Practice Model questions set 1 - quantitative aptitude Online Quiz (set-1) For All Competitive Exams
Q-1) If sin θ + cos θ = 1, then what is the value of sin θ cos θ ?
(a)
(b)
(c)
(d)
Given, $\text"sin θ + cos θ"$ = 1
Squaring both sides,
$(sin^2 θ + cos^2 θ) + 2 \text"sin θ cos θ" = 1$
⇒ 1 + 2 sin θ cos θ = 1 ⇒ sin θ cos θ = 0.
Q-2) If tan x = 1, 0 < x < 90°, then what is the value of 2 sin x cos x ?
(a)
(b)
(c)
(d)
tan x = 1 = tan 45°
∴ x = 45°
2 sin x. cos x = 2 sin (45°) . cos (45°)
= 2 × $1/{√2} × 1/{√2}$ = 1
Q-3) $({sin 35°}/{cos 55°})^2 - ({cos 55°}/{sin 35°})^2$ + 2sin 30° is equal to
(a)
(b)
(c)
(d)
$({sin 35°}/{cos 55°})^2 - ({cos 55°}/{sin 35°})^2$ + 2 sin 30° ....(1)
We know that
sin $({π}/2 - θ)$ = cos θ
⇒ sin (90 - 55°) = cos 55°
⇒ sin 35° = cos 55°
So from (1) we get
$({sin 35°}/{sin 35°})^2 - ({cos 55°}/{cos 55°})^2 + 2 × 1/2$
= $(1)^2 - (1)^2 + 1$
∴ Option (c) is correct.
Q-4) What is sin 25° sin 35° sec 65° sec 55° equal to?
(a)
(b)
(c)
(d)
sin 25° sin 35° sec 65° sec 55°
= sin 25° . sin 35° . $1/{cos 65°} . 1/{cos 55°}$
= sin 25° . sin 35° . $1/{cos (90 - 25°)} . 1/{cos (90 - 35°)}$
= sin 25° . sin 35° . $1/{sin 25°} . 1/{sin 35°}$ = 1
Q-5) If $sin^2$ x + sin x = 1, then what is the value of $cos^{12} x + 3 cosx^{10} x + 3 cos^8 x + cosx^6$ x?
(a)
(b)
(c)
(d)
We have,
$sin^2x$ + sin x = 1...(1)
∴ sin x = 1 - $sin^2 x = cos^2 x$
On cubing equation (1), we get
$(sin^2 x + sin x)^3 = {1}^3$
$sin^6 x + sin^3 x + 3 sin^2 x. sin x (sin^2 \text"x + sin x")$ = 1
$sin^6x + sin^3 x + 3 sin^5x + 3 sin^4x$ = 1
∴ $cos^{12}x + 3 cos^{10}x + 3 cos^8 x + co^6 x$ = 1
Q-6) If $cos^2$ x + cosx = 1, then what is the value of $sin^{12}x + 3sin^{l0}x + 3sin^8 x +sin^6$ x
(a)
(b)
(c)
(d)
$cos^2 \text"x + cos x = 1"$
⇒ $\text"cos x = 1" - cos^2 x = sin^2 x$
= $sin^{12}x + 3 sin^{10}x + 3 sin^8x + sin^6x$
= $sin^6x[sin^6x + 3 sin^4x + 3 sin^2 x + 1]$
= $sin^6x[sin^2 x + 1]^3$
= $[sin^4x + sin^2 x]^3$
$(∴ sin^4 x = cos^2 x)$
= $(sin^2 x + cos^2 x)$ = 1
Q-7) What is the value of tan1° tan2° tan3° tan4° ... tan89° ?
(a)
(b)
(c)
(d)
tan1° tan2° tan3° tan4°.......tan 89°
tan1° tan2° ......tan45°......tan 89°
= 1
Q-8) If $a^2 = {1 + 2 sin θ cos θ}/{1 - 2 sin θ cos θ}$, then what is the value of ${a + 1}/{a - 1}$ ?
(a)
(b)
(c)
(d)
Given that:
$a^2 = {\text"1 + 2 sin θ cos θ"}/{\text"1 - 2 sin θ cos θ"}$
⇒ $a^2 = {(sin^2 θ + cos^2 θ) + \text"2 sin θ . cos θ"}/{(sin^2 θ + cos^2 θ) - \text"2 sin θ . cos θ"}$
⇒ $a^2 = {(\text"sin θ + cos θ")^2}/{(\text"sin θ - cos θ")^2} ⇒ a/1 = {\text"sin θ + cos θ"}/{\text"sin θ - cos θ"}$
(applying componendo dividendo formula)
⇒ ${a + 1}/{a - 1} = {(\text"sin θ + cos θ") + (\text"sin θ - cos θ")}/{(\text"sin θ + cos θ") - (\text"sin θ - cos θ")}$
⇒ ${a + 1}/{a - 1} = {\text"2 sin θ"}/{\text"2 cos θ"}$ = tan θ
Q-9) What is the least value of (25 $cosec^2 x + sec^2$ x) ?
(a)
(b)
(c)
(d)
Min Value = $(√{25} + 1)^2$ = 36
Q-10) If 5 sin θ + 12 cos θ = 13, then what is 5 cos θ – 12 sin θ equal to?
(a)
(b)
(c)
(d)
∵ 5 sin θ + 12 cos θ = 12
Now, squaring both sides, we get
$25 sin^2 θ + 144 cos^2 θ + \text"120 sin θ cos θ"$ = 169
⇒ 25$(1 - cos^2 θ) + 144(1 - sin^2 θ) + \text"120 sin θ cos θ"$ = 169
⇒ 25 - 25 $ cos^2 θ + 144 - 144 sin^2 θ + \text"120 sin θ cos θ"$ - 169
⇒ 25 $cos^2 θ + 144 sin^2 θ - \text"120 sin θ cos θ" = 169 - 169$
⇒ $(\text"5 cos θ - 12 sin θ")^2$ = 0
∴ 5 cos θ - 12 sin θ = 0