10th Maths 8.3

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 8.4

NCERT Class 10th Maths Projects

Exercise 8.3

Q1. Evaluate:

\frac{{sin 18}^{\circ}}{{cos 72}^{\circ}}

$sin 18^circ/cos 72^circ$

Sol. :

= \frac{{sin 18}^{\circ}}{{cos 72}^{\circ}}

$= sin 18^circ/cos 72^circ$

= \frac{{sin 18}^{\circ}}{cos (90^{\circ} - 18^{\circ})}

$= sin 18^circ/cos(90^circ - 18^circ)$

= \frac{{sin 18}^{\circ}}{{sin 18}^{\circ}}

$= sin 18^circ/sin 18^circ$

= 1

$= 1$

Answer:

= 1

$= 1$ .

ii)

\frac{{tan 26}^{\circ}}{{cot 64}^{\circ}}

$tan 26^circ/cot 64^circ$

Sol. :

= \frac{{tan 26}^{\circ}}{{cot 64}^{\circ}}

$= tan 26^circ/cot 64^circ$

= \frac{{tan 26}^{\circ}}{cot (90^{\circ} - 26^{\circ})}

$= tan 26^circ/cot(90^circ - 26^circ)$

= \frac{{tan 26}^{\circ}}{{tan 26}^{\circ}}

$= tan 26^circ/tan 26^circ$

= 1

$= 1$

Answer:

= 1

$= 1$

iii)

{cos 48}^{\circ} - {sin 42}^{\circ}

$cos 48^circ - sin 42^circ$

Sol. :

= {cos 48}^{\circ} - {sin 42}^{\circ}

$= cos 48^circ - sin 42^circ$

= cos (90^{\circ} - 42^{\circ}) - {sin 42}^{\circ}

$= cos(90^circ - 42^circ) - sin 42^circ$

= {sin 42}^{\circ} - {sin 42}^{\circ}

$= sin 42^circ - sin 42^circ$

= 0

$= 0$

Answer:

= 0

$= 0$

iv)

cos e c 31^{\circ} - {sec 59}^{\circ}

$cosec 31^circ - sec 59^circ$

Sol. :

= cos e c 31^{\circ} - {sec 59}^{\circ}

$= cosec 31^circ - sec 59^circ$

= cos e c (90^{\circ} - 59^{\circ}) - {sec 59}^{\circ}

$= cosec(90^circ - 59^circ) - sec 59^circ$

= {sec 59}^{\circ} - {sec 59}^{\circ}

$= sec 59^circ - sec 59^circ$

= 0

$= 0$

Answer:

= 0

$= 0$

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Q2. Show that:

{tan 48}^{\circ} {tan 23}^{\circ} {tan 42}^{\circ} {tan 67}^{\circ} = 1

$tan 48^circ tan23^circ tan42^circ tan 67^circ = 1$

Sol. :

L . H . S .

$L.H.S.$

= {tan 48}^{\circ} {tan 23}^{\circ} {tan 67}^{\circ}

$= tan 48^circ tan 23^circ tan 67^circ$

= tan (90^{\circ} - 42^{\circ}) {tan 23}^{\circ} {tan 42}^{\circ} tan (90^{\circ} - 23^{\circ})

$= tan(90^circ - 42^circ) tan 23^circ tan 42^circ tan(90^circ - 23^circ)$

= {cot 42}^{\circ} {tan 23}^{\circ} {tan 42}^{\circ} {cot 23}^{\circ}

$= cot 42^circ tan 23^circ tan 42^circ cot 23^circ$

= (\frac{{cos 42}^{\circ}}{{sin 42}^{\circ}}) \times (\frac{{sin 23}^{\circ}}{{cos 23}^{\circ}}) \times (\frac{{cos 23}^{\circ}}{{sin 23}^{\circ}})

$= (cos 42^circ/sin 42^circ) times (sin 23^circ/cos 23^circ) times (cos 23^circ/sin 23^circ)$

= 1

$= 1$

= R . H . S .

$= R.H.S.$

L . H . S . = R . H . S .

$L.H.S. = R.H.S.$

Proved.

ii)

{cos 38}^{\circ} {cos 52}^{\circ} - {sin 38}^{\circ} {sin 52}^{\circ} = 0

$cos 38^circ cos 52^circ - sin 38^circ sin 52^circ =0$

Sol. :

L . H . S .

$L.H.S.$

= {cos 38}^{\circ} {cos 52}^{\circ} - {sin 38}^{\circ} {sin 52}^{\circ}

$= cos 38^circ cos 52^circ - sin 38^circ sin 52^circ$

= {cos 38}^{\circ} cos (90^{\circ} - 38^{\circ})

$= cos 38^circ cos(90^circ - 38^circ)$

- {sin 38}^{\circ} sin (90^{\circ} - 38^{\circ})

$- sin 38^circ sin(90^circ - 38^circ)$

= {sin 38}^{\circ} {cos 38}^{\circ} - {sin 38}^{\circ} {cos 38}^{\circ}

$= sin 38^circ cos38^circ - sin 38^circ cos 38^circ$

= 0

$= 0$

Answer:

= 0

$= 0$

Q3. If

tan 2 A = cot (A - 18^{\circ}),

$tan 2A = cot(A - 18^circ),$ where

$2A$ is an acute, angle find the value of

$A.$

Sol. :

$tan 2A = cot(A - 18^circ)$

$cot(90^circ - 2A) = cot(A - 18^circ)$

$90^circ - 2A = A - 18^circ$

$90^circ + 18^circ =A + 2A$

$108^circ = 3A$

$A = 108^circ/3$

$A = 36^circ$

Answer:

$A = 36^circ$ .

Q4. If

$tan A = cot B,$ prove that

$A + B = 90^circ.$

Sol. :

$tan A = cot B$

$tan A = tan(90^circ - B)$

$A = 90^circ - B$

$A + B = 90^circ$

Proved.

Q5. If

$sec 4A = cosec (A - 20^circ),$ where

$4A$ is an acute angle, find the value of

$A.$

Sol. :

$sec 4A = cosec(A - 20^circ)$

$cosec(90^circ - 4A) = cosec(A - 20^circ)$

$90^circ - 4A = A - 20^circ$

$4A + A = 90^circ + 20^circ$

$5A = 110^circ$

$A = 110^circ/5$

$A = 22^circ$

Answer:

$A = 22^circ$

Q6. If

$A, B$ and

$C$ are interior angles of a triangle

$ABC,$ then show that

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

Sol. :

Sum of interior angles of triangles

$ABC$

$A + B + C = 180^circ$

$B + C = 180^circ - A$

Divide by

$2$ both sides

$(B + C)/2 = (180^circ - A)/2$

$(B + C)/2 = (90^circ - A/2)$

$L.H.S.$

$= sin((B + C)/2)$

$= sin(90^circ - A/2)$

$= cos A/2$

$= R.H.S.$

$L.H.S. = R.H.S.$

Proved.

Q7. Express

$sin 67^circ + cos 75^circ$ in terms of trigonometric ratios of angles between

$0^circ$ and

$45^circ.$

Sol. :

$= sin 67^circ + cos 75^circ$

$= sin(90^circ - 23^circ) + cos(90^circ - 15^circ)$

$= cos 23^circ + sin 15^circ$

Answer:

$= cos 23^circ + sin 15^circ$ .

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10th Maths 8.3

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 8.4

NCERT Class 10th Maths Projects

Exercise 8.3

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