Practice Trigonometric ratios and identities - 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
Q-11) If tan (x + 40)° tan (x + 20)° tan (3x)° tan (70 – x)° tan (50 – x)° = 1, then the value of x is equal to
(a)
(b)
(c)
(d)
tan(x + 40)° tan (x + 20)° tan 3x° tan (70 - x)° tan (50 - x)° = 1
⇒ tan(x + 40) tan(x + 20) tan 3x cot [90 - (70 - x)] cot [90 - (50 - x)] = 1
⇒ tan(x + 40) tan (x + 20) tan 3x cot (x + 20) cot (x + 40) = 1
⇒ tan 3x = tan 45°
⇒ 3x = 45°
⇒ x = 15°
So, option (c) is correct.
Q-12) ABC is a right triangle with right angle at A. If the value of tan B = $1/{√3}$, then for any real k the length of the hypotenuse is of the form
(a)
(b)
(c)
(d)
Given,
In ΔABC,
tan B = $k/{√3 k}$
By Pythogaros theorem,
$AB^2 + AC^2 = BC^2$
⇒ $(√3 k)^2 + (1k)^2 = BC^2$
⇒ $BC^2 = 4k^2$
⇒ BC = 2k
Q-13) If the given figure, BC = 15 cm and sin B = $4/5$. What is the value of AB?
(a)
(b)
(c)
(d)
BC = 15 cm and sin B = $4/5$
sin B = ${AC}/{AB} = 4/5$
then BC = 3m
But, BC = 15 (given)
then AC = 4 × 5 = 20
AB = 5 × 5 = 25.
Hence, the value of AB is 25 cm.
Q-14) If tan θ + sec θ = 2, then tan θ is equal to
(a)
(b)
(c)
(d)
tan θ + sec θ = 2 ......(i)
As we know
⇒ $sec^2 θ - tan^2 θ$ = 1
⇒ (sec θ - tan θ)(sec θ + tan θ) = 1
⇒ sec θ - tan θ = $1/2$ ....(ii)
equation (i) - eq (ii)-
2 tan θ = 2 - 1/2
2 tan θ = $3/2 ⇒ tan θ = 3/4$
So, option (b) is correct.
Q-15) If α is the angle of first quadrant such that $cosec^4 α = 17 + cot^4$ α, then what is the value of sin α?
(a)
(b)
(c)
(d)
$cosec^4 α - cot^4$ α = 17 (Given)
⇒ $(cosec^2 α - cot^2 α) (cosec^2 α + cot^2 α)$ = 17
⇒ 1.$({1 + cos^2 α}/{sin^2 α})$ = 17
⇒ 2 - $sin^2 α = 17 sin^2 α$
⇒ $18 sin^2 α = 2 ⇒ sin^2 α = 1/9$
∴ sin α = $1/3$ (since, α lie in first quadrant)
Q-16) What is ${\text"sec x"}/{\text"cot x + tan x"}$ equal to ?
(a)
(b)
(c)
(d)
${\text"sec x"}/{\text"cot x + tan x"}$
⇒ ${1/{\text"cos x"}}/{{\text"cos x"}/{\text"sin x"} + {\text"sin x"}/{\text"cos x"}}$
⇒ ${1/{\text"cos x"}}/{{cos^2 x + sin^2 x}/{\text"sin x cos x"}}$
⇒ $1/{(\text"cos x")}$ × cos x × sin x
⇒ sin x
Q-17) What is the value of x in the equation $x {cosec^2 30 ° sec^2 45 °}/{8 cos^2 45 ° sin^2 60 °} = tan^2 60° – tan^2 30° $ ?
(a)
(b)
(c)
(d)
Given, x ${cosec^2 30° sec^2 45°}/{8 cos^2 45° sin^2 60°} = tan^2 60° - tan^2 30°$
⇒ ${x × (2)^2 × (√2)^2}/{8 × (1/{√2})^2 × ({√3}/2)^2} = (√3)^2 - (1/{√3})^2$
⇒ ${x × 4 × 2 × 4}/{8 × 1/2 × 3} = 3 - 1/3$
⇒ ${8x}/3 = 8/3 ⇒$ x = 1
Q-18) If p = $√{{1 - \text"sin x"}/{1 + \text"sin x"}}, q = {1 - \text"sin x"}/{\text"cos x"}, r = {\text"cos x"}/{1 + \text"sin x"}$
then which of the following is/are correct ?
1. p = q = r
2. $p^2$ = qr
Select the correct answer using the code given below.
(a)
(b)
(c)
(d)
Statement 1
1 = ${√{\text"1 - sin n"}}/{√{\text"1 + sin x"}} = √{{(\text"1 - sin x")(\text"1 - sin x")}/{(\text"1 - sin x")(\text"1 + sin x")}}$
= ${\text"1 - sin x"}/{√{1 - sin^2 x}} = {\text"1 - sin x"}/{√{cos^2 x}} = {\text"1 - sin x"}/{\text"cos x"}$
P = q
r = ${\text"cos x"}/{\text"1 + sin x"} = {\text"cos x" (\text"1 - sin x")}/{(\text"1 + sin x")(\text"1 - sin x")} = {\text"cos x" (\text"1 - sin x")}/{1 - sin^2 x}$
= ${\text"cos x" (\text"1 - sin x")}/{cos^2 x} = {\text"1 - sin x"}/{\text"cos x"}$
P = q = r
Now,
Statement 2
$p^2$ = qr
= ${\text"1 - sin x"}/{\text"cos x"} . {\text"cos x"}/{\text"1 + sin x"} = {\text"1 - sin x"}/{\text"1 + sin x"} = P^2$
So, Both are correct.
Q-19) If A = ${π}/6$ and B = ${π}/3$, then which of the following is/ are correct?
I. sin A + sin B = cos A + cos B
II. tan A + tan B = cot A + cot B
Select the correct answer using the codes given below.
(a)
(b)
(c)
(d)
Given, A = ${π}/6 \text"and B" = {π}/3$
I. L.H.S = sin A + sin B = sin ${π}/6 + sin {π}/3$
= $1/2 + {√3}/2 = {1 + √3}/2$
R.H.S = cos A + cos B = cos ${π}/6 + {cos π}/3$
${√3}/2 + 1/2 = {√3 + 1}/2$
⇒ sin A + sin B = cos A + cos B
II. L.H.S = tan A + tan B = tan ${π}/6 + tan {π}/3$
= $1/{√3} + √3 = 4/{√3}$
R.H.S = cot A + cot B = cot ${π}/6 + cot {π}/3$
= $√3 + 1/{√3} = 4/{√3}$
⇒ tan A + tan B = cot A + cot B
Both statements are true.
Alternate Method:
A + B = ${π}/6 + {π}/3 = {π}/2$
I. sin A + sin B = sin $({π}/2 - B) + sin ({π}/2 - A)$
= cos B + cos A = cos A + cos B
II. tan A + tan B = tan $({π}/2 - B) + tan ({π}/2 - A)$
= cot B + cot A = cot A + cot B
Hence, both statements are true.
Q-20) If 0 ≤ θ ≤ ${π}/2$ and cos θ + $√3$ sin θ = 2, then what is the value of θ ?
(a)
(b)
(c)
(d)
Given that, cos θ + $√3$ sin θ = 2
⇒ $1/2 cos θ + {√3}/2$ sin θ = 1
⇒ sin 30° cos θ + cos 30° sin θ = 1
⇒ sin(30° + θ) = sin 90°
30° + θ = 90°
∴ θ = 60°