Trigonometrical Identities — Multiple Choice Questions

In ΔABC, if AC = 17 m and BC = 8 m, then tan A =

1. $\Big(\dfrac{8}{15}\Big)$

2. $\Big(\dfrac{15}{8}\Big)$

3. $\Big(\dfrac{8}{17}\Big)$

4. $\Big(\dfrac{15}{17}\Big)$

Answer

tan A = $\dfrac{\text{opposite}}{\text{adjacent}} = \dfrac{BC}{AB} = \dfrac{8}{15}$

Hence, option 1 is the correct option.

If sin θ = $\Big(\dfrac{5}{13}\Big)$, then the value of tan θ is:

1. $\Big(\dfrac{5}{12}\Big)$

2. $\Big(\dfrac{12}{13}\Big)$

3. $\Big(\dfrac{12}{5}\Big)$

4. $\Big(\dfrac{13}{12}\Big)$

Answer

Given,

sin θ = $\Big(\dfrac{5}{13}\Big) = \dfrac{\text{opposite}}{\text{hypotenuse}}$

Opposite = 5, Hypotenuse = 13

tan A = $\dfrac{\text{opposite}}{\text{adjacent}} = \dfrac{5}{12}$

Hence, option 1 is the correct option.

If sec θ = $\Big(\dfrac{25}{7}\Big)$, then the value of cot θ is:

1. $\Big(\dfrac{25}{24}\Big)$

2. $\Big(\dfrac{24}{7}\Big)$

3. $\Big(\dfrac{7}{24}\Big)$

4. $\Big(\dfrac{24}{25}\Big)$

Answer

sec θ = $\Big(\dfrac{25}{7}\Big) = \dfrac{\text{hypotenuse}}{\text{base}}$

Perpendicular = $\sqrt{25^2 - 7^2} = \sqrt{576}$ = 24.

cot θ = $\dfrac{\text{Base}}{\text{perpendicular}} = \Big(\dfrac{7}{24}\Big)$.

Hence, option 3 is the correct option.

The value of (1 + tan²θ)(1 − sin θ)(1 + sin θ) is:

1. 0

2. 1

3. sec²θ sin²θ

4. cot²θ

Answer

Given,

⇒ (1 + tan²θ)(1 − sin θ)(1 + sin θ)

⇒ sec²θ (1 - sin²θ)

⇒ sec²θ cos²θ

⇒ 1

Hence, option 2 is the correct option.

Given that sin θ = $\Big(\dfrac{a}{b}\Big)$, then cos θ is equal to:

1. $\Big(\dfrac{b}{a}\Big)$

2. $\Big(\dfrac{a}{\sqrt{b^2 - a^2}}\Big)$

3. $\Big(\dfrac{b}{\sqrt{b^2 - a^2}}\Big)$

4. $\Big(\dfrac{\sqrt{b^2 - a^2}}{b}\Big)$

Answer

Let ABC be a right angle triangle with ∠B = 90° and ∠C = θ.

By formula,

$\sin \theta = \dfrac{\text{perpendicular}}{\text{hypotenuse}}$

Substituting values we get :

$\dfrac{a}{b} = \dfrac{AB}{AC}$

Let AB = ak and AC = bk.

In right angle triangle ABC,

⇒ AC² = AB² + BC²

⇒ (bk)² = (ak)² + BC²

⇒ b²k² = a²k² + BC²

⇒ BC² = b²k² - a²k²

⇒ BC = k$\sqrt{b^2 - a^2}$

By formula,

$\cos \theta = \dfrac{\text{base}}{\text{hypotenuse}} \\[1em] = \dfrac{BC}{AC} \\[1em] = \dfrac{k\sqrt{b^2 - a^2}}{bk} \\[1em] = \dfrac{\sqrt{b^2 - a^2}}{b} .$

Hence, option 4 is the correct option.

In the adjoining figure, D is the mid-point of BC. Then the value of $\Big(\dfrac{\cot y}{\cot x}\Big)$ is:

1. $\Big(\dfrac{1}{2}\Big)$

2. $\Big(\dfrac{1}{3}\Big)$

3. $\Big(\dfrac{1}{4}\Big)$

4. 2

Answer

We know that,

$cot θ = \dfrac{\text{base}}{\text{perpendicular}} \\[1em] \Rightarrow \dfrac{\cot y}{\cot x} = \dfrac{\dfrac{AC}{BC}}{\dfrac{AC}{CD}} \\[1em] \Rightarrow \dfrac{\cot y}{\cot x} = \dfrac{CD}{BC} \\[1em] \Rightarrow \dfrac{\cot y}{\cot x} = \dfrac{CD}{2CD} \\[1em] \Rightarrow \dfrac{\cot y}{\cot x} = \dfrac{1}{2} .$

Hence, option 1 is the correct option.

If tan A = $\Big(\dfrac{5}{12}\Big)$, then the value of (sin A + cos A) sec A is:

1. $\Big(\dfrac{5}{12}\Big)$

2. $\Big(\dfrac{7}{12}\Big)$

3. $\Big(\dfrac{17}{12}\Big)$

4. $\Big(\dfrac{5}{13}\Big)$

Answer

Given,

(sin A + cos A) sec A

$\Rightarrow (\sin A + \cos A)\dfrac{1}{\cos A} \\[1em] \Rightarrow \dfrac{\sin A}{\cos A} + \dfrac{\cos A}{\cos A} \\[1em] \Rightarrow \tan A + 1 \\[1em] \Rightarrow \dfrac{5}{12} + 1 \\[1em] \Rightarrow \dfrac{5 + 12}{12} \\[1em] \Rightarrow \dfrac{17}{12}.$

Hence, option 3 is the correct option.

If 3 cos θ = 1, then the value of cosec θ is:

1. $2\sqrt{2}$

2. $\Big(\dfrac{3}{2\sqrt{2}}\Big)$

3. $\Big(\dfrac{2\sqrt{3}}{3}\Big)$

4. $\Big(\dfrac{4}{3\sqrt{2}}\Big)$

Answer

Given,

3 cos θ = 1

We know that,

⇒ sin² θ = 1 - cos² θ

$\Rightarrow \sin^2 \theta = 1 - \dfrac{1}{9} \\[1em] \Rightarrow \sin^2 \theta = \dfrac{9 - 1}{9} \\[1em] \Rightarrow \sin^2 \theta = \dfrac{8}{9} \\[1em] \Rightarrow \sin \theta = \dfrac{2\sqrt{2}}{3} \\[1em] \Rightarrow \cosec \theta = \dfrac{1}{\sin \theta} = \dfrac{3}{2\sqrt{2}}.$

Hence, option 2 is the correct option.

If x cos A = 1 and tan A = y, then x² − y² is equal to:

1. 0

2. 1

3. −tan A

4. tan A

Answer

Given,

x cos A = 1

cos A = $\dfrac{1}{x}$

⇒ sec A = x

tan A = y

We know that

sec² A - tan² A = 1

∴ x² − y² = 1

Hence, option 2 is the correct option.

In the adjoining figure, if PS = 14 cm, then the value of tan α is equal to:

1. $\Big(\dfrac{4}{3}\Big)$

2. $\Big(\dfrac{5}{3}\Big)$

3. $\Big(\dfrac{13}{3}\Big)$

4. $\Big(\dfrac{14}{3}\Big)$

Answer

ST = PS − RQ = 14 − 5 = 9 cm

In ΔSTR,

$TR = \sqrt{13^2 - 5^2} = \sqrt{169 - 25} = \sqrt{144} = 12.$

We know that,

$\tan \alpha = \dfrac{\text{opposite}}{\text{adjacent}} \\[1em] \tan \alpha = \dfrac{TR}{ST} \\[1em] \tan \alpha = \dfrac{12}{9} \\[1em] \tan \alpha = \dfrac{4}{3}.$

Hence, option 1 is the correct option.

The value of (1 + tan θ + sec θ)(1 + cot θ − cosec θ) is:

1. −4

2. −1

3. 1

4. 2

Answer

Given,

(1 + tan θ + sec θ)(1 + cot θ − cosec θ)

$\Rightarrow \Big(1 + \dfrac{\sin \theta}{\cos \theta} + \dfrac{1}{\cos \theta}\Big) \Big(1 + \dfrac{\cos \theta}{\sin \theta} - \dfrac{1}{\sin \theta}\Big)\\[1em] \Rightarrow \Big( \dfrac{\cos \theta + \sin \theta + 1}{\cos \theta} \Big) \Big(\dfrac{\sin \theta + \cos \theta - 1}{\sin \theta} \Big)\\[1em] \Rightarrow \dfrac{(\cos \theta + \sin \theta)^2 - 1^2}{\cos \theta \sin \theta} \\[1em] \Rightarrow \dfrac{\cos^2 \theta + \sin^2 \theta + 2\sin \theta \cos \theta - 1}{\cos \theta \sin \theta} \\[1em] \Rightarrow \dfrac{1 + 2\sin \theta \cos \theta - 1}{\cos \theta \sin \theta} \\[1em] \Rightarrow \dfrac{ 2\sin \theta \cos \theta}{\cos \theta \sin \theta} \\[1em] \Rightarrow 2.$

Hence, option 4 is the correct option.

If sin θ = $\Big(\dfrac{1}{2}\Big)$, then the value of $\Big(\dfrac{1}{5} \cot^2\theta + \dfrac{1}{5}\Big)$ is:

1. $\Big(\dfrac{1}{5}\Big)$

2. $\Big(\dfrac{4}{5}\Big)$

3. $\Big(\dfrac{1}{125}\Big)$

4. 25

Answer

Solving,

$\Rightarrow \Big(\dfrac{1}{5} \cot^2\theta + \dfrac{1}{5}\Big) \\[1em] \Rightarrow \dfrac{1}{5} \Big(\cot^2\theta + 1 \Big) \\[1em] \Rightarrow \dfrac{1}{5} (\cosec^2 \theta).$

We know that,

$\cosec \theta = \dfrac{1}{\sin \theta } \\[1em] \Rightarrow \cosec \theta = \dfrac{1}{\dfrac{1}{2}} = 2\\[1em] \Rightarrow \cosec^2 \theta = 2^2 = 4 \\[1em] \Rightarrow \dfrac{1}{5}\cosec^2 \theta \\[1em] \Rightarrow \dfrac{1}{5}(4) \\[1em] \Rightarrow \dfrac{4}{5}.$

Hence, option 2 is the correct option.

If cos θ = $\Big(\dfrac{2}{3}\Big)$, then 2 sec²θ + 2 tan²θ − 7 is equal to:

1. 0

2. 1

3. 3

4. 4

Answer

cos θ = $\dfrac{2}{3}$

sec θ = $\dfrac{3}{2}$

sec² θ = $\dfrac{9}{4}$

tan² θ = sec² θ - 1 = $\dfrac{9}{4} - 1 = \dfrac{5}{4}$

Given,

⇒ 2 sec²θ + 2 tan²θ − 7

$\Rightarrow 2\Big(\dfrac{9}{4}\Big) + 2\Big(\dfrac{5}{4}\Big) - 7 \\[1em] \Rightarrow \Big(\dfrac{18}{4}\Big) + \Big(\dfrac{10}{4}\Big) - 7 \\[1em] \Rightarrow \Big(\dfrac{28}{4}\Big) - 7 \\[1em] \Rightarrow 7 - 7 \\[1em] \Rightarrow 0.$

Hence, option 1 is the correct option.

If 24 cot θ = 7, then sin θ is equal to:

1. $\Big(\dfrac{24}{7}\Big)$

2. $\Big(\dfrac{24}{25}\Big)$

3. $\Big(\dfrac{7}{25}\Big)$

4. $\Big(\dfrac{25}{24}\Big)$

Answer

Given,

24 cot θ = 7

cot θ = $\dfrac{7}{24} = \dfrac{\text{adjacent}}{\text{opposite}}$

We know that,

$\Rightarrow \text{Hypotenuse} = \sqrt{\text{opposite}^2 + \text{adjacent}^2 } \\[1em] = \sqrt{(24)^2 + (7)^2} \\[1em] = \sqrt{576 + 49} \\[1em] = \sqrt{625} \\[1em] = 25.$

Now,

sin θ = $\dfrac{\text{opposite}}{\text{hypotenuse}} = \dfrac{24}{25}$

Hence, option 2 is the correct option.

If tan θ + cot θ = 2, then tan²θ + cot²θ is equal to:

1. 1

2. 2

3. 4

4. 8

Answer

Given,

tan θ + cot θ = 2

Square on both sides,

⇒ (tan θ + cot θ)² = 2²

⇒ tan² θ + cot² θ + 2 tan θ cot θ = 4

⇒ tan² θ + cot² θ + 2 tan θ $\dfrac{1}{\tan \theta}$ = 4

⇒ tan² θ + cot² θ + 2 = 4

⇒ tan² θ + cot² θ = 4 - 2

⇒ tan² θ + cot² θ = 2

Hence, option 2 is the correct option.

If cot A + $\Big(\dfrac{1}{\cot A}\Big)$ = 2, then cot²A + $\Big(\dfrac{1}{\cot^2 A}\Big)$ equals:

1. 0

2. 1

3. 2

4. 4

Answer

Given,

cot A + $\Big(\dfrac{1}{\cot A}\Big)$ = 2

Square on both sides,

$\Rightarrow \Big(\cot A + \dfrac{1}{\cot A}\Big)^2 = 2^2 \\[1em] \Rightarrow \cot^2 A + \dfrac{1}{\cot^2 A} + 2 = 4 \\[1em] \Rightarrow \cot^2 A + \dfrac{1}{\cot^2 A} = 4 - 2 \\[1em] \Rightarrow \cot^2 A + \dfrac{1}{\cot^2 A} = 2 .$

Hence, option 3 is the correct option.

If 4 tan θ = 3, then $\Big(\dfrac{4\sin\theta - 3\cos\theta}{4\sin\theta + 3\cos\theta}\Big)$ = ?

1. 0

2. $\Big(\dfrac{1}{3}\Big)$

3. $\Big(\dfrac{2}{3}\Big)$

4. $\Big(\dfrac{3}{4}\Big)$

Answer

tan θ = $\dfrac{3}{4}$

Then,

sin θ = $\dfrac{3}{5}$, cos θ = $\dfrac{4}{5}$

Substitute,

$\Rightarrow \dfrac{4\Big(\dfrac{3}{5}\Big) - 3\Big(\dfrac{4}{5}\Big)}{4\Big(\dfrac{3}{5}\Big) + 3\Big(\dfrac{4}{5}\Big)} \\[1em] \Rightarrow \dfrac{\Big(\dfrac{12}{5}\Big) - \Big(\dfrac{12}{5}\Big)}{\Big(\dfrac{12}{5}\Big) + \Big(\dfrac{12}{5}\Big)} \\[1em] \Rightarrow \dfrac{0}{\dfrac{24}{5}} \\[1em] \Rightarrow 0.$

Hence, option 1 is the correct option.

If tan θ = $\Big(\dfrac{1}{\sqrt{7}}\Big)$, then the value of $\Big(\dfrac{\cosec^2\theta + \sec^2\theta}{\cosec^2\theta - \sec^2\theta}\Big)$ is:

1. $\Big(\dfrac{3}{4}\Big)$

2. $\Big(\dfrac{4}{3}\Big)$

3. $\Big(\dfrac{3}{7}\Big)$

4. $\Big(\dfrac{4}{7}\Big)$

Answer

Given,

tan θ = $\Big(\dfrac{1}{\sqrt{7}}\Big)$

tan² θ = $\Big(\dfrac{1}{7}\Big)$

We know that,

$\sec^2 \theta = 1 + \tan^2 \theta \\[1em] \sec^2 \theta = 1 + \dfrac{1}{7} \\[1em] \sec^2 \theta = \dfrac{8}{7}. \\[1em] \cosec^2 \theta = 1 + \cot^2 \theta \\[1em] \cosec^2 \theta = 1 + 7 \\[1em] \cosec^2 \theta = 8.$

Given expression,

$\Rightarrow \Big(\dfrac{\cosec^2\theta + \sec^2\theta}{\cosec^2\theta - \sec^2\theta}\Big) \\[1em] \Rightarrow \dfrac{8 + \dfrac{8}{7}}{8 - \dfrac{8}{7}} \\[1em] \Rightarrow \dfrac{\dfrac{56 + 8}{7}}{\dfrac{56 - 8}{7}} \\[1em] \Rightarrow \dfrac{64}{48} = \dfrac{4}{3}.$

Hence, option 2 is the correct option.

(1 + sin A)(1 − sin A) is equal to:

1. cosec²A

2. sin²A

3. sec²A

4. cos²A

Answer

⇒ (1 + sin A)(1 − sin A)

⇒ 1 − sin² A

⇒ cos²A

Hence, option 4 is the correct option.

sin A expressed in terms of cot A is:

1. $\Big(\dfrac{1}{\sqrt{1 + \cot^2 A}}\Big)$

2. $\Big(\dfrac{\sqrt{1 + \cot^2 A}}{\cot A}\Big)$

3. $\Big(\dfrac{\sqrt{1 + \cot^2 A}}{1}\Big)$

4. $\Big(\dfrac{\sqrt{1 - \cot^2 A}}{\cot A}\Big)$

Answer

1 + cot² A = cosec² A

$\sqrt{1 + \cot^2 A}$ = cosec A

sin A = $\dfrac{1}{\cosec A}$

sin A = $\dfrac{1}{\sqrt{1 + \cot^2 A}}$

Hence, option 1 is the correct option.

If sec θ = 2x and y tan θ = 2, then the value of $2\Big(x^2 - \dfrac{1}{y^2}\Big)$ is:

1. $\Big(\dfrac{1}{2}\Big)$

2. $\Big(\dfrac{1}{3}\Big)$

3. $\Big(\dfrac{1}{4}\Big)$

4. 1

Answer

sec θ = 2x

x = $\dfrac{\sec \theta}{2}$

y tan θ = 2

$\dfrac{1}{y} = \dfrac{\tan \theta}{2}$

We have,

$\Rightarrow 2\Big(x^2 - \dfrac{1}{y^2}\Big) \\[1em] \Rightarrow 2\Big(\dfrac{\sec^2 \theta}{4} - \dfrac{\tan^2 \theta}{4}\Big) \\[1em] \Rightarrow \dfrac{2}{4}\Big(\sec^2 \theta - \tan^2 \theta\Big) \\[1em] \Rightarrow \dfrac{1}{2}\Big(\sec^2 \theta - \tan^2 \theta\Big) \\[1em] \Rightarrow \dfrac{1}{2}.$

Hence, option 1 is the correct option.

Given a = 3 sec² θ and b = 3 tan² θ - 2. The value of (a - b) is :

1. 1

2. 2

3. 3

4. 5

Answer

    a - b = 3 sec² θ - (3 tan² θ - 2)

= 3 sec² θ - 3 tan² θ + 2

= 3(sec² θ - tan² θ) + 2

= 3(1) + 2

= 3 + 2

= 5.

Hence, option 4 is the correct option.

(cos⁴ θ − sin⁴ θ) is equal to :

1. 2 cos² θ + 1

2. 2 cos² θ − 1

3. 2 sin² θ + 1

4. 2 sin² θ − 1

Answer

⇒ (cos⁴ θ − sin⁴ θ)

⇒ (cos² θ − sin² θ)(cos² θ + sin² θ)

⇒ (cos² θ - sin² θ)

⇒ (cos² θ - 1 + cos² θ)

⇒ (2cos² θ - 1)

Hence, option 2 is the correct option.

If cosec θ − cot θ = $\dfrac{1}{3}$, then the value of cosec θ + cot θ is :

1. 1

2. 2

3. 3

4. 4

Answer

We know that,

cosec² θ − cot² θ = 1

cosec θ − cot θ (cosec θ + cot θ)= 1

$\dfrac{1}{3}$ (cosec θ + cot θ)= 1

(cosec θ + cot θ)= 3

Hence, option 3 is the correct option.

If sin θ − cos θ = 0, then the value of sin θ + cos θ is :

1. $\dfrac{1}{\sqrt{2}}$

2. $\sqrt{2}$

3. $\dfrac{1}{4}$

4. $\dfrac{3}{4}$

Answer

sin θ = cos θ

Divide both sides by cos θ

$\dfrac{\sin \theta}{\cos \theta} = \dfrac{\cos \theta}{\cos \theta}$

tan θ = 1

For acute angles, tan θ = 1 when θ = 45°.

⇒ sin 45° + cos 45°

$\dfrac{1}{\sqrt2} + \dfrac{1}{\sqrt2}$

⇒ $\dfrac{2}{\sqrt2}$

⇒ $\sqrt2$

Hence, option 2 is the correct option.

If sec θ + tan θ + 1 = 0, then sec θ − tan θ is equal to :

1. −1

2. 0

3. 1

4. 2

Answer

⇒ sec θ + tan θ + 1 = 0

⇒ sec θ + tan θ = -1

We know that,

⇒ sec² θ - tan² θ = 1

⇒ (sec θ + tan θ)(sec θ - tan θ) = 1

⇒ (-1)(sec θ - tan θ) = 1

⇒ sec θ - tan θ = -1

Hence, option 1 is the correct option.

If sec θ + tan θ = x, then sec θ is equal to :

1. $\dfrac{x^2 + 1}{x}$

2. $\dfrac{x^2 − 1}{x}$

3. $\dfrac{x^2 + 1}{2x}$

4. $\dfrac{x^2 − 1}{2x}$

Answer

⇒ sec θ + tan θ = x ....(1)

We know that,

⇒ sec² θ − tan² θ = 1

⇒ (sec θ + tan θ)(sec θ − tan θ) = 1

⇒ x (sec θ − tan θ) = 1

⇒ sec θ − tan θ = $\dfrac{1}{x}$ ....(2)

Adding eqn (1) and (2):

⇒ sec θ + tan θ + sec θ − tan θ = x + $\dfrac{1}{x}$

⇒ 2 sec θ = $x + \dfrac{1}{x}$

⇒ sec θ = $\dfrac{1}{2}\Big(x + \dfrac{1}{x}\Big)$

⇒ sec θ = $\dfrac{1}{2}\Big(\dfrac{x^2 + 1}{x}\Big)$

⇒ sec θ = $\dfrac{x^2 + 1}{2x}$

Hence, option 3 is the correct option.

If sin θ − cos θ = 0, then the value of (sin⁴ θ + cos⁴ θ) is :

1. $\dfrac{1}{4}$

2. $\dfrac{1}{2}$

3. $\dfrac{3}{4}$

4. 1

Answer

⇒ sin θ − cos θ = 0

⇒ sin θ = cos θ

Divide by cos θ

⇒ $\dfrac{sin \theta}{\cos \theta}$ = 1

tan 45° = 1

Given expression,

(sin⁴ 45° + cos⁴ 45°)

$\Rightarrow \Big(\dfrac{1}{\sqrt2}\Big)^4 + \Big(\dfrac{1}{\sqrt2}\Big)^4 \\[1em] \Rightarrow \Big(\dfrac{1}{4}\Big) + \Big(\dfrac{1}{4}\Big) \\[1em] \Rightarrow \dfrac{1}{2}.$

Hence, option 2 is the correct option.

If a cot θ + b cosec θ = p and b cot θ + a cosec θ = q, then p² − q² is equal to :

1. a² − b²

2. b² − a²

3. a² + b²

4. b − a

Answer

⇒ a cot θ + b cosec θ = p

p² = (a cot θ + b cosec θ)²

p² = (a² cot² θ + b² cosec² θ + 2ab cot θ cosec θ)

⇒ b cot θ + a cosec θ = q

q² = (b cot θ + a cosec θ)²

q² = (b² cot² θ + a² cosec² θ + 2ab cot θ cosec θ)

⇒ p² − q² = (a² cot² θ + b² cosec² θ + 2ab cot θ cosec θ) - (b² cot² θ + a² cosec² θ + 2ab cot θ cosec θ)

= (a² cot² θ + b² cosec² θ + 2ab cot θ cosec θ - b² cot² θ - a² cosec² θ - 2ab cot θ cosec θ)

= (a² cot² θ + b² cosec² θ - b² cot² θ - a² cosec² θ )

= a² (cot² θ - cosec²) + b² (cosec² θ - cot² θ )

= a² (-1) + b² (1)

= b² - a².

Hence, option 2 is the correct option.

The expression equivalent to sec² θ + cosec² θ is :

1. sec² θ · cosec² θ

2. tan² θ + cot² θ

3. $\dfrac{1}{\sec^2 θ \times \cosec^2 θ}$

4. 2 sec² θ + 1

Answer

sec² θ + cosec² θ

$\Rightarrow \dfrac{1}{\cos^2 \theta} + \dfrac{1}{\sin^2 \theta} \\[1em] \Rightarrow \dfrac{\sin^2 \theta + \cos^2 \theta}{\cos^2 \theta\sin^2 \theta} \\[1em] \Rightarrow \dfrac{1}{\cos^2 \theta\sin^2 \theta} \\[1em] \Rightarrow \sec^2 \theta (\cosec^2 \theta)$

Hence, option 1 is the correct option.

If x = a cos³ θ and y = b sin³ θ, then $\Big(\dfrac{x}{a}\Big)^{\dfrac{2}{3}} + \Big(\dfrac{y}{b}\Big)^{\dfrac{2}{3}}$ is equal to :

1. a

2. b

3. 1

4. 2

Answer

Given,

x = a cos³ θ and y = b sin³ θ

$\Rightarrow \dfrac{x}{a} = \cos^3 θ \\[1em] \Rightarrow \Big(\dfrac{x}{a}\Big)^{\dfrac{2}{3}} = \cos^2 θ \\[1em] \Rightarrow \dfrac{y}{b} = \sin^3 θ \\[1em] \Rightarrow \Big(\dfrac{y}{b}\Big)^{\dfrac{2}{3}} = \sin^2 θ$

Add the expressions,

$\Big(\dfrac{x}{a}\Big)^{\dfrac{2}{3}} + \Big(\dfrac{y}{b}\Big)^{\dfrac{2}{3}}$ = cos² θ + sin² θ

$\Big(\dfrac{x}{a}\Big)^{\dfrac{2}{3}} + \Big(\dfrac{y}{b}\Big)^{\dfrac{2}{3}}$ = 1

Hence, option 3 is the correct option.

If sin θ + cosec θ = 2, then sin³ θ + cosec³ θ is equal to :

1. 2

2. 2 sin θ

3. −2 sin θ

4. 2 cos θ

Answer

sin θ + cosec θ = 2

Let,

⇒ sin θ = x

⇒ cosec θ = $\dfrac{1}{x}$

$\Rightarrow x + \dfrac{1}{x} = 2 \\[1em] \Rightarrow \dfrac{x^2 + 1}{x} = 2 \\[1em] \Rightarrow x^2 + 1 = 2x \\[1em] \Rightarrow x^2 + 1 - 2x = 0 \\[1em] \Rightarrow (x - 1)^2 = 0 \\[1em] \Rightarrow x = 1$

Hence,

sin θ = 1 and cosec θ = 1

⇒ sin³ θ + cosec³ θ

⇒ 1³ + 1³

⇒ 2

Hence, option 1 is the correct option.

The expression equivalent to sec x, is :

1. $\tan x + \dfrac{\sin x}{\sec x}$

2. $\cos x + \dfrac{\tan x}{\cos x}$

3. cos x + tan x sin x

4. tan x − cos x sin x

Answer

Solving for option 3,

cos x + tan x sin x

$\Rightarrow \cos x + \dfrac{\sin x}{\cos x} \times \sin x \\[1em] \Rightarrow \cos x + \dfrac{\sin^2 x}{\cos x} \\[1em] \Rightarrow \dfrac{\cos^2 x + \sin^2 x}{\cos x} \\[1em] \Rightarrow \dfrac{1}{\cos x} \\[1em] \Rightarrow \sec x$

Hence, option 3 is the correct option.

$\dfrac{\sin^4 A − \cos^4 A}{1 − \sin^2 A}$ is equal to :

1. cot² A − 1

2. tan² A − 1

3. 1 − cot² A

4. 1 − tan² A

Answer

$\Rightarrow \dfrac{\sin^4 A − \cos^4 A}{1 − \sin^2 A} \\[1em] \Rightarrow \dfrac{(\sin^2 A − \cos^2 A)(\sin^2 A + \cos^2 A)}{\cos^2 A} \\[1em] \Rightarrow \dfrac{\sin^2 A − \cos^2 A}{\cos^2 A} \\[1em] \Rightarrow \dfrac{\sin^2 A }{\cos^2 A} - 1 \\[1em] \Rightarrow \tan^2 A - 1 .$

Hence, option 2 is the correct option.

If sin A + sin² A = 1, then cos² A + cos⁴ A is :

1. $\dfrac{1}{2}$

2. 1

3. 2

4. 3

Answer

Given,

sin A + sin² A = 1

It can be written as

sin A = 1 - sin² A …. (1)

We have to find the value of (cos² A + cos⁴ A)

Using the trigonometric identities,

cos² A = 1 - sin² A ….. (2)

From both the equations

sin A = cos² A

Now, (cos² A + cos⁴ A) = (cos² A + (sin A)²)

= cos² A + sin²A

cos² A + sin² A = 1

Therefore, (cos² A + cos4 A) = 1

Hence, option 2 is the correct option.

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

1. 1

2. 2

3. 3

4. 4

Answer

cos A + cos² A = 1

⇒ 1 - cos² A = cos A

⇒ sin² A = cos A

Given,

⇒ sin² A + sin⁴ A

⇒ sin² A + (sin² A)²

⇒ sin² A + cos² A [∵ sin² A = cos A]

⇒ 1.

Hence, option 1 is the correct option.

$\sqrt{\Big(\dfrac{1 − \sin A}{1 + \sin A}\Big)}$ = ?

1. sec A + tan A

2. sec A − tan A

3. sec A tan A

4. none of these

Answer

Rationalize the expression,

$\Rightarrow \sqrt{\Big(\dfrac{1 − \sin A}{1 + \sin A}\Big)} \\[1em] \Rightarrow \sqrt{\Big(\dfrac{1 − \sin A}{1 + \sin A}\Big) \times \dfrac{1 - \sin A}{1 - \sin A}} \\[1em] \Rightarrow \sqrt{\Big(\dfrac{(1 − \sin A)^2}{1 - \sin^2 A}\Big)} \\[1em] \Rightarrow \sqrt{\Big(\dfrac{(1 − \sin A)^2}{\cos^2 A}\Big)} \\[1em] \Rightarrow \dfrac{(1 − \sin A)}{\cos A} \\[1em] \Rightarrow \dfrac{1}{\cos A} − \dfrac{\sin A}{\cos A} \\[1em] \Rightarrow \sec A − \tan A$

Hence, option 2 is the correct option.

Statement 1 : sin² θ + cos² θ = 1

Statement 2 : cosec² θ + cot² θ = 1

Which of the following is valid ?

1. only (1)

2. only (2)

3. both (1) and (2)

4. neither (1) nor (2)

Answer

Trigonometry identity :

sin² θ + cos² θ = 1

cosec² θ - cot² θ = 1

∴ Only statement (i) is correct.

Hence, Option 1 is the correct option.