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

10th Grade > Mathematics

INTRODUCTION TO TRIGONOMETRY MCQs

Total Questions : 52 | Page 6 of 6 pages

Question 51.

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

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

Discuss Question

Answer: Option B. ->

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

:
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 52.

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

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

Discuss Question

Answer: Option B. ->

\frac{s i n θ}{1 - c o s θ}

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