Trigonometric Ratios & Identity Model Questions Set 1 Practice Questions Answers Test with Solutions & More Shortcuts
Trigonometric Ratios & Identities PRACTICE TEST [1 - EXERCISES]
Trigonometric Ratios & Identity Model Questions Set 1
Question : 21
Suppose 0 < θ < 90°, then for every θ, 4 $sin^2$ θ + 1 is greater than or equal to
a) 4 sin θ
b) 2
c) 4 cos θ
d) 4 tan θ
Answer »Answer: (a)
Let 4 $sin^2$ θ + 1 ≥ 4 sin θ
4 $sin^2$ θ - 4 sin θ + 1 ≥ 0
$(2 sin θ - 1)^2 ≥ 0$
sin θ ≥ $1/2$
Question : 22
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) x = 2
b) x = 1
c) x = $1/2$
d) x = $3/2$
Answer »Answer: (b)
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
Question : 23
If x lies in the first quadrant and cos x = $5/{13}$, what is the value of tan x – cot x?
a) ${139}/{60}$
b) ${–139}/{60}$
c) ${119}/{60}$
d) None of these
Answer »Answer: (c)
Given that, cos x = $5/{13} = {Base}/{Hypotenuse}$
P = $√{h^2 - b^2} = √{13^2 - 5^2}$
= $√{169 - 25} = √{144}$ = 12
∴ tan x - cot x = $p/b - b/p$
= ${12}/5 - 5/{12} = {144 - 25}/{60} = {119}/{6}$
Question : 24
What is ${cos θ}/{1 + sin θ} + 1/{cot θ}$ equal to?
a) sec θ
b) cosec θ
c) sec θ + cosec θ
d) cosec θ – cot θ
Answer »Answer: (a)
${\text"cos θ"}/{\text"1 + sin θ"} + {\text"sin θ"}/{\text"cos θ"} = {cos^2 + sin θ^2 \text"+ sin θ"}/{(\text"1 + sin θ)cos θ"}$
= $1/{\text"cos θ"}$ = sec θ
Question : 25
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) 20
b) 30
c) 15
d) 10
Answer »Answer: (c)
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.
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