Free Introduction to Trigonometry 02 Practice Test - 10th Grade

Question 1

$If c o s 3 θ = \frac{\sqrt{3}}{2}$ , 0° < $3 θ$ < 90°, then find the value of $θ$ .

15°

10°

0°

12°

SOLUTION

Solution : B

Given: $cos 3 θ = \frac{\sqrt{3}}{2}$
We know that $cos 30^{\circ} = \frac{\sqrt{3}}{2}$
Comparing the two we get,
$3 θ = 30^{\circ}$ .... (given 0 $^{\circ}$ < $3 θ$ < 90 $^{\circ}$ )
$\Rightarrow θ = 10^{\circ}$

Question 2

$\frac{s i n 18^{\circ}}{c o s 72^{\circ}} =$

SOLUTION

Solution : $\frac{sin 18^{\circ}}{cos 72^{\circ}}$
$= \frac{sin (90^{\circ} - 72^{\circ})}{cos 72^{\circ}}$
$= \frac{cos 72^{\circ}}{cos 72^{\circ}} = 1$ as $s i n (90^{\circ} - A) = c o s A$

Question 3

$(1 + tan θ + sec θ) (1 + cot θ - c o s e c θ) =$

A. 0

B. 1

C. 2

D. -1

SOLUTION

Solution : C

$(1 + tan θ + sec θ) (1 + cot θ - c o s e c θ)$
$= (1 + \frac{sin θ}{cos θ} + \frac{1}{cos θ}) (1 + \frac{cos θ}{sin θ} - \frac{1}{sin θ})$
$= \frac{(cos θ + sin θ + 1)}{cos θ} \times \frac{(sin θ + cos θ - 1)}{sin θ}$
$= \frac{(cos θ + sin θ)^{2} - 1^{2}}{cos θ sin θ}$
$= \frac{({cos}^{2} θ + {sin}^{2} θ + 2 cos θ sin θ - 1)}{cos θ sin θ}$
$= \frac{(1 + 2 cos θ sin θ - 1)}{cos θ sin θ}$
$= \frac{2 cos θ sin θ}{cos θ sin θ} = 2$

Question 4

$\frac{(1 + t a n A t a n B)^{2} + (t a n A - t a n B)^{2}}{{s e c}^{2} A {s e c}^{2} B} =$ ___

$1$

$-tan A$

$2$

$cot A$

SOLUTION

Solution : A

$\frac{(1 + tan A tan B)^{2} + (tan A-tan B)^{2}}{{sec}^{2} A {sec}^{2} B}$

$= \frac{1 + 2 tan A tan B + {tan}^{2} A {tan}^{2} B + {tan}^{2} A - 2 tan A tan B + {tan}^{2} B}{{sec}^{2} A {sec}^{2} B}$

$= \frac{1 + {tan}^{2} A {tan}^{2} B + {tan}^{2} A + {tan}^{2} B}{{sec}^{2} A {sec}^{2} B}$

$= \frac{1 + {tan}^{2} A + {tan}^{2} B + {tan}^{2} A {tan}^{2} B}{{sec}^{2} A {sec}^{2} B}$

$= \frac{1 (1 + {tan}^{2} A) + {tan}^{2} B (1 + {tan}^{2} A)}{{sec}^{2} A {sec}^{2} B}$

$= \frac{(1 + {tan}^{2} A) (1 + {tan}^{2} B)}{{sec}^{2} A {sec}^{2} B}$

$= \frac{({sec}^{2} A) ({sec}^{2} B)}{{sec}^{2} A {sec}^{2} B} . \dots (1 + {tan}^{2} θ = {sec}^{2} θ)$

$= 1$

Question 5

Which of the following trigonometric expressions is equal to ${sec}^{6} θ$ ?

${tan}^{6} θ + 3 {tan}^{2} θ {sec}^{2} θ + 1$

${tan}^{6} θ - 3 {tan}^{2} θ {sec}^{2} θ + 1$

${tan}^{6} θ - 1$

${tan}^{6} θ + 1$

SOLUTION

Solution : A

${sec}^{6} θ = ({sec}^{2} θ)^{3}$

$= ({tan}^{2} θ + 1)^{3} \dots \dots (1 + {tan}^{2} θ = {sec}^{2} θ)$

$= {tan}^{6} θ + 3 {tan}^{4} θ + 3 {tan}^{2} θ + 1$

$= {tan}^{6} θ + 3 {tan}^{2} θ ({tan}^{2} θ) + 3 {tan}^{2} θ + 1$

$= {tan}^{6} θ + 3 {tan}^{2} θ ({sec}^{2} θ - 1) + 3 {tan}^{2} θ + 1$

$= {tan}^{6} θ + 3 {tan}^{2} θ {sec}^{2} θ - 3 {tan}^{2} θ + 3 {tan}^{2} θ + 1$

$= {tan}^{6} θ + 3 {tan}^{2} θ {sec}^{2} θ + 1$

Question 6

Which of the following options are equal to $\frac{tan A}{sec A - 1} + \frac{tan A}{sec A + 1}$ ?

$2 c o s e c A$

$\frac{1}{sin A}$

$\frac{2}{sin A}$

$2$

SOLUTION

Solution : A and C

$\frac{tan A}{sec A - 1} + \frac{tan A}{sec A + 1}$

$= tan A (\frac{1}{sec A - 1} + \frac{1}{sec A + 1})$

$= tan A (\frac{sec A + 1 + sec A - 1}{{sec}^{2} A - 1})$

$= t a n A (\frac{2 sec A}{{tan}^{2} A}) . . . (1 + {tan}^{2} A = {sec}^{2} A)$

$= \frac{2 sec A}{tan A}$

$= \frac{2}{cos A (\frac{sin A}{cos A})} . . . . (tan A = \frac{sin A}{cos A})$

$= \frac{2}{sin A} = 2 c o s e c A$ $. . . (\frac{1}{sin A} = c o s e c A)$

Question 7

Which of the following options are equal to $\sqrt{\frac{1 + c o s A}{1 - c o s A}} ?$

$\frac{1 - c o s A}{s i n A}$

$\frac{1 + c o s A}{s i n A}$

$c o s e c A + c o t A$

$s e c A + t a n A$

SOLUTION

Solution : B and C

$\sqrt{\frac{1 + c o s A}{1 - c o s A}} = \sqrt{(\frac{1 + c o s A}{1 - c o s A}) (\frac{1 + c o s A}{1 + c o s A})}$

$= \sqrt{\frac{(1 + c o s A)^{2}}{1 - c o s^{2} A}}$

$= \sqrt{\frac{(1 + c o s A)^{2}}{s i n^{2} A}} . . . . (s i n^{2} A + c o s^{2} A = 1)$

$= \frac{1 + c o s A}{s i n A}$

$= \frac{1}{s i n A} + \frac{c o s A}{s i n A}$

$= c o s e c A + c o t A$

Question 8

If $x = a sec A cos B$ , $y = b sec A sin B$ and $z = c tan A$ , then $\frac{x^{2}}{a^{2}} + \frac{y^{2}}{b^{2}} - \frac{z^{2}}{c^{2}} =$

SOLUTION

Solution : D

$x = a sec A cos B \Rightarrow \frac{x}{a} = sec A cos B$

$y = b sec A sin B \Rightarrow \frac{y}{b} = sec A sin B$

$z = c tan A \Rightarrow \frac{z}{c} = tan A$

Squaring each of the equations we get,

$\frac{x^{2}}{a^{2}} = {sec}^{2} A {cos}^{2} B, \frac{y^{2}}{b^{2}} = {sec}^{2} A {sin}^{2} B, \frac{z^{2}}{c^{2}} = {tan}^{2} A$

$∴ \frac{x^{2}}{a^{2}} + \frac{y^{2}}{b^{2}} - \frac{z^{2}}{c^{2}} = {sec}^{2} A {cos}^{2} B + {sec}^{2} A {sin}^{2} B - {tan}^{2} A$

$= {sec}^{2} A ({cos}^{2} B + {sin}^{2} B) - {tan}^{2} A$

$= {sec}^{2} A - {tan}^{2} A \dots \dots ({cos}^{2} B + {sin}^{2} B = 1)$

$= 1 \dots \dots ({sec}^{2} A - {tan}^{2} A = 1)$

Question 9

If $x = a c o s e c θ$ and $y = b c o t θ$ , then which of the following equations is true ?

$x^{2} - y^{2} = a^{2} - b^{2}$

$\frac{x^{2}}{a^{2}} - \frac{y^{2}}{b^{2}} = 1$

$x^{2} + y^{2} = a^{2} + b^{2}$

$\frac{x^{2}}{a^{2}} + \frac{y^{2}}{b^{2}} = 1$

SOLUTION

Solution : B

$x = a c o s e c θ \Rightarrow \frac{x}{a} = c o s e c θ$ .....(1)

$y = b c o t θ \Rightarrow \frac{y}{b} = c o t θ$ .....(2)

Squaring the equations and subtracting (2) from (1) we get,

$\frac{x^{2}}{a^{2}} - \frac{y^{2}}{b^{2}} = c o s e c^{2} θ - c o t^{2} θ$

$∴ \frac{x^{2}}{a^{2}} - \frac{y^{2}}{b^{2}} = 1.... (1 + c o t^{2} θ = c o s e c^{2} θ)$

Question 10

Which of the following options is equal to $\frac{c o t θ + c o s e c θ - 1}{c o t θ - c o s e c θ + 1}$

$1$

$\frac{s i n θ}{1 - c o s θ}$

$\frac{1 + c o s θ}{s i n θ}$

$cot θ + c o s e c θ$

SOLUTION

Solution : B, C, and D

$= \frac{cot θ + c o s e c θ - 1}{cot θ - c o s e c θ + 1}$

$= \frac{c o t θ + c o s e c θ - (c o s e c^{2} θ - c o t^{2} θ)}{c o t θ - c o s e c θ + 1}$

$= \frac{(c o t θ + c o s e c θ) - (c o s e c θ - c o t θ) (c o s e c θ + c o t θ)}{c o t θ - c o s e c θ + 1}$

$= \frac{(c o t θ + c o s e c θ) (1 - c o s e c θ + c o t θ)}{(c o t θ - c o s e c θ + 1)}$

$= c o t θ + c o s e c θ$

$= \frac{c o s θ}{s i n θ} + \frac{1}{s i n θ}$

$= \frac{1 + c o s θ}{s i n θ}$

$= \frac{(1 + c o s θ) (1 - c o s θ)}{s i n θ (1 - c o s θ)}$ ....[Multiplying numerator and denominator by $1 - cos θ$ ]

$= \frac{1 - c o s^{2} θ}{s i n θ (1 - c o s θ)}$

$= \frac{s i n^{2} θ}{s i n θ (1 - c o s θ)}$

$= \frac{s i n θ}{1 - c o s θ}$

Free Introduction to Trigonometry 02 Practice Test - 10th Grade

Question 1

SOLUTION

Question 2

SOLUTION

Question 3

SOLUTION

Question 4

SOLUTION

Question 5

SOLUTION

Question 6

SOLUTION

Question 7

SOLUTION

Question 8

SOLUTION

Question 9

SOLUTION

Question 10

SOLUTION

Related ExamsArithmetic Progressions 02Real Numbers 02Statistics 03

Related Exams
Arithmetic Progressions 02
Real Numbers 02
Statistics 03