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This mock test of Test: Complementary Angles of Trigonometry for Class 10 helps you for every Class 10 entrance exam.
This contains 15 Multiple Choice Questions for Class 10 Test: Complementary Angles of Trigonometry (mcq) to study with solutions a complete question bank.
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QUESTION: 1

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

Solution:

QUESTION: 2

The value of cos^{2} 17° – sin^{2} 73° is

Solution:

cos^{2}17-sin^{2}73

=cos^{2}17-sin^{2}(90-17)

=cos^{2}17-co^{s}217 (because sin(90-x)=cos x)

=0

QUESTION: 3

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

Solution:

QUESTION: 4

=

Solution:

Taking LCM

QUESTION: 5

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

Solution:

QUESTION: 6

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

Solution:

QUESTION: 7

The value of is

Solution:

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

QUESTION: 8

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

Solution:

QUESTION: 9

Solution:

Taking LCM

QUESTION: 10

The value of is

Solution:

QUESTION: 11

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

Solution:

QUESTION: 12

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

Solution:

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

QUESTION: 13

Using the ratio of complementary angles, the value of is

Solution:

QUESTION: 14

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

Solution:

QUESTION: 15

If A and B are the angles of a right angled triangle ABC, right angled at C, then 1+cot^{2}A =

Solution:

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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