JEE Advanced Level Test: Trigonometric Ratio- 1 - JEE MCQ

# JEE Advanced Level Test: Trigonometric Ratio- 1 - JEE MCQ

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## 30 Questions MCQ Test Chapter-wise Tests for JEE Main & Advanced - JEE Advanced Level Test: Trigonometric Ratio- 1

JEE Advanced Level Test: Trigonometric Ratio- 1 for JEE 2024 is part of Chapter-wise Tests for JEE Main & Advanced preparation. The JEE Advanced Level Test: Trigonometric Ratio- 1 questions and answers have been prepared according to the JEE exam syllabus.The JEE Advanced Level Test: Trigonometric Ratio- 1 MCQs are made for JEE 2024 Exam. Find important definitions, questions, notes, meanings, examples, exercises, MCQs and online tests for JEE Advanced Level Test: Trigonometric Ratio- 1 below.
JEE Advanced Level Test: Trigonometric Ratio- 1 - Question 1

### If tan a+cot a = a then the value of tan4a+cot4a =

JEE Advanced Level Test: Trigonometric Ratio- 1 - Question 2

### If a cos q + b sin q = 3 & a sin q – b cos q = 4 then a2 + b2 has the value =

Detailed Solution for JEE Advanced Level Test: Trigonometric Ratio- 1 - Question 2

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JEE Advanced Level Test: Trigonometric Ratio- 1 - Question 3

### The value of tan 1º tan 2º tan 3º ..... tan 89º is

Detailed Solution for JEE Advanced Level Test: Trigonometric Ratio- 1 - Question 3

tan1° tan2° .........tan89°
= (tan1° tan89°)(tan2° tan85°)........tan45°
= (tan1° cot1°)(tan2° cot°)......tan45°
= 1•1 = 1

JEE Advanced Level Test: Trigonometric Ratio- 1 - Question 4

when simplified reduces to :

Detailed Solution for JEE Advanced Level Test: Trigonometric Ratio- 1 - Question 4

{tan(x-π/2).cos(3π/2+x)-sin³(7π/2-x)}/{cos(x-π/2)tan(3π/2+x)}
= [tan{-(π/2-x)}.cos{(π/2×3)+x} - sin³{(π/2×7)-x}]/[cos{-(π/2-x)}.tan{(π/2×3)+x}
= [-tan(π/2-x).sinx - (-cosx)³]/[cos(π/2-x).(-cotx)]
= (-cotx.sinx+cos³x)/[sinx.(-cotx)]
= [-(cosx/sinx).sinx+cos³x]/[sinx.(-cosx/sinx)]
= {-cosx(1 - cos²x)}/(-cosx)
= 1 - cos²x
= sin²x

JEE Advanced Level Test: Trigonometric Ratio- 1 - Question 5

The expression3–2 is equal to

JEE Advanced Level Test: Trigonometric Ratio- 1 - Question 6

cos (540º – q) – sin (630º – q) is equal to

JEE Advanced Level Test: Trigonometric Ratio- 1 - Question 7

The value of sin(p + q) sin (p - q) cosec2q is equal to

Detailed Solution for JEE Advanced Level Test: Trigonometric Ratio- 1 - Question 7

Explanation : sin(p + q) sin(p -q)cosec2q

= (sin2p –sin2q)1/sin2q

= sin2p/sin2q - 1

JEE Advanced Level Test: Trigonometric Ratio- 1 - Question 8

If sin a sin b – cos a cos b + 1 = 0, then the value of 1 + cot a tan b is

Detailed Solution for JEE Advanced Level Test: Trigonometric Ratio- 1 - Question 8

sin A sin B - cos A cos B + 1 = 0
=> cos A cos B -sin A sin B = 1
⇒ cos(A+B) = 1 = cos0
⇒ A = 2nΠ − B , where n∈Z
Now 1+cotAtanB
⇒ 1+cot(2nπ−B)tanB
⇒ 1 − cot(B)tanB
= 1−1   = 0

JEE Advanced Level Test: Trigonometric Ratio- 1 - Question 9

The value of  is

JEE Advanced Level Test: Trigonometric Ratio- 1 - Question 10

If 3 sina = 5 sinb, then  is equal to

JEE Advanced Level Test: Trigonometric Ratio- 1 - Question 11

In a triangle ABC if tan A < 0 then :

JEE Advanced Level Test: Trigonometric Ratio- 1 - Question 12

If tan A – tan B = x and cot B – cot A = y, then cot (A – B) is equal to

Detailed Solution for JEE Advanced Level Test: Trigonometric Ratio- 1 - Question 12

JEE Advanced Level Test: Trigonometric Ratio- 1 - Question 13

If tan 25º = x, thenis equal to

JEE Advanced Level Test: Trigonometric Ratio- 1 - Question 14

If A + B = 225º, then the value of  is

JEE Advanced Level Test: Trigonometric Ratio- 1 - Question 15

The value of tan 3A – tan 2A – tan A is equal to

Detailed Solution for JEE Advanced Level Test: Trigonometric Ratio- 1 - Question 15

3A= A+ 2A
⇒ tan 3A = tan (A + 2A)
⇒ tan 3 A = tanA + tan2A/ 1 – tan A . tan 2A
⇒ tan A + tan 2A = tan 3A – tan 3A. tan 2A . tan A
⇒ tan 3 A – tan 2A – tan A = tan 3A . tan 2A . tan A

JEE Advanced Level Test: Trigonometric Ratio- 1 - Question 16

tan 203º + tan 22º + tan 203º tan 22º =

Detailed Solution for JEE Advanced Level Test: Trigonometric Ratio- 1 - Question 16

tan225=+1
tan(203+22)=+1
= tan203+tan22/1−tan203tan22 = +1
⇒ tan203+tan22=+1−tan203tan22
⇒ tan203+tan22+tan203tan22=1

JEE Advanced Level Test: Trigonometric Ratio- 1 - Question 17

The value of  is

Detailed Solution for JEE Advanced Level Test: Trigonometric Ratio- 1 - Question 17

JEE Advanced Level Test: Trigonometric Ratio- 1 - Question 18

If A lies in the third quadrant and 3 tan A – 4 = 0, then 5 sin 2A + 3 sinA + 4 cos A is equal to

Detailed Solution for JEE Advanced Level Test: Trigonometric Ratio- 1 - Question 18

A lies in third quadrant = 180degree < A ≤270degree
3tan A - 4 = 0
tan A = 4/3
sinA = -4/5 ; cos A = -3/5
A = 5sin(2A) + 3sinA + 4cosA
= 10(-4/5)(-3/5) + 3(4/5) + 4(-3/5)
= 0

JEE Advanced Level Test: Trigonometric Ratio- 1 - Question 19

is equal to

Detailed Solution for JEE Advanced Level Test: Trigonometric Ratio- 1 - Question 19

So, (cos20+8sin10sin50sin70)/sin2(80)
= cos20+8(1/4*sin3(10))/sin2(80)
= cos20+1/sin2(80)
= 2cos2(10)/sin2(80)
= 2sin2(80)/sin2(80)
= 2

JEE Advanced Level Test: Trigonometric Ratio- 1 - Question 20

If cos A = 3/4, then the value of 16cos2 (A/2) – 32 sin (A/2) sin (5A/2) is

JEE Advanced Level Test: Trigonometric Ratio- 1 - Question 21

The value of the expression  is

Detailed Solution for JEE Advanced Level Test: Trigonometric Ratio- 1 - Question 21

JEE Advanced Level Test: Trigonometric Ratio- 1 - Question 22

The numerical value of sin 12º . sin 48º . sin 54º is equal to

Detailed Solution for JEE Advanced Level Test: Trigonometric Ratio- 1 - Question 22

sin 12°sin 48°sin 54°

JEE Advanced Level Test: Trigonometric Ratio- 1 - Question 23

If α + β + γ = 2π, then

Detailed Solution for JEE Advanced Level Test: Trigonometric Ratio- 1 - Question 23

α+β+γ=2π
⇒(α+β+γ)/2 =π
⇒ [tanα/2 + tanβ/2+  tanγ/2 − tanα/2 tanβ/2 tanγ/2]/[1 −tanα/2 tanβ/2 − tanβ/2 tanγ/2 - −tanα/2 tanγ/2] = 0
= tanα/2 tanβ/2 tanγ/2

JEE Advanced Level Test: Trigonometric Ratio- 1 - Question 24

cosθ+cos(pi/7)+cos+cos+cos+cos+cos=

JEE Advanced Level Test: Trigonometric Ratio- 1 - Question 25

A regular hexagon & a regular dodecagon are inscribed in the same circle. If the side of the dodecagon is (√3–1), then the side of the hexagon is

Detailed Solution for JEE Advanced Level Test: Trigonometric Ratio- 1 - Question 25

JEE Advanced Level Test: Trigonometric Ratio- 1 - Question 26

In a right angled triangle the hypotenuse is 2√2 times the perpendicular drawn from the opposite vertex. Then the other acute angles of the triangle are

Detailed Solution for JEE Advanced Level Test: Trigonometric Ratio- 1 - Question 26

AC = x/(sin∠ACD) = x/sinC
AB = x/sinB
x2(1/sin2C + 1/sin2B) = 8x2
sin2B + sin2C = 8sin2C sin2B

1 = 8sin2C cos2C
(sin2C) = 1/√2 ⇒ C = π/8 ⇒ B = 3π/8

JEE Advanced Level Test: Trigonometric Ratio- 1 - Question 27

If a ∈  then the value of –  is equal to

Detailed Solution for JEE Advanced Level Test: Trigonometric Ratio- 1 - Question 27

JEE Advanced Level Test: Trigonometric Ratio- 1 - Question 28

The value of cot x + cot(60º + x) + cot (120º + x) is equal to

Detailed Solution for JEE Advanced Level Test: Trigonometric Ratio- 1 - Question 28

JEE Advanced Level Test: Trigonometric Ratio- 1 - Question 29

If x ∈  then 4 cos2  is always equal to

JEE Advanced Level Test: Trigonometric Ratio- 1 - Question 30

The expression  is equal to

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