Trigonometric Ratios & Identity Model Questions Set 1 Section-Wise Topic Notes With Detailed Explanation And Example Questions
MOST IMPORTANT quantitative aptitude - 1 EXERCISES
The following question based on Trigonometric Ratios & Identities topic of quantitative aptitude
(a) 0
(b) &1
(c) 1
(d) 8
The correct answers to the above question in:
Answer: (c)
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
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Question : 1
If from the top of a post a string twice the length of the post is stretched tight to a point on the ground, then what angle will the string make with the post?
a) ${π}/4$
b) ${π}/6$
c) ${5 π}/{12}$
d) ${π}/3$
Answer »Answer: (d)
Let AB be the height, AC be the string and the angle made by string with the post be θ.
From figure cos θ = ${AB}/{AC} = h/{2h} = 1/2 = cos {π}/3$
θ = ${π}/3$
Question : 2
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) 2
b) 1
c) 4
d) 8
Answer »Answer: (b)
$cos^2 \text"x + cos x = 1"$
⇒ $\text"cos x = 1" - cos^2 x = sin^2 x$
= $sin^{12}x + 3 sin^{10}x + 3 sin^8x + sin^6x$
= $sin^6x[sin^6x + 3 sin^4x + 3 sin^2 x + 1]$
= $sin^6x[sin^2 x + 1]^3$
= $[sin^4x + sin^2 x]^3$
$(∴ sin^4 x = cos^2 x)$
= $(sin^2 x + cos^2 x)$ = 1
Question : 3
If 5 sin θ + 12 cos θ = 13, then what is 5 cos θ – 12 sin θ equal to?
a) –1
b) –2
c) 0
d) 1
Answer »Answer: (c)
∵ 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
Question : 4
What is sin 25° sin 35° sec 65° sec 55° equal to?
a) 0
b) –1
c) $1/2$
d) 1
Answer »Answer: (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
Question : 5
Which one of the following is correct?
a) There is more than one θ with 0° < θ < 90° such that sin θ = a, where a is a real number.
b) There is only one q with 0° < θ < 90° such that sin θ = a, where a is a real number.
c) There is no θ with 0° < θ < 90° such that sin θ = a, where a is a real number.
d) There are exactly θ's with 0° < θ < 90° such that sin θ = a, where a is a real number.
Answer »Answer: (b)
It is true, for 0° < θ < 90°, there exist only one θ such sin θ = a.
Question : 6
$({sin 35°}/{cos 55°})^2 - ({cos 55°}/{sin 35°})^2$ + 2sin 30° is equal to
a) 0
b) –1
c) 1
d) 2
Answer »Answer: (c)
$({sin 35°}/{cos 55°})^2 - ({cos 55°}/{sin 35°})^2$ + 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.
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