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

10th Grade > Mathematics

INTRODUCTION TO TRIGONOMETRY MCQs

Total Questions : 52 | Page 3 of 6 pages

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

Discuss Question

\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 22.

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

0
1
2
-1

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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 23.

cos $1^{\circ}$ $\times$ cos $2^{\circ}$ $\times$ cos $3^{\circ}$ $\times$ …….. $\times$ cos $180^{\circ}$ is equal to:

1
0
$\frac{1}{2}$
-1

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Answer: Option B. -> 0
:
B

We know that, $c o s 90^{\circ} = 0$
The given expression
$c o s 1^{\circ}$ $\times$ $c o s 2^{\circ}$ $\times$ $c o s 3^{\circ}$ $\times$ .... $\times$ $c o s 90^{\circ}$ $\times$ …….. $\times$ $c o s 180^{\circ}$ reduces to zero as it contains $c o s 90^{\circ}$ which is equal to 0.

Question 24.

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

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Answer: Option B. -> 0
:
B

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

= $\frac{(s i n A - s i n B) (s i n A + s i n B) + (c o s A - c o s B) (c o s A + c o s B)}{(c o s A + c o s B) (s i n A + s i n B)}$
= $\frac{s i n^{2} A - s i n^{2} B + c o s^{2} A - c o s^{2} B}{(c o s A + c o s B) (s i n A + s i n B)}$
= $\frac{s i n^{2} A + c o s^{2} A - (s i n^{2} B ‘ + c o s^{2} B)}{(c o s A + c o s B) (s i n A + s i n B)}$
= $\frac{1 - 1}{(c o s A + c o s B) (s i n A + s i n B)}$
= 0

Question 25.

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

$1 + cosec A sec A$
$1 + tan A + cot A$
$cosec A sec A$
$tan A + cot A$

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Answer: Option A. ->

1 + cosec A sec A

:
A and B

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

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

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

$= \frac{{tan}^{2} A}{tan A - 1} - \frac{1}{tan A (tan A - 1)} \dots \dots (a - b = - (b - a))$

$= \frac{{tan}^{3} A - 1}{tan A (tan A - 1)}$

$= \frac{(/ tan A - 1) ({tan}^{2} A + tan A + 1)}{tan A (/ tan A - 1)} \dots \dots [a^{3} - b^{3} = (a - b) (a^{2} + a b + b^{2})]$

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

$= tan A + 1 + cot A$

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

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

$= 1 + \frac{1}{sin A cos A}$

$= 1 + cosec A sec A$

Question 26.

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

0
$tan A$
1
2

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Answer: Option D. -> 2
:
D

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

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

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

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

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

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

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

$= 2$

Question 27.

If $x = a sec θ + b tan θ$ and $y = a tan θ + b sec θ$ , then $\frac{x^{2} - y^{2}}{a^{2} - b^{2}} =$ ___

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Answer: Option B. -> 1
:
B

$x = a sec θ + b tan θ, y = a tan θ + b sec θ$
Squaring both sides we get,
$x^{2} = (a sec θ + b tan θ)^{2}, y^{2} = (a tan θ + b sec θ)^{2}$
$\begin{matrix} ∴ & x^{2} = a^{2} {sec}^{2} θ + 2 a b sec θ tan θ + b^{2} {tan}^{2} θ y^{2} = a^{2} {tan}^{2} θ + 2 a b sec θ tan θ + b^{2} {sec}^{2} θ - - - - ------------------------------------------- - Subtracting, & x^{2} - y^{2} = a^{2} ({sec}^{2} θ - {tan}^{2} θ) + b^{2} ({tan}^{2} θ - {sec}^{2} θ) \end{matrix}$
$x^{2} - y^{2} = a^{2} ({sec}^{2} θ - {tan}^{2} θ) - b^{2} {(sec}^{2} θ - {tan}^{2} θ)$
$x^{2} - y^{2} = a^{2} - b^{2} \dots \dots ({sec}^{2} θ - {tan}^{2} θ = 1)$
$∴ \frac{x^{2} - y^{2}}{a^{2} - b^{2}} = 1$

Question 28.

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

$\frac{1 + sin θ}{1 - sin θ}$
$1$
$tan θ$
$\frac{1 + cos θ}{1 - cos θ}$

Discuss Question

Answer: Option D. ->

\frac{1 + cos θ}{1 - cos θ}

:
D

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

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

$= \frac{\frac{1}{cos θ} + 1}{\frac{1}{cos θ} - 1}$

$= \frac{\frac{1 + cos θ}{cos θ}}{\frac{1 - cos θ}{cos θ}}$

$= \frac{1 + cos θ}{1 - cos θ}$

Question 29.

$\frac{sec 17^{\circ}}{cosec 73^{\circ}} + \frac{tan 68^{\circ}}{cot 22^{\circ}} - [{cos}^{2} 44^{\circ} + {cos}^{2} 46^{\circ}] =$ ___.

0
-1
2
1

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Answer: Option D. -> 1
:
D

$\frac{sec 17^{\circ}}{cosec 73^{\circ}} + \frac{tan 68^{\circ}}{cot 22^{\circ}} - [{cos}^{2} 44^{\circ} + {cos}^{2} 46^{\circ}]$
$= \frac{sec 17^{\circ}}{cosec (90^{\circ} - 17^{\circ})} + \frac{tan 68^{\circ}}{cot (90^{\circ} - 68^{\circ})} - [{cos}^{2} 44^{\circ} + {cos}^{2} (90^{\circ} - 44^{\circ})]$
$= \frac{sec 17^{\circ}}{sec 17^{\circ}} + \frac{tan 68^{\circ}}{tan 68^{\circ}} - [{cos}^{2} 44^{\circ} + {sin}^{2} 44^{\circ}]$ ....(complementary angles)
$= 1 + 1 - 1$
$= 1$

Question 30.

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

$2$
$1$
$2 {sec}^{2} A$
$2 {cos}^{2} A$

Discuss Question

Answer: Option C. ->

2 {sec}^{2} A

:
C

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