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RS Aggarwal Test: Introduction to Trigonometry - Class 10 MCQ


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15 Questions MCQ Test - RS Aggarwal Test: Introduction to Trigonometry

RS Aggarwal Test: Introduction to Trigonometry for Class 10 2024 is part of Class 10 preparation. The RS Aggarwal Test: Introduction to Trigonometry questions and answers have been prepared according to the Class 10 exam syllabus.The RS Aggarwal Test: Introduction to Trigonometry MCQs are made for Class 10 2024 Exam. Find important definitions, questions, notes, meanings, examples, exercises, MCQs and online tests for RS Aggarwal Test: Introduction to Trigonometry below.
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RS Aggarwal Test: Introduction to Trigonometry - Question 1

The value of 

Detailed Solution for RS Aggarwal Test: Introduction to Trigonometry - Question 1

we know

tan 30= 1/√3
cot60= 1/√3

tan30/cot60= 1/√3 ×√3/1= 1

RS Aggarwal Test: Introduction to Trigonometry - Question 2

If ΔABC is right angled at C, then the value of cos (A + B) is

Detailed Solution for RS Aggarwal Test: Introduction to Trigonometry - Question 2

Δ ABC is right angled at C, A + B + C = 180°
∴ A + B = 180° - 90° = 90° (∵ ∠C = 90°)
cos (A + B) = cos 90° = 0

RS Aggarwal Test: Introduction to Trigonometry - Question 3

If cos A + cos2 A = 1, then sin2 A + sin4 A is

Detailed Solution for RS Aggarwal Test: Introduction to Trigonometry - Question 3

RS Aggarwal Test: Introduction to Trigonometry - Question 4

If cos (α + β) = 0, then sin (α - β) can be reduced to

Detailed Solution for RS Aggarwal Test: Introduction to Trigonometry - Question 4

cos (a + β) = 0 = cos 90°
⇒ α + β = 90°
sin (α - β) = sin (α - β + β - β)
= sin (α + β - 2β)
= sin (90° - 2β) = cos 2β

RS Aggarwal Test: Introduction to Trigonometry - Question 5

The value of 

RS Aggarwal Test: Introduction to Trigonometry - Question 6

(1 + tanθ + secθ) (1 + cotθ - cosecθ) is equal to

Detailed Solution for RS Aggarwal Test: Introduction to Trigonometry - Question 6

= 2
= R.H.S.

RS Aggarwal Test: Introduction to Trigonometry - Question 7

If cos 9α = sin α and 9α < 90°, then the value of tan 5α is

Detailed Solution for RS Aggarwal Test: Introduction to Trigonometry - Question 7

RS Aggarwal Test: Introduction to Trigonometry - Question 8

sin (45° + θ) - cos (45° - θ) is equal to

Detailed Solution for RS Aggarwal Test: Introduction to Trigonometry - Question 8

sin (45° + θ) - cos (45° - θ)
= sin {90° - (45° - θ)} - cos (45° - θ)
= cos (45° - θ) - cos (45° - θ) = 0

RS Aggarwal Test: Introduction to Trigonometry - Question 9

If sin A = 1/2,  then the value of cot A is

RS Aggarwal Test: Introduction to Trigonometry - Question 10

tan A =

Detailed Solution for RS Aggarwal Test: Introduction to Trigonometry - Question 10

RS Aggarwal Test: Introduction to Trigonometry - Question 11

Given that sin θ = a/b, then cos θ is equal to

RS Aggarwal Test: Introduction to Trigonometry - Question 12

If cosec A - cot A = 4/5, then cosec A = 

Detailed Solution for RS Aggarwal Test: Introduction to Trigonometry - Question 12

cosec A - cot A = 4/5  ........(i)
Also cosec2 A - cot2 A = 1
⇒ (cosec A - cot A) (cosec A - cot A) = 1

⇒ cosec A + cot A = 4/5   ......(ii)
From (i) and (ii), cosec A = 41/40

RS Aggarwal Test: Introduction to Trigonometry - Question 13

The value of (tan1° tan2° tan3°... tan89°) is

Detailed Solution for RS Aggarwal Test: Introduction to Trigonometry - Question 13

tan 1 . tan 2 . tan 3 ... tan 87 . tan 88 . tan 89 = LHS 
tan 1 . tan 2 . tan 3 ... tan (90 - 3 ) . tan ( 90 - 2 ) . tan ( 90 - 1)
tan 1 . tan 2  . tan 3 ... cot 3 . cot 2 . cot 1 

tan 1 . cot 1 . tan 2 . cot 2 . tan 3 . cot 3 ... tan 89 . cot 89 
1 x 1 x 1 x 1 x ... x 1 
As 1ⁿ = 1  = RHS 

RS Aggarwal Test: Introduction to Trigonometry - Question 14

If sin x + cosec x = 2, then sin19x + cosec20 x =

Detailed Solution for RS Aggarwal Test: Introduction to Trigonometry - Question 14

sin x + cosec x = 2

RS Aggarwal Test: Introduction to Trigonometry - Question 15

The value of the expression  is

Detailed Solution for RS Aggarwal Test: Introduction to Trigonometry - Question 15

We know that sin 60 =√3/2 and cos 30 = √3/2.
Therefore , Sin 60/cos 30= (√3/2)/(√3/2) = 1

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