10th Maths 8.4

Chapter 8 Introduction to Trigonometry

NCERT Class 10th solution of Exercise 8.1

NCERT Class 10th solution of Exercise 8.2

NCERT Class 10th solution of Exercise 8.3

NCERT Class 10th Maths Projects

Exercise 8.4

Q1. Express the trigonometric ratios

sin A, sec A

$sin A, sec A$ and

tan A

$tan A$ in terms of

cot A

$cot A$ .

Sol. :

sin A = \frac{1}{cos e c A} = \frac{1}{\sqrt{cos e c^{2} A}} = \frac{1}{\sqrt{1 + {cot}^{2} A}}

$sin A = 1/(cosec A) = 1/(sqrt (cosec^2 A)) = 1/sqrt(1 + cot^2 A)$

sec A = \frac{1}{cos A} = \frac{\frac{1}{sin A}}{\frac{cos A}{sin A}} = \frac{cos e c A}{cot A}

$sec A = 1/(cos A) = (1/sin A)/(cos A/sin A) = (cosec A)/(cot A)$

sec A = \frac{\sqrt{cos e c^{2} A}}{Cot A} = \frac{\sqrt{1 + {cot}^{2} A}}{cot A}

$sec A = sqrt(cosec^2 A)/(Cot A) = sqrt(1 + cot^2 A)/cot A$

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

$tan A = 1/cot A$

Answer:

sin A = \frac{1}{\sqrt{1 + {cot}^{2} A}}

$sin A = 1/sqrt(1 + cot^2 A)$ ,

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

$sec A = sqrt(1 + cot^2 A)/cot A$ and

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

$tan A = 1/cot A$

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Q2. Write all the other trigonometric ratios of

∠ A

$angle A$ in terms of

sec A

$sec A$ .

Sol. :

Let

△ A B C

$triangle ABC$ is right-angled at

∠ B

$angleB$ and

sec A = x

$sec A = x$

sec A = x = \frac{A C}{A B}

$sec A = x = (AC)/(AB)$

By Pythagoras theorem

B C^{2} = A C^{2} - A B^{2}

$BC^2 = AC^2 - AB^2$

B C^{2} = x^{2} - 1

$BC^2 = x^2 - 1$

B C = \sqrt{x^{2} - 1}

$BC = sqrt(x^2 -1)$

sin A = \frac{B C}{A C} = \frac{\sqrt{x^{2} - 1}}{x} = \frac{\sqrt{{sec}^{2} - 1}}{sec A}

$sin A = (BC)/(AC) = sqrt(x^2 - 1)/x = sqrt(sec^2 - 1)/(sec A)$

cos A = \frac{A B}{A C} = \frac{1}{x} = \frac{1}{sec A}

$cos A = (AB)/(AC) = 1/x = 1/sec A$

tan A = \frac{B C}{A B} = \frac{\sqrt{x^{2} - 1}}{1} = \sqrt{{sec}^{2} A - 1}

$tan A = (BC)/(AB) = sqrt(x^2 - 1)/1 = sqrt(sec^2 A - 1)$

cos e c A = \frac{A C}{B C} = \frac{x}{\sqrt{x^{2} - 1}} = \frac{sec A}{\sqrt{sec A - 1}}

$cosec A = (AC)/(BC) = x/sqrt(x^2 - 1) = (sec A)/sqrt(sec A - 1)$

cot A = \frac{A B}{B C} = \frac{1}{\sqrt{x^{2} - 1}} = \frac{1}{\sqrt{{sec}^{2} A - 1}}

$cot A = (AB)/(BC) = 1/sqrt(x^2 - 1) = 1/sqrt(sec^2 A - 1)$

Answer:

sin A = \frac{\sqrt{{sec}^{2} - 1}}{sec A}

$sin A = sqrt(sec^2 - 1)/(sec A)$

cos A = \frac{1}{sec A}

$cos A = 1/sec A$

tan A = \sqrt{{sec}^{2} A - 1}

$tan A = sqrt(sec^2 A - 1)$

cos e c A = \frac{sec A}{\sqrt{sec A - 1}}

$cosec A = (sec A)/sqrt(secA - 1)$

cot A = \frac{1}{\sqrt{{sec}^{2} A - 1}}

$cot A = 1/sqrt(sec^2 A - 1)$

Q3. Evaluate:

\frac{{sin}^{2} 63^{\circ} + {sin}^{2} 27^{\circ}}{{cos}^{2} 17^{\circ} + {cos}^{2} 73^{\circ}}

$(sin^2 63^circ + sin^2 27^circ)/(cos^2 17^circ + cos^2 73^circ)$

Sol. :

= \frac{{sin}^{2} (90^{\circ} - 27^{\circ}) + {sin}^{2} 27^{\circ}}{{cos}^{2} 17^{\circ} + {cos}^{2} (90^{\circ} - 17^{\circ})}

$= (sin^2(90^circ - 27^circ) + sin^2 27^circ)/(cos^2 17^circ + cos^2(90^circ - 17^circ)$

$= (cos^2 27^circ + sin^2 27^circ)/(cos^2 17^circ + sin^2 17^circ)$

$= 1/1$

$= 1$

Answer:

$= 1$

ii)

$sin 25^circ cos 65^circ + cos 25^circ sin 65^circ$

Sol. :

$= sin 25^circ cos(90^circ - 25^circ) + cos 25^circ sin(90^circ - 25^circ)$

$= sin 25^circ sin 25^circ + cos 25^circ cos 25^circ$

$= sin^2 25^circ + cos^2 25^circ$

$= 1$

Answer:

$= 1$

Q4. Choose the correct option. Justify your choice:

$9 sec^2 A - 9 tan^2 A =$

$1$ B)

$9$ C)

$8$ D)

$0$

Sol. :

$= 9 sec^2 A - 9 tan^2 A$

$= 9(sec^2 - tan^2A)$

$= 9 times 1$

$= 9$

Answer:

$B) 9$ .

ii)

$(1 + tan theta + sec theta)(1 + cot theta - cosec theta) =$

$0$ B)

$1$ C)

$2$ D)

$-1$

Sol. :

$= (1 + tan theta + sec theta)(1 + cot theta - cosec theta)$

$= (1 + (sin theta)/(cos theta) + 1/(cos theta)) times (1 + (cos theta)/(sin theta) - 1/(sin theta))$

$= (cos theta + sin theta + 1)/(cos theta) times (sin theta + cos theta - 1)/(sin theta)$

$= ([(sin theta + cos theta) +1][(sin theta + cos theta) - 1])/(sin theta cos theta)$

$= ((sin theta + cos theta)^2 -1^2)/(sin theta cos theta)$

$= (sin^2 theta + cos^2 theta + 2 sin theta cos theta - 1)/(sin theta cos theta)$

$= (1 + 2 sin theta cos theta - 1)/(sin theta cos theta)$

$= (2sin theta cos theta)/(sin theta cos theta)$

$= 2$

Answer:

$D) 2$ .

iii)

$(sec A + tan A)(1 - sin A) =$

$sec A$ B) sin A C)

$cosec A$ D)

$cos A$

Sol. :

$= (sec A + tan A)(1 - sin A)$

$=(1/cos A + sin A/cos A)(1 - sin A)$

$= ((1 + sin A) (1 - sin A))/cos A$

$= (1 - sin^2 A)/cos A$

$= cos^2/cos A$

$= cos A$

Answer:

$D) cos A$

iv)

$(1 + tan^2 A)/(1 + cot^2 A) =$

$sec^2 A$ B)

$- 1$ C)

$cot^2 A$ D)

$tan^2 A$

Sol. :

$= (1+ tan^2 A)/(1 + cot^2 A)$

$= (sec^2 A)/(cosec^2 A)$

$= 1/cos^2 A times sin^2 A/1$

$= (sin A/cos A)^2$

$= tan^2 A$

Answer:

$D) tan^2 A$ .

Q5. Prove the following identities, where the angles involved are acute angles for which the expressions are defined.

$(cosec theta - cot theta)^2 = (1 - cos theta)/(1 + cos theta)$

Sol. :

$LHS.$

$= (cosec theta - cot theta)^2$

$= (1/sin theta - cos theta/sin theta)^2$

$= ((1 - cos theta)/sin theta)^2$

$= (1 - cos theta)^2/sin^2 theta$

$= (1 - cos theta) ^2/(1 - cos^2 theta)$

$= (1 - cos theta)^2/((1 - cos theta)(1 + cos theta))$

$= (1 - cos theta)/(1 + cos theta)$

$= RHS.$

$LHS. = RHS.$

Proved.

ii)

$cos A/(1 + sin A) + (1 + sin A)/cos A = 2 sec A$

Sol. :

$= LHS.$

$= cos A/(1 + sin A) + (1 + sin A)/cos A$

$= (cos^2 A + (1 + sin A)^2)/(cosA(1 + sin A))$

$= (cos^2 A + 1 + sin^2 A + 2 sin A)/(cos A(1 + sin A))$

$= (sin^2 A + cos^2 A + 1 + 2 sin A)/(cos A(1 + sin A))$

$= (1 + 1 + 2 sin A)/(cos A(1 + sin A))$

$= (2 + 2 sin A)/(cos A(1 + sin A))$

$= (2(1 + sin A))/(cos A(1 + sin A))$

$= 2/cos A$

$= 2 sec A$

$= RHS.$

$LHS. = RHS.$

Proved.

iii)

$tan theta/(1 - cot theta) + cot theta/(1 - tan theta) = 1 + sec theta cosec theta$

[Hint: Write the expression in terms of

$sin theta$ and

$cos theta$ ]

Sol. :

$= LHS.$

$= tan theta/(1 - cot theta) + cot theta/(1 - tan theta)$

$= (sin theta/cos theta)/(1 - cos theta/sin theta) + (cos theta/sin theta)/(1 - sin theta/cos theta)$

$= sin^2 theta/(cos theta(sin theta - cos theta)) + cos^2 theta/(sin theta(cos theta - sin theta))$

$= (sin^3 theta - cos^3 theta)/(sin theta cos theta(sin theta - cos theta))$

$= (1 + sin theta cos theta)/(sin theta cos theta)$

$= 1/(sin theta cos theta) + 1$

$= cosec theta sec theta$

$= 1 + sec theta cosec theta$

$= RHS$

$LHS. = RHS.$

Proved.

iv)

$(1 + sec A)/sec A = sin^2 A/(1 - cos A)$ [Hint: Simplify

$LHS$ and

$RHS$ separately]

Sol. :

$= LHS.$

$= (1+ sec A)/sec A$

$= (1 + 1/cos A)/(1/cos A)$

$= ((cos A + 1)/cos A)/(1/cos A)$

$= (cos A + 1)/cos A times (cos A)/1$

$= (cos A + 1)/1$

$= (1+ cos A)/1 times (1 - cos A)/(1 - cos A)$

$= (1 - cos^2 A)/(1 - cos A)$

$= sin^2 A/(1 - cos A)$

$= RHS$

$LHS. = RHS.$

Proved.

$(cos A - sin A + 1)/(cosA + sinA - 1) = cosec A + cot A,$ using the identity

$cosec^2 A = 1 + cot A$

Sol. :

$= LHS.$

$= (cos A - sin A + 1)/(cos A + sin A - 1)$

$= (cot A - 1 + cosec A)/(cot A + 1 - cosec A)$

$= ((cot A - cosec A) - 1)/(cot A - cosec A + 1)$

$= ([(cot A + cosec A) - 1](cot A - cosec A))/([cot A - cosec A + 1](cot A - cosec A))$

$= ((cot^2 A - cosec^2 A) - (cot A - cosec A))/((cot A - cosec A +1)(cot A - cosec A))$

$= (- 1 - cot A + cosec A)/((cot A - cosec A +1)(cot A - cosec A))$

$= (cot A - cosec A + 1)/((cot A - cosec A + 1)(cosec A - cot A))$

$= 1/(cosec A - cot A)$

$= (cosec^2 A - cot^2 A)/(cosec A - cot A)$

$= ((cosec A - cot A)(cosec A + cot A))/(cosec A - cot A)$

$= cosec A + cot A$

$= RHS.$

$LHS. = RHS.$

Proved.

vi)

$sqrt((1 + sin A)/(1 - sin A)) = sec A + tan A$

Sol. :

$= LHS.$

$= sqrt((1 + sin A)/(1 - sin A))$

$= sqrt((1 + sin A)/(1 - sin A)) times sqrt((1+ sin A)/(1 + sin A))$

$= sqrt((1 + sin A)^2/(1 - sin^2 A))$

$= sqrt((1 + sin A)^2/(cos^2 A))$

$= (1 + sin A)/cos A$

$= 1/cos A + sin A/cos A$

$= sec A + tan A$

$= RHS.$

$LHS. = RHS.$

Proved.

vii)

$(sin theta - 2 sin^3 theta)/(2 cos^3 theta - cos theta) = tan theta$

Sol. :

$= LHS.$

$= (sin theta - 2sin^3 theta)/(2 cos^3 theta - cos theta)$

$= (sin theta(1 - 2 sin^2theta))/(cos theta (2 cos^2 theta - 1))$

$= (sin theta(1 - 2 sin^2 theta))/(cos theta[2(1 - sin^2 theta) - 1])$

$= (sin theta (1 - 2 sin^2 theta))/(cos theta(1 - 2 sin^2 theta)$

$= sin theta/cos theta$

$= tan theta$

$= LHS.$

$RHS. = LHS.$

Proved.

viii)

$(sin A + cosec A)^2 + (cos A + sec A)^2 = 7 + tan^2 A + cot^2 A$

Sol. :

$= LHS.$

$= (sin A + cosec A)^2 + (cos A + sec A)^2$

$= sin^2 A + 2sin A cosec A + cosec^2 A + cos^2 A + 2cos A sec A + sec^2 A$

$= sin^2 A + cos^2 A + 2sin A times 1/sin A + cosec^2 A + sec^2 A + 2cos A times 1/cos A$

$= 1 + 2 + cosec^2 A + sec ^2 A + 2$

$= 1 + 2 + 2 + (1 + cot^2 A) + (1 + tan^2 A)$

$= 1 + 2 + 2 + 1 + 1 + cot^2 A + tan^2 A$

$= 7 + tan^2 A + cot^2 A$

$= RHS.$

$LHS. = RHS.$

ix)

$(cosec A - sin A)(sec A - cos A) = 1/(tan A + cot A)$

[Hint: Simplify

$LHS$ and

$RHS$ separately]

Sol. :

$= LHS.$

$= (cosec A - sin A)(sec A - cos A)$

$= (1/sin A - sin A)(1/cos A - cos A)$

$= ((1- sin^2 A)/sin A)((1 - cos^2 A)/cos A)$

$= cos^2 A/sin A sin^2 A/cos A$

$= sin A cos A$

$= RHS.$

$= 1/(tan A + cot A)$

$= 1/(sin A/cos A + cos A/sin A)$

$= 1/((sin^2 A + cos^2 A)/(cos A sin A))$

$= 1/(1/(cosAsin A))$

$= sin A cos A$

$RHS. = LHS.$

Proved.

$((1 + tan^2 A)/(1 + cot^2 A)) = ((1 - tan A)/(1 - cot A))^2 = tan^2 A$

Sol. :

$= ((1 + tan^2 A)/(1 + cot^2 A))$

$= \frac{sec^2 A}{cosec^2 A}$

$= 1/cos^2 A times sin^2 A/1$

$= sin^2 A/cos^2 A$

$= tan^2 A$

$= ((1 - tan A)/(1 - cot A))^2$

$= ((1 - sin A/cos A)/(1 - cos A/sin A))^2$

$= [((cos A - sin A)/cos A)/((sin A - cos A)/sin A)]^2$

$= [(-((sin A - cos A))/cos A)/((sin A - cos A)/sin A)]^2$

$= (- sin A/cos A)^2$

$= (- tan A)^2$

$= tan^2$

$LHS. = RHS.$

Proved.

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10th Maths 8.4

Chapter 8

Introduction to Trigonometry

NCERT Class 10th solution of Exercise 8.1

NCERT Class 10th solution of Exercise 8.2

NCERT Class 10th solution of Exercise 8.3

NCERT Class 10th Maths Projects

Exercise 8.4

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