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• The same part of animals used by different ways by different animals. 
• Ex: Tongue used by dog in a different manner as compared to frog. 
• The dog licks with its tongue while the frog captures and swallows it.

• The same type of food is taken by different animals by using their different body parts. 
• Ex: Insects are the food for hen and frog, but the body parts are different to take in. Hens use their beak to pick insects, while frogs use their tongue to grab the insect into the mouth.

Reptiles: The vertebrates (having vertebral column) that crawl or move on their belly or on short legs. Ex: Snake, Lizard.

Jaws, teeth, and tongue are involved in eating its food.

Natural scavenger: An organism or animal that feeds on waste, dead organic matter. They keep their surroundings clean in this manner. Ex: Crow, Vultures, Fungi, Microorganisms.

The correct answer is C) Crow

Crows, Vultures are examples for natural scavengers.

Given if x < 0, y < 0 such that xy = 1 

Also given tan^-1 x + tan^-1 y 

We know that tan^-1 x+ tan^-1 y = tan-1 \((\frac{x+y}{1-xy})\)

\(=-\pi+tan^{-1}(\frac{x+y}{1-xy})\)

\(=-\pi+tan^{-1}(\frac{x+y}{1-1})\)

= -π + tan^-1(∞)

\(=-\pi+\frac{\pi}{2}\)

=\(-\frac{\pi}{2}\)

\(\therefore tan^{-1}x+tan^{-1}y=-\frac{\pi}{2}\)

Given cos^-1(cos 6)

We know that

cos^-1 (cos θ) = 2π – θ, if θ ∈ [π, 2π]

= 2π – 6

∴ cos^-1 (cos 6) = 2π - 6

Given sin^-1 (cos π/9)

We know that cos θ = sin (π/2 – θ)

= sin^-1 (sin (π/2 - π/9))

= sin^-1 (sin 7π/18)

We know that sin^-1 (sin θ) = θ = 7π/18

∴ sin^-1 (cos π/9) = 7π/18

Given sin^-1 (1/3) + cos^-1 x = π/2

⇒ sin^-1 (1/3) = π/2 – cos^-1 x

We know that sin^-1 x + cos^-1 x = π/2

⇒ sin^-1 (1/3) = sin^-1 x

∴ x = 1/3

Given sin (π/3 – sin^-1 (-1/2))

We know that sin^-1 (-θ) = -sin^-1 θ

= sin (π/3 + sin^-1 (1/2) 0)

= sin (π/3 + π/6)

= sin (π/2)

= 1

∴ sin (π/3 – sin^-1 (-1/2)) = 1

(i) Given as tan^-1(tan 2)

As tan^-1(tan x) = x if x ϵ [-π/2, π/2]

Here x = 2 which does not belongs to above range

We have tan (π – θ) = –tan (θ)

So, tan (θ – π) = tan (θ)

tan (2 – π) = tan (2)

Now 2 – π is in the given range

Hence, tan^–1 (tan 2) = 2 – π

(ii) Given as tan^-1(tan 4)

As tan^-1(tan x) = x if x ϵ [-π/2, π/2]

But here x = 4 which does not belongs to above range

We also have tan (π – θ) = –tan (θ)

So, tan (θ – π) = tan (θ)

tan (4 – π) = tan (4)

Now 4 – π is in the given range

Thus, tan^–1 (tan 2) = 4 – π

(iii) Given as tan^-1(tan 12)

As tan^-1(tan x) = x if x ϵ [-π/2, π/2]

Here x = 12 which does not belongs to above range

As we know that tan (nπ – θ) = –tan (θ)

tan (θ – 2nπ) = tan (θ)

Here n = 4

tan(12 – 4π) = tan (12)

Now 12 – 4π is in the given range

Therefore, tan^–1(tan 12) = 12 – 4π.

As cos^–1(cos x) = x

Provided x ∈ [0,π] 

∴ we can write cos^–1(cos 3) as 3.

sin x = sin^-1 x is possible only when x = 0 (∵ x ∈ R)

Let cos^-1 (1/2) = π

⇒ cos π = (1/2) = cos π/3

⇒ π = π/3

⇒ cos^-1 (1/2) = π/3

cos^–1(cos x) = x

Provided x ϵ [0,π] ≈ [0,3.14]

And in our equation x is 4 which does not lie in the above range.

We know cos[2π – x] = cos[x]

∴ cos(2π – 4) = cos(4)

Also 2π–4 belongs in [0,π]

∴ cos^–1(cos 4) = 2π–4

d) Receipts and Payment A/c

cos^–1(cos x) = x

Provided x ϵ [0,π] ≈ [0,3.14]

And in our equation x is 5 which does not lie in the above range.

We know cos[2π – x] = cos[x]

∴ cos(2π – 5) = cos(5)

Also 2π–5 belongs in [0,π]

∴ cos^–1(cos 5) = 2π–5

d) A liability

Question:-

Subscription received in advance is ………… 

a) Income  b) Expense  c) An asset  d) A liability

Answer:-

Subscription recieved in advance is a liability.

sin^–1(sin x) = x

Provided x ∈ \([\frac{-\pi}2,\frac{\pi}2]\) ≈ [–1.57,1.57]

And in our equation x is 4 which does not lie in the above range.

We know sin[π – x] = sin[–x]

∴ sin(π – 4) = sin(–4)

Also π–4 belongs in \([\frac{-\pi}2,\frac{\pi}2]\)

∴ sin^–1(sin 4) = π – 4

a) Capital Receipt, 

b) Capital Receipt

Corected Question:-

State the following items as Revenue Receipt or Capital Receipt  

a) Specific Donation

b)Legacies

Answer:-

Specific Donation:-Specific Donation is a capital reciept.

Legacies:-Legacy is also a capital receipt.

(π/2) < cos^-1 (3x -1) < π 

⇒ cos (π/2) < 3x – 1 < cos π 

⇒ 0 < 3x – 1 < -1 

⇒ 1 < 3x < 0 

⇒ (1/3) < x < 0 

This inequality holds only if x < 0 or x > (1/3)

cosec \((\frac {6\pi}5)\) can be written as cosec (π + \(\frac{\pi}5\))

cosec  (π + \(\frac{\pi}5\)) = - cosec\((\frac \pi 5)\)

Also,

–cosec(θ) = cosec(–θ)

⇒ –cosec\((\frac{\pi}5)\)= cosec\((\frac{-\pi}5)\)

Now the question becomes cosec^-1(cosec\((\frac{-\pi}5)\))

cosec^-1(cosec x) = x

Provided x ∈ \([\frac{-\pi}2,\frac{\pi}2]\) - {0}

∴ we can write cosec –1(cosec\((\frac{-\pi}5)\)) = \(\frac{-\pi}5.\)

cos^–1(cos x) = x

Provided x ϵ [0,π] ≈ [0,3.14]

And in our equation x is 4 which does not lie in the above range.

We know cos[2nπ – x] = cos[x]

∴ cos(2nπ – 12) = cos(12)

Here n = 2.

Also 4π–12 belongs in [0,π]

∴ cos^–1(cos 12) = 4π–12

(a) 0

The maximum value of sin^-1 x is π/2 and sin^-1 1 = π/2

Here it is given that 

sin^-1 x + sin^-1 y + sin^-1 z = 3π/2

⇒ x = y = z = 1

and so 1 + 1 + 1 – 9/(1 + 1+ 1) = 3 - 3 = 0

(i) cos x = 0 

⇒ x = (2n + 1) ± π/2

n = 0, ±1, ±2, ±3, ±4, ±5 

(ii) cos x = 1 = cos 0 

⇒ x = 2nπ ± 0 

n = 0, ±1, ±2

Tan\(\frac{6\pi}7\) can be written as tan\((\pi-\frac{\pi}7)\)

tan\((\pi-\frac{\pi}7)\) = –tan\(\frac{\pi}7\)

∴ As, tan^–1(tan x) = x

Provided x ϵ \((\frac{-\pi}2,\frac{\pi}2)\)

tan ^–1(tan\(\frac{6\pi}7\)) = –\(\frac{\pi}7\)

sin^–1(sin x) = x

Provided x ∈ \([\frac{-\pi}2,\frac{\pi}2]\) ≈ [–1.57,1.57]

And in our equation x is 3 which does not lie in the above range.

We know sin[π – x] = sin[x]

∴ sin(π – 3) = sin(3)

Also π–3 belongs in \([\frac{-\pi}2,\frac{\pi}2]\)

∴ sin^–1(sin3) = π–3

\(sin^ -1(sin-3)=0.11874839215\)

(b) 0 ≤ x ≤ π

The value of tan\(\frac{7\pi}6\) = \(\frac{1}{\sqrt3}\)

∴ The question becomes tan^–1\((\frac{1}{\sqrt3})\)

Let,

tan^–1\((\frac{1}{\sqrt3})\) = y

⇒ tan y =\((\frac{1}{\sqrt3})\)

⇒ tan\((\frac \pi{6})=(\frac{1}{\sqrt3})\)

The range of the principal value of tan^–1 is \((\frac{-\pi}2,\frac{\pi}2)\) and tan\((\frac \pi{6})=(\frac{1}{\sqrt3})\).

∴ The value of tan^–1(tan\(\frac{7\pi}6\)) is \(\frac{\pi}6\).

(i) Given as tan^-1(tan π/3)

As tan^-1(tan x) = x if x ϵ [-π/2, π/2]

On, applying this condition in the given question we get,

tan^-1(tan π/3) = π/3

(ii) Given as tan^-1(tan 6π/7)

As we know that tan 6π/7 can be written as (π – π/7)

tan(π – π/7) = – tan π/7

As we know that tan^-1(tan x) = x if x ϵ [-π/2, π/2]

tan^-1(tan 6π/7) = – π/7

(iii) Given as tan^-1(tan 7π/6)

As we know that tan 7π/6 = 1/√3

On, substituting this value in tan^-1(tan 7π/6) we get,

tan^-1 (1/√3)

Let tan^-1 (1/√3) = y

tan y = 1/√3

tan (π/6) = 1/√3

So, the range of the principal value of tan^-1 is (-π/2, π/2) and tan (π/6) = 1/√3

So, tan^-1(tan 7π/6) = π/6

(iv) Given as tan^-1(tan 9π/4)

As we know that tan 9π/4 = 1

On, substituting this value in tan^-1(tan 9π/4) we get,

tan^-1 (1)

Let tan^-1 (1) = y

tan y = 1

tan (π/4) = 1

So, the range of the principal value of tan^-1 is (-π/2, π/2) and tan (π/4) = 1

So, tan^-1(tan 9π/4) = π/4

(v) Given as tan^-1(tan 1)

we have tan^-1(tan x) = x if x ϵ [-π/2, π/2]

On, substituting this condition in given question

tan^-1(tan 1) = 1

As we know sin(–θ) is –sin(θ )

∴ We can write \((sin\frac{-17\pi}8)\) as \(-sin(\frac{17\pi}8)\)

Now \(-sin(\frac{17\pi}8)\) = \(-sin(2\pi+\frac{\pi}8)\)

As we know sin(2π +θ) = sin(θ )

So \(-sin(2\pi+\frac{\pi}8)\) can be written as \(-sin(\frac{\pi}8)\)

And \(-sin(\frac{\pi}8)\) = \(sin(\frac{-\pi}8)\)

As sin^–1(sin x) = x

Provided x ∈ \([\frac{-\pi}2,\frac{\pi}2]\)

∴ we can write \(sin^{-1}(sin \frac{-\pi}8)=\frac{-\pi}8\) 

(c) [-2,-√2] ∪ [√2,2]

f(x) = sin^-1 (x² – 3) 

Domain of sin^-1 (x) is [-1, 1] 

⇒ -1 ≤ x² – 3 ≤ 1 ⇒ 2 ≤ x² ≤ 4

⇒ √2 ≤ x ≤ 2 ⇒ √2 ≤ |x| ≤ 2

x ∈ [-2,-√2] ∪ [√2,2]

sin^–1(sin x) = x

Provided x ∈\([\frac{-\pi}2,\frac{\pi}2]\)≈ [–1.57,1.57]

And in our equation x is 4 which does not lie in the above range.

We know sin[2nπ – x] = sin[–x]

∴ sin(2nπ – 12) = sin(–12)

Here n = 2

Also 2π–12 belongs in \([\frac{-\pi}2,\frac{\pi}2]\)

∴ sin^–1(sin12) = 2π – 12

We can write \((sin\frac{13\pi}7)\) as \(sin(2\pi-\frac\pi{7})\)

As we know sin(2π –θ) = sin(–θ )

So, \(sin(2\pi-\frac\pi{7})\) can be written as \(sin(\frac\pi{7})\)

∴ The equation becomes sin^–1\((sin\frac{\pi}7)\)

As sin^-1(sin x) = x

Provided x ∈ \([\frac{-\pi}2,\frac{\pi}2]\)

∴ we can write \(sin^{-1}(sin\frac{\pi}7)=\frac{\pi}7\)

sin^–1(sin x) = x

Provided x ∈\([\frac{-\pi}2,\frac{\pi}2]\) ≈ [–1.57,1.57] 

And in our equation x is 3 which does not lie in the above range.

We know sin[π – x] = sin[x]

∴ sin(π – 2) = sin(2)

Also π–2 belongs in \([\frac{-\pi}2,\frac{\pi}2]\)

∴ sin^–1(sin2) = π–2

The principal value of sin-1(1/2) and sin-1(-1/√2) are

π/6 and -π/4

(i) Given as sin^-1(sin π/6)

As we know that the value of sin π/6 is ½

On, substituting these value in sin^-1(sin π/6)

We get, sin^-1 (1/2)

Let y = sin^-1 (1/2)

sin (π/6) = ½

So, the range of principal value of sin^-1(-π/2, π/2) and sin (π/6) = ½

So, sin^-1(sin π/6) = π/6

(ii) Given as sin^-1(sin 7π/6)

We know that sin 7π/6 = – ½

On, substituting these in sin^-1(sin 7π/6) we get,

sin^-1 (-1/2)

Let y = sin^-1 (-1/2)

– sin y = ½

– sin (π/6) = ½

– sin (π/6) = sin (- π/6)

So, the range of principal value of sin^-1(-π/2, π/2) and sin (- π/6) = – ½

So, sin^-1(sin 7π/6) = – π/6

(iii) Given as sin^-1(sin 5π/6)

As we know that the value of sin 5π/6 is ½

Substitute this value in sin^-1(sin π/6)

We get, sin^-1 (1/2)

Let y = sin^-1 (1/2)

sin (π/6) = ½

So, the range of principal value of sin^-1(-π/2, π/2) and sin (π/6) = ½

So, sin^-1(sin 5π/6) = π/6

(iv) Given as sin^-1(sin 13π/7)

The given trigonometry form can be written as sin (2π – π/7)

sin (2π – π/7) can be written as sin (π/7) [since sin (2π – θ) = sin (-θ)]

On, substituting these values in sin^-1(sin 13π/7) we get sin^-1(sin – π/7)

As sin^-1(sin x) = x with x ∈ [-π/2, π/2]

So, sin^-1(sin 13π/7) = – π/7

(v) Given as sin^-1(sin 17π/8)

Given can be written as sin (2π + π/8)

sin (2π + π/8) can be written as sin (π/8)

On, substituting these values in sin^-1(sin 17π/8) we get sin^-1(sin π/8)

As sin^-1(sin x) = x with x ∈ [-π/2, π/2]

So, sin^-1(sin 17π/8) = π/8

The principal value of cos^-1(-√3/2) and cos^-1(-1/2) are

5π/6 and 2π/3

(D) constant angle

A radian is a constant angle

The answer is 3π/4

(A) \(\frac{5\pi^c}{12}\) 

1° = \((\frac{\pi}{180})^c\) 

∴ 75° = \((75\times\frac{\pi}{180})^c\)

= \(\frac{5\pi^c}{12}\)

(i) y = sin 7x 

Period of the function sin x is 2π 

Period of the function sin 7x is 2π/7

The amplitude of sin 7x is 1 

(ii) y = -sin (1/3)x 

Period of sin x is 2π 

So, period of sin (1/3)x is 6π and the amplitude is 1.

(iii) y = 4 sin(-2x) = -4 sin 2x 

Period of sin x is 2π 

π Period of sin 2x is π and the amplitude is 4.

= cot (\(\frac{\pi}2\))

(\(\because\) tan^-1 θ + cot^-1 θ = \(\frac{\pi}2\) )

= 0

(A) \((\frac{4\pi}3)^c\) 

240° = \((240\times\frac{\pi}{180})^c\)

= \((\frac{4\pi}3)^c\)

(i) sin x = 0 

⇒ x = nπ 

where n = 0, ±1, ±2, ±3, ……., ±10 

(ii) sin x = -1 

⇒ x = (4n – 1) (π/2), n = 0, ±1, ±2, ±3, 4

Correct Answer is \(\frac{2\pi}{5}\)

Now, let x = \(sec^{-1}\left(sec\frac{8\pi}{5}\right)\)

⇒ sec x =sec ( \(\frac{8\pi}{5}\)) 

Here range of principle value of sec is [0, π] 

⇒ x = \(\frac{8\pi}{5}\)∉ [0, π] 

Hence for all values of x in range [0, π] ,the value of 

\(sec^{-1}\left(sec\frac{8\pi}{5}\right)\) is 

⇒ sec x =sec (2π - \(\frac{8\pi}{5}\) ) ( \(\because\)sec ( \(\frac{8\pi}{5}\))= sec (2π -  \(\frac{2\pi}{5}\)) ) 

⇒ sec x =sec ( \(\frac{2\pi}{5}\)) ( \(\because\)sec (2π - θ)= sec θ) 

⇒ x= \(\frac{2\pi}{5}\)

cot \(\frac{4\pi}3\) can be written as cot\((\pi+\frac{\pi}3)\)

We know cot(π + θ) = cot (θ)

∴  cot\((\pi+\frac{\pi}3)\) = cot \((\frac{\pi}3)\)

Now the question becomes cot^-1(cot \(\frac{\pi}3\))

cot^-1(cot x) = x

Provided x ∈ (0, π)

∴ cot^–1(cot \(\frac{4\pi}3\)) = \(\frac{\pi}3.\)

(B) \((\frac{-13\pi}{9})^c\) 

-260° = \((-260\times\frac{\pi}{180})^c\) 

= \(\frac{-13\pi}{9}^c\)