}{(1 + \sin A) \cos A}$.

5.Cancel the common factor $(1 + \sin A)$:
LHS $= \frac{2}{\cos A}$.

6.Since $\sec A = \frac{1}{\cos A}$: LHS $= 2 \sec A$.

7.LHS = RHS. Hence proved.

Question 4: Prove the following identities, where the angles involved are acute angles for which the expressions are defined.
(iii) $\frac{\tan \theta}{1 - \cot \theta} + \frac{\cot \theta}{1 - \tan \theta} = 1 + \sec \theta \operatorname{cosec} \theta$ [Hint : Write the expression in terms of $\sin \theta$ and $\cos \theta$]

Solution-
1.Start with LHS, converting all ratios to $\sin \theta$ and $\cos \theta$:
LHS $= \frac{\frac{\sin \theta}{\cos \theta}}{1 - \frac{\cos \theta}{\sin \theta}} + \frac{\frac{\cos \theta}{\sin \theta}}{1 - \frac{\sin \theta}{\cos \theta}}$.

2.Simplify the denominators:
LHS $= \frac{\frac{\sin \theta}{\cos \theta}}{\frac{\sin \theta - \cos \theta}{\sin \theta}} + \frac{\frac{\cos \theta}{\sin \theta}}{\frac{\cos \theta - \sin \theta}{\cos \theta}}$.

3.Invert and multiply:
LHS $= \frac{\sin^2 \theta}{\cos \theta (\sin \theta - \cos \theta)} + \frac{\cos^2 \theta}{\sin \theta (\cos \theta - \sin \theta)}$.

4.To obtain a common denominator, change the sign of the second term's denominator: $\cos \theta - \sin \theta = -(\sin \theta - \cos \theta)$.
LHS $= \frac{\sin^2 \theta}{\cos \theta (\sin \theta - \cos \theta)} - \frac{\cos^2 \theta}{\sin \theta (\sin \theta - \cos \theta)}$.

5.Combine the terms:
LHS $= \frac{\sin^3 \theta - \cos^3 \theta}{\sin \theta \cos \theta (\sin \theta - \cos \theta)}$.

6.Use the algebraic identity $a^3 - b^3 = (a - b)(a^2 + ab + b^2)$, where $a=\sin \theta$ and $b=\cos \theta$:
LHS $= \frac{(\sin \theta - \cos \theta)(\sin^2 \theta + \cos^2 \theta + \sin \theta \cos \theta)}{\sin \theta \cos \theta (\sin \theta - \cos \theta)}$.

7.Cancel $(\sin \theta - \cos \theta)$ and use $\sin^2 \theta + \cos^2 \theta = 1$:
LHS $= \frac{1 + \sin \theta \cos \theta}{\sin \theta \cos \theta} = \frac{1}{\sin \theta \cos \theta} + \frac{\sin \theta \cos \theta}{\sin \theta \cos \theta}$.

8.LHS $= \operatorname{cosec} \theta \sec \theta + 1$.

9.LHS = RHS. Hence proved.

Question 4: Prove the following identities, where the angles involved are acute angles for which the expressions are defined.
(v) $\frac{\cos A – \sin A + 1}{\cos A + \sin A – 1} = \operatorname{cosec} A + \cot A$, using the identity $\operatorname{cosec}^2 A = 1 + \cot^2 A$.

Solution-
1.Since the required RHS contains $\operatorname{cosec} A$ and $\cot A$, divide the numerator (N) and denominator (D) of the LHS by $\sin A$:
LHS $= \frac{\frac{\cos A}{\sin A} - \frac{\sin A}{\sin A} + \frac{1}{\sin A}}{\frac{\cos A}{\sin A} + \frac{\sin A}{\sin A} - \frac{1}{\sin A}} = \frac{\cot A - 1 + \operatorname{cosec} A}{\cot A + 1 - \operatorname{cosec} A}$.

2.Use the required identity $\operatorname{cosec}^2 A - \cot^2 A = 1$ to replace '1' only in the numerator:
$N = (\cot A + \operatorname{cosec} A) - 1 = (\cot A + \operatorname{cosec} A) - (\operatorname{cosec}^2 A - \cot^2 A)$.

3.Factor the difference of squares:
$N = (\cot A + \operatorname{cosec} A) - (\operatorname{cosec} A - \cot A)(\operatorname{cosec} A + \cot A)$.

4.Factor out $(\cot A + \operatorname{cosec} A)$:
$N = (\cot A + \operatorname{cosec} A) [1 - (\operatorname{cosec} A - \cot A)]$.
$N = (\cot A + \operatorname{cosec} A) [1 - \operatorname{cosec} A + \cot A]$.

5.Substitute N back into LHS:
LHS $= \frac{(\cot A + \operatorname{cosec} A) (1 + \cot A - \operatorname{cosec} A)}{(\cot A + 1 - \operatorname{cosec} A)}$.

6.Cancel the common factor $(1 + \cot A - \operatorname{cosec} A)$:
LHS $= \cot A + \operatorname{cosec} A$.

7.LHS = RHS. Hence proved.

Question 4: Prove the following identities, where the angles involved are acute angles for which the expressions are defined.
(vi) $\sqrt{\frac{1 + \sin A}{1 – \sin A}} = \sec A + \tan A$

Solution-
1.Start with LHS, rationalize the expression by multiplying the numerator and denominator under the square root by $(1 + \sin A)$:
LHS $= \sqrt{\frac{1 + \sin A}{1 - \sin A} \times \frac{1 + \sin A}{1 + \sin A}}$.

2.LHS $= \sqrt{\frac{(1 + \sin A)^2}{1 - \sin^2 A}}$.

3.Use $1 - \sin^2 A = \cos^2 A$:
LHS $= \sqrt{\frac{(1 + \sin A)^2}{\cos^2 A}}$.

4.Take the square root:
LHS $= \frac{1 + \sin A}{\cos A}$.

5.Split the fraction:
LHS $= \frac{1}{\cos A} + \frac{\sin A}{\cos A}$.

6.Use reciprocal and quotient identities:
LHS $= \sec A + \tan A$.

7.LHS = RHS. Hence proved.

Question 4: Prove the following identities, where the angles involved are acute angles for which the expressions are defined.
(ix) $\frac{1}{(\operatorname{cosec} A – \sin A) (\sec A – \cos A)} = \tan A + \cot A$ [Hint : Simplify LHS and RHS separately]

Solution-
1.Simplify LHS, converting to $\sin A$ and $\cos A$:
LHS $= (\operatorname{cosec} A – \sin A) (\sec A – \cos A)$.
LHS $= \left(\frac{1}{\sin A} - \sin A\right) \left(\frac{1}{\cos A} - \cos A\right)$.
LHS $= \left(\frac{1 - \sin^2 A}{\sin A}\right) \left(\frac{1 - \cos^2 A}{\cos A}\right)$.

2.Use identities $1 - \sin^2 A = \cos^2 A$ and $1 - \cos^2 A = \sin^2 A$:
LHS $= \left(\frac{\cos^2 A}{\sin A}\right) \left(\frac{\sin^2 A}{\cos A}\right) = \cos A \sin A$. (Wait, the question is $\frac{1}{LHS}$)

3.Re-evaluating the LHS based on the displayed question structure in:
LHS of the equality to be proven $= \frac{1}{(\operatorname{cosec} A – \sin A) (\sec A – \cos A)}$.
LHS $= \frac{1}{\cos A \sin A}$.

4.Simplify RHS, converting to $\sin A$ and $\cos A$:
RHS $= \tan A + \cot A = \frac{\sin A}{\cos A} + \frac{\cos A}{\sin A}$.
RHS $= \frac{\sin^2 A + \cos^2 A}{\cos A \sin A}$.

5.Use $\sin^2 A + \cos^2 A = 1$:
RHS $= \frac{1}{\cos A \sin A}$.

6.Since LHS = RHS, the identity is proved.

Question 4: Prove the following identities, where the angles involved are acute angles for which the expressions are defined.
(x) $\left(\frac{1 + \tan^2 A}{1 + \cot^2 A}\right) = \left(\frac{1 - \tan A}{1 - \cot A}\right)^2 = \tan^2 A$

Solution-
1.Prove the first part: $\frac{1 + \tan^2 A}{1 + \cot^2 A} = \tan^2 A$.
LHS $= \frac{1 + \tan^2 A}{1 + \cot^2 A}$.
Use identities $1 + \tan^2 A = \sec^2 A$ and $1 + \cot^2 A = \operatorname{cosec}^2 A$.
LHS $= \frac{\sec^2 A}{\operatorname{cosec}^2 A} = \frac{1/\cos^2 A}{1/\sin^2 A} = \frac{\sin^2 A}{\cos^2 A} = \tan^2 A$.

2.Prove the second part: $\left(\frac{1 - \tan A}{1 - \cot A}\right)^2 = \tan^2 A$.
Simplify the base of the square, converting $\cot A$ to $\tan A$:
$\frac{1 - \tan A}{1 - \cot A} = \frac{1 - \tan A}{1 - \frac{1}{\tan A}}$.
$= \frac{1 - \tan A}{\frac{\tan A - 1}{\tan A}} = (1 - \tan A) \times \frac{\tan A}{\tan A - 1}$.
$= -( \tan A - 1) \times \frac{\tan A}{\tan A - 1} = -\tan A$.

3.Square the result:
$\left(\frac{1 - \tan A}{1 - \cot A}\right)^2 = (-\tan A)^2 = \tan^2 A$.

4.Since both expressions equal $\tan^2 A$, the identity is proved.