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Test: Complementary Angles of Trigonometry - Class 10 MCQ


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15 Questions MCQ Test Mathematics (Maths) Class 10 - Test: Complementary Angles of Trigonometry

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

If tan 2A = cot (A – 18°), then the value of A is

Detailed Solution for Test: Complementary Angles of Trigonometry - Question 1

 

Test: Complementary Angles of Trigonometry - Question 2

The value of cos2 17° – sin2 73° is

Detailed Solution for Test: Complementary Angles of Trigonometry - Question 2

cos217-sin273
=cos217-sin2(90-17)
=cos217-cos217   (because sin(90-x)=cos x)
=0

Test: Complementary Angles of Trigonometry - Question 3

If sec 4A = cosec (A-20°),where 4A is an acute angle, find the value of A

Detailed Solution for Test: Complementary Angles of Trigonometry - Question 3

 

Test: Complementary Angles of Trigonometry - Question 4

​=

Detailed Solution for Test: Complementary Angles of Trigonometry - Question 4


Taking LCM

Test: Complementary Angles of Trigonometry - Question 5

The value of cos θ cos(90° - θ) – sin θ sin (90° - θ) is:

Test: Complementary Angles of Trigonometry - Question 6

Out of the following options, the two angles that are together classified as complementary angles are

Test: Complementary Angles of Trigonometry - Question 7

The value of    is

Detailed Solution for Test: Complementary Angles of Trigonometry - Question 7

we know sin(90 - a) = cos(a) 

cos(90 - a) = sin(a)

sin(a) = 1/cosec(a)

sec(a) = 1/cos(a)

 

cos40 = cos(90-50) = sin50

cosec40 = cosec(90-50) = sec50

so our expression becomes

sin50/sin50 + sec50/sec50 - 4cos50 / sin40

= 1 + 1 - 4(1)   since cos50 = sin40

= -2

Test: Complementary Angles of Trigonometry - Question 8

If cos (40° + A) = sin 30°, the value of A is:​

Detailed Solution for Test: Complementary Angles of Trigonometry - Question 8

cos(θ)=sin(90-θ)
so 40+A+30=90
A=20

Test: Complementary Angles of Trigonometry - Question 9

Detailed Solution for Test: Complementary Angles of Trigonometry - Question 9


Taking LCM

Test: Complementary Angles of Trigonometry - Question 10

The value of   is

Test: Complementary Angles of Trigonometry - Question 11

sin (60° + θ) – cos (30° – θ) is equal to (where (60° + θ) and (30° - θ) are both acute angles):

Test: Complementary Angles of Trigonometry - Question 12

The value of tan1°.tan2°.tan3°………. tan89° is :

Detailed Solution for Test: Complementary Angles of Trigonometry - Question 12

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 =1

Test: Complementary Angles of Trigonometry - Question 13

Using the ratio of complementary angles, the value of   is

Test: Complementary Angles of Trigonometry - Question 14

If x sin (90° – θ) cot (90°- θ) = cos (90° – θ) then x is

Test: Complementary Angles of Trigonometry - Question 15

If A and B are the angles of a right angled triangle ABC, right angled at C, then 1+cot2A =​

Detailed Solution for Test: Complementary Angles of Trigonometry - Question 15
ABC is a Δ, right angle at c.
1 +cot^2 =?........ 
we know that.....
Cosec^2 - cot^2= 1...
So,
=> 1+ cot^2
=> cosec^2 A
=> (AB)^2/( CB)^2 
= sec ^2B.
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