# Trigonometric Ratios & Identity Model Questions Set 1 Practice Questions Answers Test With Solutions & More Shortcuts

#### TRIGONOMETRIC RATIOS & IDENTITIES PRACTICE TEST [1 - EXERCISES]

Question : 1

Consider the following :

1. ${cos 75°}/{sin 15 °} + {sin 12°}/{cos 78°} - {cos 18°}/{sin 72°}$ = 1
2. ${cos 35°}/{sin 55°} - {sin 11°}/{cos 79°}$ + cos 28° cosec 62°= l
3. ${sin 80°}/{cos 10°}$ - sin 59° sec 31° = 0
Which of the above are correct?

a) 2 and 3 only

b) 1 and 2 only

c) 1 and 3 only

d) 1, 2 and 3

1. ${cos 75°}/{sin 15°} + {sin 12°}/{cos 78°} - {cos 18°}/{sin 72°}$ = 1

${cos 75°}/{cos 75°} + {sin 12°}/{sin 12°} - {cos 18°}/{cos 18°}$ = 1

2 - 1 = 1

Statement is correct.

2. ${cos 35°}/{sin 55°} - {sin 11°}/{cos 79°}$ + cos 28° cosec 62° = 1

${cos 35°}/{cos 35°} - {sin 11°}/{sin 11°}$ + cos 28° sec 28° = 1

1 - 1 + 1 = 1

2nd Statement is also true.

3. ${sin 80°}/{cos 10°}$ - sin 59° sec 31° = 0

${sin 80°}/{sin 80°}$ - sin 59° cosec 59° = 0

1 - 1 = 0

3rd Statement is correct.

Question : 2

If cos A = tan B, cos B = tan C and cos C = tan A then sin A is equal to

a) ${√5 - 1}/2$

b) ${√5 - 1}/4$

c) ${√3 - 1}/4$

d) None of these

Cos A = tan B

Squaring on both sides

$cos^2 A = tan^2 B$

⇒ $tan^2 B = {sin^2 B}/{cos^2 B} = {1 - cos^2 B}/{cos^2 B}$

∴ $cos^2 A = {1 - cos^2 B}/{cos^2 B}$

$cos^2 A = {1 - tan^2 C}/{tan^2 C}$

(∵ cos B = tan C)

⇒ $cos^2 A tan^2 C = 1 - tan^2 C = 1 - {sin^2 C}/{cos^2 C}$

⇒ $cos^2 A {sin^2 C}/{cos^2 C} ={cos^2 C - sin^2 C}/{cos^2 C}$

⇒ $cos^2 A (1 - cos^2 C) = 2 cos^2 C - 1$

⇒ $cos^2 A (1 - tan^2 A) = 2 tan^2 A - 1$

$cos^2 A - sin^2 A = {2 sin^2 A}/{cos^2 A} - 1$

⇒ 1 - $2sin^2 A = {2 sin^2 A - cos^2 A}/{cos^2 A}$

⇒ $cos^2 A (1 - sin^2 A) = 2 sin^2 A - cos^2 A$

⇒ $cos^2 A (1 - 2 sin^2 A) = 2 sin^2 A - 1 + sin^2 A$

⇒ $(1 - sin^2 A) (1 - 2 sin^2 A) = 3 sin^2 A - 1$

⇒ $1 - 2 sin^2 A - sin^2 A + 2 sin^4A = 3 sin^2 A - 1$

⇒ $1 - 3 sin^2 A + 2 sin^4 A = 3 sin^2 A - 1$

⇒ $2 sin^4 A - 6 sin^2 A + 2 = 0$

⇒ $sin^4 A - 3 sin^2$ A + 1 = 0

This is quadratic equation in $sin^2$ A

$(sin^2 A)^2 - 3 (sin^2 A) + 1 = 0$

$sin^2 A = {3 ± √{(- 3)^2 - 4 (1) (1)}}/2$

= ${3 ± √5}/2$

$sin^2 A = {3 + √5}/2$ not possible because $sin^2 A$ ≯ 1

So $sin^2 A = {3 - √5}/2$

So none of the options are correct.

Question : 3

If sin θ + cos θ = 1, then what is the value of sin θ cos θ ?

a) 0

b) 2

c) 1

d) $1/2$

Given, $\text"sin θ + cos θ"$ = 1

Squaring both sides,

$(sin^2 θ + cos^2 θ) + 2 \text"sin θ cos θ" = 1$

⇒ 1 + 2 sin θ cos θ = 1 ⇒ sin θ cos θ = 0.

Question : 4

If 0 ≤ θ ≤ ${π}/2$ and cos θ + $√3$ sin θ = 2, then what is the value of θ ?

a) ${π}/4$

b) ${π}/3$

c) ${π}/6$

d) ${π}/2$

Given that, cos θ + $√3$ sin θ = 2

⇒ $1/2 cos θ + {√3}/2$ sin θ = 1

⇒ sin 30° cos θ + cos 30° sin θ = 1

⇒ sin(30° + θ) = sin 90°

30° + θ = 90°

∴ θ = 60°

Question : 5

The value of cos 25° – sin 25° is

a) positive but greater than 1

b) positive but less than 1

c) negative

d) 0

Since, value of cos θ decreases, from 0° to 90° and at 45°, it is equal to the value of sin θ.

Similarly, value of sin θ increases from 0° to 90° and at 45°, it is equal to the value of cos θ

For 0° < 45°, cos θ > sin θ

So, value of cos 25° is always positive but less than 1.

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