Introduction To Trigonometry(10th Grade > Mathematics ) Questions and answers for exam Preparation

10th Grade > Mathematics

INTRODUCTION TO TRIGONOMETRY MCQs

Total Questions : 52 | Page 5 of 6 pages

$\frac{1 + s i n (90^{\circ} - θ) - {c o s}^{2} (90^{\circ} - θ)}{c o s (90^{\circ} - θ) [1 + s i n (90^{\circ} - θ)]} =$

$cot θ$
$tan θ$
1
0

Discuss Question

Answer: Option A. ->

cot θ

:
A

$\frac{1 + sin (90^{\circ} - θ) - {cos}^{2} (90^{\circ} - θ)}{cos (90^{\circ} - θ) [1 + sin (90^{\circ} - θ)]}$

$= \frac{1 + cos θ - {sin}^{2} θ}{sin θ (1 + cos θ)}$

$= \frac{(1 - {sin}^{2} θ) + cos θ}{sin θ (1 + cos θ)}$

$= \frac{{cos}^{2} θ + cos θ}{sin θ (1 + cos θ)}$

$= \frac{cos θ (cos θ + 1)}{sin θ (1 + cos θ)} = cot θ$

Question 42.

Which of the following options is equal to $\sqrt{{sec}^{2} θ + c o s e c^{2} θ}$ ?

$tan θ + cot θ$
$1$
$c o s e c θ + sec θ$
$cosec θ sec θ$

Discuss Question

Answer: Option A. ->

tan θ + cot θ

:
A and D

$\sqrt{{sec}^{2} θ + {cosec}^{2} θ} = \sqrt{1 + {tan}^{2} θ + 1 + {cot}^{2} θ}$

$= \sqrt{{tan}^{2} θ + 2 + {cot}^{2} θ}$

$= \sqrt{{tan}^{2} θ + 2 (tan θ) (cot θ) + {cot}^{2} θ} \dots \dots (tan θ cot θ = 1)$

$= \sqrt{(tan θ + cot θ)^{2}}$

$= tan θ + cot θ$

$= \frac{sin θ}{cos θ} + \frac{cos θ}{sin θ}$

$= \frac{{sin}^{2} θ + {cos}^{2} θ}{sin θ cos θ}$

$= \frac{1}{s i n θ c o s θ}$

$= cosec θ sec θ$

Question 43.

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

15°
10°
0°
12°

Discuss Question

Answer: Option B. -> 10°
:
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 44.

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

Discuss Question

Answer: Option B. -> 10°
:

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

= \frac{sin (90^{\circ} - 72^{\circ})}{cos 72^{\circ}}

= \frac{cos 72^{\circ}}{cos 72^{\circ}} = 1

s i n (90^{\circ} - A) = c o s A

Question 45.

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

0
1
2
-1

Discuss Question

Answer: Option C. -> 2
:
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 46.

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

$1$
$-tan A$
$2$
$cot A$

Discuss Question

Answer: Option A. ->

1

:
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 47.

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

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

Discuss Question

Answer: Option A. ->

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

:
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 48.

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

$2 c o s e c A$
$\frac{1}{sin A}$
$\frac{2}{sin A}$
$2$

Discuss Question

Answer: Option A. ->

2 c o s e c A

:
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 49.

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

$\frac{1 - c o s A}{s i n A}$
$\frac{1 + c o s A}{s i n A}$
$c o s e c A + c o t A$
$s e c A + t a n A$

Discuss Question

Answer: Option B. ->

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

:
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 50.

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

Discuss Question

Answer: Option D. -> 1
:
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,
$\frac{x^{2}}{a^{2}} = {sec}^{2} A {cos}^{2} B, \frac{y^{2}}{b^{2}} = {sec}^{2} A {sin}^{2} B, \frac{z^{2}}{c^{2}} = {tan}^{2} A$
$∴ \frac{x^{2}}{a^{2}} + \frac{y^{2}}{b^{2}} - \frac{z^{2}}{c^{2}} = {sec}^{2} A {cos}^{2} B + {sec}^{2} A {sin}^{2} B - {tan}^{2} A$
$= {sec}^{2} A ({cos}^{2} B + {sin}^{2} B) - {tan}^{2} A$
$= {sec}^{2} A - {tan}^{2} A \dots \dots ({cos}^{2} B + {sin}^{2} B = 1)$
$= 1 \dots \dots ({sec}^{2} A - {tan}^{2} A = 1)$