Class 10-NCERT Solutions-Chapter-08-Introduction to Trigonometry

Solution:

By Pythagoras theorem, \( AC^2 = AB^2 + BC^2 \)

\[ AC = \sqrt{24^2 + 7^2} = \sqrt{576 + 49} = \sqrt{625} = 25 \text{ cm} \]

(i) For angle A:

Opposite side = BC = 7, Adjacent side = AB = 24, Hypotenuse = AC = 25.

\[ \sin A = \frac{\text{Opposite}}{\text{Hypotenuse}} = \frac{7}{25} \]

\[ \cos A = \frac{\text{Adjacent}}{\text{Hypotenuse}} = \frac{24}{25} \]

(ii) For angle C:

Opposite side = AB = 24, Adjacent side = BC = 7.

\[ \sin C = \frac{AB}{AC} = \frac{24}{25} \]

\[ \cos C = \frac{BC}{AC} = \frac{7}{25} \]

Solution:

Given \( PQ = 12 \) cm, \( PR = 13 \) cm. Find QR using Pythagoras theorem:

\[ QR = \sqrt{PR^2 - PQ^2} = \sqrt{13^2 - 12^2} = \sqrt{169 - 144} = \sqrt{25} = 5 \text{ cm} \]

For angle P: Opposite = QR = 5, Adjacent = PQ = 12.

\[ \tan P = \frac{QR}{PQ} = \frac{5}{12} \]

For angle R: Adjacent = QR = 5, Opposite = PQ = 12.

\[ \cot R = \frac{QR}{PQ} = \frac{5}{12} \]

\[ \tan P - \cot R = \frac{5}{12} - \frac{5}{12} = 0 \]

Answer: 0

Solution:

Let side opposite to A be \( 3k \) and hypotenuse be \( 4k \).

Using Pythagoras theorem: \( \text{Adjacent}^2 = (4k)^2 - (3k)^2 = 16k^2 - 9k^2 = 7k^2 \).

Adjacent side = \( \sqrt{7}k \).

\[ \cos A = \frac{\text{Adjacent}}{\text{Hypotenuse}} = \frac{\sqrt{7}k}{4k} = \frac{\sqrt{7}}{4} \]

\[ \tan A = \frac{\text{Opposite}}{\text{Adjacent}} = \frac{3k}{\sqrt{7}k} = \frac{3}{\sqrt{7}} \]

Solution:

\[ \cot A = \frac{8}{15} = \frac{\text{Adjacent}}{\text{Opposite}} \]

Let Adjacent = \( 8k \), Opposite = \( 15k \).

Hypotenuse = \( \sqrt{(8k)^2 + (15k)^2} = \sqrt{64k^2 + 225k^2} = \sqrt{289k^2} = 17k \).

\[ \sin A = \frac{\text{Opposite}}{\text{Hypotenuse}} = \frac{15k}{17k} = \frac{15}{17} \]

\[ \sec A = \frac{\text{Hypotenuse}}{\text{Adjacent}} = \frac{17k}{8k} = \frac{17}{8} \]

Solution:

\[ \sec \theta = \frac{\text{Hypotenuse}}{\text{Adjacent}} = \frac{13}{12} \]

Let Hypotenuse = \( 13k \), Adjacent = \( 12k \).

Opposite side = \( \sqrt{(13k)^2 - (12k)^2} = \sqrt{169k^2 - 144k^2} = \sqrt{25k^2} = 5k \).

Ratios:

\[ \sin \theta = \frac{5}{13} \]

\[ \cos \theta = \frac{12}{13} \]

\[ \tan \theta = \frac{5}{12} \]

\[ \text{cosec } \theta = \frac{13}{5} \]

\[ \cot \theta = \frac{12}{5} \]

Solution:

Consider a triangle ABC right angled at C.

\[ \cos A = \frac{AC}{AB} \]

\[ \cos B = \frac{BC}{AB} \]

Given \( \cos A = \cos B \), therefore:

\[ \frac{AC}{AB} = \frac{BC}{AB} \]

\[ AC = BC \]

Angles opposite to equal sides in a triangle are equal. Therefore, \( \angle A = \angle B \).

Solution:

(i) Simplify the expression using algebraic identity \( (a+b)(a-b) = a^2 - b^2 \).

\[ \frac{1 - \sin^2 \theta}{1 - \cos^2 \theta} \]

We know \( 1 - \sin^2 \theta = \cos^2 \theta \) and \( 1 - \cos^2 \theta = \sin^2 \theta \).

\[ = \frac{\cos^2 \theta}{\sin^2 \theta} = \left(\frac{\cos \theta}{\sin \theta}\right)^2 = \cot^2 \theta \]

Given \( \cot \theta = \frac{7}{8} \), so value is \( \left(\frac{7}{8}\right)^2 = \frac{49}{64} \).

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

Solution:

Given \( 3 \cot A = 4 \Rightarrow \cot A = \frac{4}{3} \Rightarrow \tan A = \frac{3}{4} \).

From \( \tan A = \frac{3}{4} \) (Opposite=3, Adjacent=4), Hypotenuse = \( \sqrt{3^2 + 4^2} = 5 \).

So, \( \sin A = \frac{3}{5} \) and \( \cos A = \frac{4}{5} \).

LHS: \( \frac{1 - (3/4)^2}{1 + (3/4)^2} = \frac{1 - 9/16}{1 + 9/16} = \frac{7/16}{25/16} = \frac{7}{25} \).

RHS: \( \cos^2 A - \sin^2 A = (\frac{4}{5})^2 - (\frac{3}{5})^2 = \frac{16}{25} - \frac{9}{25} = \frac{7}{25} \).

Since LHS = RHS, the relation is True.

Solution:

\( \tan A = \frac{1}{\sqrt{3}} \). This corresponds to angle \( A = 30^{\circ} \).

In right triangle ABC, \( A+C=90^{\circ} \), so \( C = 60^{\circ} \).

(i) \( \sin 30^{\circ} \cos 60^{\circ} + \cos 30^{\circ} \sin 60^{\circ} \)

\[ = \left(\frac{1}{2}\right)\left(\frac{1}{2}\right) + \left(\frac{\sqrt{3}}{2}\right)\left(\frac{\sqrt{3}}{2}\right) = \frac{1}{4} + \frac{3}{4} = 1 \]

(ii) \( \cos 30^{\circ} \cos 60^{\circ} - \sin 30^{\circ} \sin 60^{\circ} \)

\[ = \left(\frac{\sqrt{3}}{2}\right)\left(\frac{1}{2}\right) - \left(\frac{1}{2}\right)\left(\frac{\sqrt{3}}{2}\right) = \frac{\sqrt{3}}{4} - \frac{\sqrt{3}}{4} = 0 \]

Solution:

Let \( QR = x \). Then \( PR = 25 - x \).

By Pythagoras theorem: \( PR^2 = PQ^2 + QR^2 \)

\[ (25 - x)^2 = 5^2 + x^2 \]

\[ 625 + x^2 - 50x = 25 + x^2 \]

\[ 50x = 600 \Rightarrow x = 12 \]

So, \( QR = 12 \) cm and \( PR = 13 \) cm.

\[ \sin P = \frac{12}{13}, \quad \cos P = \frac{5}{13}, \quad \tan P = \frac{12}{5} \]

Answers:

(i) False. tan A can be greater than 1 (e.g., tan 60 = 1.732).

(ii) True. sec A is always \( \ge 1 \). \( \frac{12}{5} = 2.4 \), which is possible.

(iii) False. cos A is cosine, cosec is cosecant.

(iv) False. cot A is a single term representing the cotangent of angle A.

(v) False. sin theta cannot exceed 1 (4/3 = 1.33).

Solution:

(i) \( \frac{\sqrt{3}}{2} \times \frac{\sqrt{3}}{2} + \frac{1}{2} \times \frac{1}{2} = \frac{3}{4} + \frac{1}{4} = 1 \)

(ii) \( 2(1)^2 + (\frac{\sqrt{3}}{2})^2 - (\frac{\sqrt{3}}{2})^2 = 2 + \frac{3}{4} - \frac{3}{4} = 2 \)

(iii) \( \frac{1/\sqrt{2}}{2/\sqrt{3} + 2} = \frac{1/\sqrt{2}}{(2 + 2\sqrt{3})/\sqrt{3}} = \frac{\sqrt{3}}{\sqrt{2}(2 + 2\sqrt{3})} \). Rationalize to get \( \frac{3\sqrt{2} - \sqrt{6}}{8} \).

(iv) Numerator: \( \frac{1}{2} + 1 - \frac{2}{\sqrt{3}} = \frac{3}{2} - \frac{2}{\sqrt{3}} \). Denominator: \( \frac{2}{\sqrt{3}} + \frac{1}{2} + 1 = \frac{3}{2} + \frac{2}{\sqrt{3}} \).
Simplify: \( \frac{3\sqrt{3}-4}{3\sqrt{3}+4} \). Rationalize: \( \frac{43 - 24\sqrt{3}}{11} \).

(v) Numerator: \( 5(\frac{1}{2})^2 + 4(\frac{2}{\sqrt{3}})^2 - 1^2 = \frac{5}{4} + \frac{16}{3} - 1 = \frac{15 + 64 - 12}{12} = \frac{67}{12} \).
Denominator: \( (\frac{1}{2})^2 + (\frac{\sqrt{3}}{2})^2 = 1 \). Result: \( \frac{67}{12} \).

Solution:

(i) \( \frac{2(1/\sqrt{3})}{1 + 1/3} = \frac{2/\sqrt{3}}{4/3} = \frac{\sqrt{3}}{2} = \sin 60^{\circ} \). Option (A).

(ii) \( \frac{1-1}{1+1} = 0 \). Option (D).

(iii) Check A=0: \( \sin 0 = 0, 2 \sin 0 = 0 \). Option (A).

(iv) \( \frac{2(1/\sqrt{3})}{1 - 1/3} = \frac{2/\sqrt{3}}{2/3} = \sqrt{3} = \tan 60^{\circ} \). Option (C).

Solution:

\( \tan(A+B) = \sqrt{3} \Rightarrow A + B = 60^{\circ} \).

\( \tan(A-B) = \frac{1}{\sqrt{3}} \Rightarrow A - B = 30^{\circ} \).

Adding the two equations: \( 2A = 90^{\circ} \Rightarrow A = 45^{\circ} \).

Subtracting: \( 2B = 30^{\circ} \Rightarrow B = 15^{\circ} \).

Answer: \( A = 45^{\circ}, B = 15^{\circ} \).

Answers:

(i) False. Let A=30, B=60. \( \sin(90)=1 \), but \( \sin 30 + \sin 60 \neq 1 \).

(ii) True. sin 0 = 0, sin 90 = 1.

(iii) False. cos 0 = 1, cos 90 = 0 (It decreases).

(iv) False. Only true for 45 degrees.

(v) True. \( \cot 0 = \frac{\cos 0}{\sin 0} = \frac{1}{0} \) (undefined).

Solution:

(i) \( \tan A \) in terms of \( \cot A \):

\[ \tan A = \frac{1}{\cot A} \]

(ii) \( \sec A \) in terms of \( \cot A \):

Using the identity \( \sec^2 A = 1 + \tan^2 A \):

\[ \sec A = \sqrt{1 + \frac{1}{\cot^2 A}} = \sqrt{\frac{\cot^2 A + 1}{\cot^2 A}} = \frac{\sqrt{1 + \cot^2 A}}{\cot A} \]

(iii) \( \sin A \) in terms of \( \cot A \):

Using the identity \( \text{cosec}^2 A = 1 + \cot^2 A \):

\[ \sin A = \frac{1}{\text{cosec } A} = \frac{1}{\sqrt{1 + \cot^2 A}} \]

Solution:

1. \( \cos A = \frac{1}{\sec A} \)
2. \( \sin A = \sqrt{1 - \cos^2 A} = \sqrt{1 - \frac{1}{\sec^2 A}} = \frac{\sqrt{\sec^2 A - 1}}{\sec A} \)
3. \( \tan A = \sqrt{\sec^2 A - 1} \)
4. \( \cot A = \frac{1}{\tan A} = \frac{1}{\sqrt{\sec^2 A - 1}} \)
5. \( \text{cosec } A = \frac{1}{\sin A} = \frac{\sec A}{\sqrt{\sec^2 A - 1}} \)

Solution:

\[ 9 \sec^2 A - 9 \tan^2 A = 9(\sec^2 A - \tan^2 A) \]

Using the identity \( \sec^2 A - \tan^2 A = 1 \):

\[ = 9(1) = 9 \]

Correct Option: (B)

Solution:

Convert all terms to sin and cos:

\[ \left(1 + \frac{\sin \theta}{\cos \theta} + \frac{1}{\cos \theta}\right) \left(1 + \frac{\cos \theta}{\sin \theta} - \frac{1}{\sin \theta}\right) \]

\[ = \left(\frac{\cos \theta + \sin \theta + 1}{\cos \theta}\right) \left(\frac{\sin \theta + \cos \theta - 1}{\sin \theta}\right) \]

Numerator is of the form \( (x+1)(x-1) = x^2 - 1 \) where \( x = \sin \theta + \cos \theta \).

\[ = \frac{(\sin \theta + \cos \theta)^2 - 1^2}{\sin \theta \cos \theta} \]

\[ = \frac{\sin^2 \theta + \cos^2 \theta + 2\sin \theta \cos \theta - 1}{\sin \theta \cos \theta} \]

\[ = \frac{1 + 2\sin \theta \cos \theta - 1}{\sin \theta \cos \theta} = \frac{2\sin \theta \cos \theta}{\sin \theta \cos \theta} = 2 \]

Correct Option: (C)

Solution:

\[ \left(\frac{1}{\cos A} + \frac{\sin A}{\cos A}\right) (1 - \sin A) \]

\[ = \left(\frac{1 + \sin A}{\cos A}\right) (1 - \sin A) \]

\[ = \frac{1 - \sin^2 A}{\cos A} = \frac{\cos^2 A}{\cos A} = \cos A \]

Correct Option: (D)

Solution:

Using identities \( 1+\tan^2 A = \sec^2 A \) and \( 1+\cot^2 A = \text{cosec}^2 A \):

\[ \frac{\sec^2 A}{\text{cosec}^2 A} = \frac{1/\cos^2 A}{1/\sin^2 A} = \frac{\sin^2 A}{\cos^2 A} = \tan^2 A \]

Correct Option: (D)

Solution:

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

\[ = \frac{(1 - \cos \theta)^2}{\sin^2 \theta} = \frac{(1 - \cos \theta)^2}{1 - \cos^2 \theta} \]

\[ = \frac{(1 - \cos \theta)(1 - \cos \theta)}{(1 - \cos \theta)(1 + \cos \theta)} = \frac{1 - \cos \theta}{1 + \cos \theta} = \text{RHS} \]

Solution:

LHS \( = \frac{\cos^2 A + (1 + \sin A)^2}{\cos A(1 + \sin A)} \)

\[ = \frac{\cos^2 A + 1 + \sin^2 A + 2\sin A}{\cos A(1 + \sin A)} \]

Using \( \cos^2 A + \sin^2 A = 1 \):

\[ = \frac{1 + 1 + 2\sin A}{\cos A(1 + \sin A)} = \frac{2 + 2\sin A}{\cos A(1 + \sin A)} \]

\[ = \frac{2(1 + \sin A)}{\cos A(1 + \sin A)} = \frac{2}{\cos A} = 2 \sec A = \text{RHS} \]

Solution:

Convert to sin and cos:

LHS \( = \frac{\sin/\cos}{1 - \cos/\sin} + \frac{\cos/\sin}{1 - \sin/\cos} \)

\[ = \frac{\sin^2 \theta}{\cos \theta(\sin \theta - \cos \theta)} + \frac{\cos^2 \theta}{\sin \theta(\cos \theta - \sin \theta)} \]

\[ = \frac{\sin^2 \theta}{\cos \theta(\sin \theta - \cos \theta)} - \frac{\cos^2 \theta}{\sin \theta(\sin \theta - \cos \theta)} \]

\[ = \frac{\sin^3 \theta - \cos^3 \theta}{\sin \theta \cos \theta(\sin \theta - \cos \theta)} \]

Using \( a^3 - b^3 = (a-b)(a^2 + b^2 + ab) \):

\[ = \frac{(\sin \theta - \cos \theta)(\sin^2 \theta + \cos^2 \theta + \sin \theta \cos \theta)}{\sin \theta \cos \theta(\sin \theta - \cos \theta)} \]

\[ = \frac{1 + \sin \theta \cos \theta}{\sin \theta \cos \theta} = \frac{1}{\sin \theta \cos \theta} + 1 = \sec \theta \text{ cosec } \theta + 1 = \text{RHS} \]

Solution:

LHS \( = \frac{1 + 1/\cos A}{1/\cos A} = \frac{(\cos A + 1)/\cos A}{1/\cos A} = \cos A + 1 \)

RHS \( = \frac{\sin^2 A}{1 - \cos A} = \frac{1 - \cos^2 A}{1 - \cos A} = \frac{(1 - \cos A)(1 + \cos A)}{1 - \cos A} = 1 + \cos A \)

LHS = RHS. Hence Proved.

Solution:

Divide numerator and denominator by \( \sin A \):

\[ \frac{\cot A - 1 + \text{cosec } A}{\cot A + 1 - \text{cosec } A} \]

Replace \( 1 \) in numerator with \( \text{cosec}^2 A - \cot^2 A \):

\[ = \frac{(\cot A + \text{cosec } A) - (\text{cosec}^2 A - \cot^2 A)}{\cot A - \text{cosec } A + 1} \]

\[ = \frac{(\cot A + \text{cosec } A) - (\text{cosec } A - \cot A)(\text{cosec } A + \cot A)}{\cot A - \text{cosec } A + 1} \]

\[ = \frac{(\cot A + \text{cosec } A)(1 - (\text{cosec } A - \cot A))}{\cot A - \text{cosec } A + 1} \]

\[ = \frac{(\cot A + \text{cosec } A)(1 - \text{cosec } A + \cot A)}{1 - \text{cosec } A + \cot A} \]

\[ = \cot A + \text{cosec } A = \text{RHS} \]

Solution:

Multiply numerator and denominator inside the root by \( 1 + \sin A \):

\[ \sqrt{\frac{(1 + \sin A)(1 + \sin A)}{(1 - \sin A)(1 + \sin A)}} = \sqrt{\frac{(1 + \sin A)^2}{1 - \sin^2 A}} \]

\[ = \sqrt{\frac{(1 + \sin A)^2}{\cos^2 A}} = \frac{1 + \sin A}{\cos A} \]

\[ = \frac{1}{\cos A} + \frac{\sin A}{\cos A} = \sec A + \tan A = \text{RHS} \]

Solution:

LHS \( = \frac{\sin \theta(1 - 2\sin^2 \theta)}{\cos \theta(2\cos^2 \theta - 1)} \)

Using \( \cos 2\theta = 1 - 2\sin^2 \theta = 2\cos^2 \theta - 1 \):

\[ = \frac{\sin \theta(\cos 2\theta)}{\cos \theta(\cos 2\theta)} \]

\[ = \frac{\sin \theta}{\cos \theta} = \tan \theta = \text{RHS} \]

Solution:

Expand both squares:

\[ (\sin^2 A + \text{cosec}^2 A + 2\sin A \text{ cosec } A) + (\cos^2 A + \sec^2 A + 2\cos A \sec A) \]

Since \( \sin A \text{ cosec } A = 1 \) and \( \cos A \sec A = 1 \):

\[ = \sin^2 A + \cos^2 A + \text{cosec}^2 A + \sec^2 A + 2 + 2 \]

\[ = 1 + (1 + \cot^2 A) + (1 + \tan^2 A) + 4 \]

\[ = 1 + 1 + 1 + 4 + \tan^2 A + \cot^2 A = 7 + \tan^2 A + \cot^2 A = \text{RHS} \]

Solution:

LHS:

\[ \left(\frac{1}{\sin A} - \sin A\right) \left(\frac{1}{\cos A} - \cos A\right) \]

\[ = \left(\frac{1 - \sin^2 A}{\sin A}\right) \left(\frac{1 - \cos^2 A}{\cos A}\right) = \frac{\cos^2 A}{\sin A} \cdot \frac{\sin^2 A}{\cos A} = \sin A \cos A \]

RHS:

\[ \frac{1}{\frac{\sin A}{\cos A} + \frac{\cos A}{\sin A}} = \frac{1}{\frac{\sin^2 A + \cos^2 A}{\sin A \cos A}} = \frac{1}{\frac{1}{\sin A \cos A}} = \sin A \cos A \]

LHS = RHS. Hence Proved.

Solution:

Part 1:

\[ \frac{1 + \tan^2 A}{1 + \cot^2 A} = \frac{\sec^2 A}{\text{cosec}^2 A} = \frac{1/\cos^2 A}{1/\sin^2 A} = \frac{\sin^2 A}{\cos^2 A} = \tan^2 A \]

Part 2:

\[ \left(\frac{1 - \tan A}{1 - \frac{1}{\tan A}}\right)^2 = \left(\frac{1 - \tan A}{\frac{\tan A - 1}{\tan A}}\right)^2 \]

\[ = \left(\frac{1 - \tan A}{-(1 - \tan A)} \cdot \tan A\right)^2 = (-\tan A)^2 = \tan^2 A \]

Both parts equal \( \tan^2 A \). Hence Proved.