Inverse Trigonometric Functions

We know that, and

[since, cos ]

[since, tan^-1 (-x)=- tan^-1 x, x ∈ R and cos^-1 (-x)=π-cos^-1 (x), x ∈(-1, 1)]

[since, cos ]

We have

[Since, ]

[Since, ]

=cos π

We have to prove, = 7

[Since, ]

[Since, ]

LHS=RHS

Hence Proved.

We have,

=

=

[Since, ; ;

and ]

=

=

=

=

We have

=

[ Since, , ]

=

[ Since, , ]

=

We have to prove, .

LHS, [ Since, , ]

= [ Since, ]

=

=

= [ Since, ]

=

= =

=

[ Since, ]

=

= = RHS

Hence Proved.

We have, ……(i)

Let =

[Since, ]

On putting the value of in Eq. (i), we get

……(ii)

We know that,

, xy

So, (ii) becomes,

=

=

or

or

or

or x = -1

or

For real solution, we have x = 0, -1.

We have,

[Since, ]

[Since,

[Since, ]

We have,

[Since, ]

Hence Proved.

We have,

[Since,]

[Since, ]

Since, LHS=RHS

Hence Proved.

We have,

-----(i)

Let

-----(a)

----(c)

And,

---(b)

----(d)

From (c), (d) ; (i) becomes

[from (a), (b)]

On squaring both sides, we get

We have,

LHS, -------(i)

[let ]

And,

LHS

[Since, tan(x+y) =]

=RHS

LHS=RHS

Hence Proved

Let cos y =

[Since, cos(A-B)= cos A. Cos B + sin A. sin B]

[Since,

[ Since, ]

We have

LHS

Let

And,

[Since, ]

Let

Hence Proved.

Solving LHS,

Let

And,

----(i)

Again, let

-----(ii)

We know that,

[from(i), (ii)]

= RHS

Since, LHS=RHS

Hence Proved.

Solving LHS,

Let

Squaring both sides,

Since,

Again,

Let

Squaring both sides,

Since,

We know that, sin(x+y) = sin x. cos y + cos x. sin y

=RHS

Since, LHS=RHS

Hence Proved.

We have,

[since, ]

[since, ]

[Since, ]

.

Hence,

Solving LHS,

Let

Let

{but as we can see, , since }

= RHS

NOTE: Since,

We have

And,

Given that,

[Since, ]

[Since, ]

We know that the principal value branch of is

We know that the principal value branch of is

Given that,

[Since, ]

[Since, ]

Cross multiplying,

Here only x=1 satisfies the given equation.

NOTE: Here, putting x=-1 in the given equation we get,

Hence, x=-1 does not satisfy the given equation.

We have,

[Since, ]

[Since, ]

[Since, ]

We have

Since,

[Since, ]

Given,

Let,

So, cos θ = 0 … (1)

Principal value cos^-1 x is [0, π] … (2)

Also, we know that … (3)

From (1), (2) and (3), we have

But

So,

We know that

As,

So,

sin (2tan^–1 (.75))

Let, tan^–1 (.75) = θ

As, , so

Now,

sin (2tan^–1 (.75)) = sin 2θ

= 2 sin θ cos θ

So, sin (2tan^–1 (.75)) = 0.96.

We have,

We know that,

So,

Let, cos^-1 0 = θ

⇒ cos θ = 0

Principal value of cos^-1 x is [0, π]

For, cos θ = 0

So,

We have,

Principal value of sin^-1 x is

Principal value of sec^-1 x is [0, π]-

Let

So, … (1)

Let sec^-1 2 = B

⇒ sec B = 2

So, 2 sec^-1 2 = 2B

So, the value of from (1) and (2) is

So,

We know that,

We have,

tan^–1 x + tan^–1 y = 4π/5 … (1)

Let, cot^–1x + cot^–1 y = k … (2)

Adding (1) and (2) –

Now, tan^–1 A + cot^–1 A = π/2 for all real numbers.

So, (tan^–1 x + cot^–1 x) + (tan^–1y + cot^–1 y) = π … (4)

From (3) and (4), we get,

We have,

We know that,

From (1) and (2) we have,

L.H.S-

From (3), R.H.S-

So, we have 4 tan^-1 a = 2 tan^-1 x

⇒ 2 tan^-1 a = tan^-1 x

But from (3)

So,

We have to find,

Let,

Also,

As,

So,

We need to find cot A

So,

We need to find,

Let,

Also, we need to find

We know that,

So,

On rationalizing,

Again rationalizing,

We need to find,

We know that,

So,

=4 tan^-1x

Given, cos^–1α + cos^–1β + cos^–1γ = 3π … (1)

Principal value of cos^-1 x is [0, π]

So, maximum value which cos^-1 x can have is π.

So, if (1) is correct then all the three terms i.e,

cos^–1α, cos^–1β, cos^–1γ should be equal to π

So, cos^–1α = π

cos^–1β = π

cos^–1γ = π

So, α = β = γ = -1

So, α(β + γ) + β (γ + α) + γ (α + β)

= (-1)(-1-1) + (-1)(-1-1) + (-1)(-1-1)

= 3(-1)(-2)

= 6

We have,

R.H.S-

So,

Squaring both sides, we get,

(1 + cos 2x) = 2x²

⇒ cos 2x = 2x² – 1

Now, plotting cos 2x and 2x² – 1, we get,

As, there is no point of intersection in , so there is no

solution of the given equation in .

Plotting cos^-1 x and sin^-1 x, we get,

As, graph of cos^-1 x is above graph of sin^-1 x in .

So, cos^–1x > sin^–1 x for all x in .

The principal value of is .

Principal value of cos^-1 x is [0, π]

Let,

As,

So,

The value of is .

Principal value of sin^-1 x is

Now, should be .

is outside the range

As, sin (π – x) = sin x

So,

If cos (tan^–1x + cot^–1 √3) = 0, then value of x is

Given, cos (tan^–1x + cot^–1 √3) = 0

We know that,

So,

The set of values of is .

Domain of sec^-1 x is R – (-1,1).

As, is outside domain of sec^-1 x.

Which means there is no set of value of .

So, the solution set of is null set or

The principal value of tan^–1 √3 is .

Principal value of tan^-1 x is

Let,

As,

So,

The value of is

We need,

Principal value of cos^-1 x is [0,π]

Also, cos (2nπ + θ) = cos θ for all n Є N

So,

The value of cos (sin^–1 x + cos^–1 x) for |x| ≤ 1 is 0.

cos (sin^–1 x + cos^–1 x), |x| ≤ 1

We know that, (sin^–1 x + cos^–1 x), |x| ≤ 1 is

So,

= 0

The value of expression when is 1.

when

We know that, (sin^–1 x + cos^–1 x) for all |x| ≤ 1 is

As, lies in domain.

So,

= 1

If y = 2 tan^–1 x + sin^–1 for all x, then -2π < y < 2π.

We know that,

So,

=4 tan^-1 x

So, y = 4 tan^-1 x

As, principal value of tan^-1 x is

So,

Hence, -2π < y < 2π

The result tan^–1 x – tan^–1 is true when value of xy is > -1.

We have,

Principal range of tan^-1a is

Let tan^-1x = A and tan^-1y = B … (1)

So, A,B ϵ

We know that, … (2)

From (1) and (2), we get,

Applying, tan^-1 both sides, we get,

As, principal range of tan^-1a is .

So, for tan^-1tan(A-B) to be equal to A-B,

A-B must lie in – (3)

Now, if both A,B < 0, then A, B ϵ

∴ A ϵ and -B ϵ

So, A – B ϵ

So, from (3),

tan^-1tan(A-B) = A-B

Now, if both A,B > 0, then A, B ϵ

∴ A ϵ and -B ϵ

So, A – B ϵ

So, from (3),

tan^-1tan(A-B) = A-B

Now, if A > 0 and B < 0,

Then, A ϵ and B ϵ

∴ A ϵ and -B ϵ

So, A – B ϵ (0,π)

But, required condition is A – B ϵ

As, here A – B ϵ (0,π), so we must have A – B ϵ

Applying tan on both sides,

As,

So, tan A < - cot B

Again,

So,

⇒ tan A tan B < -1

As, tan B < 0

xy > -1

Now, if A < 0 and B > 0,

Then, A ϵ and B ϵ

∴ A ϵ and -B ϵ

So, A – B ϵ (-π,0)

But, required condition is A – B ϵ

As, here A – B ϵ (0,π), so we must have A – B ϵ

Applying tan on both sides,

As,

So, tan B > - cot A

Again,

So,

⇒ tan A tan B > -1

⇒xy > -1

The value of cot^–1(–x) for all x ϵ R in terms of cot^–1 x is

π – cot^-1 x.

Let cot^–1(–x) = A

⇒ cot A = -x

⇒ -cot A = x

⇒ cot (π – A) = x

⇒ (π – A) = cot^-1 x

⇒ A = π – cot^-1 x

So, cot^–1(–x) = π – cot^-1 x

True.

It is well known that, all trigonometric functions have inverse

over their respective domains.

False

As, cos^-1 x is not equal to sec x. So, (cos^–1x)² is not equal to

sec² x.

True

As, all trigonometric and their corresponding inverse functions

are periodic so, we can obtain the inverse of a trigonometric

ratio in any branch in which it is one-one and onto.

True

We know that the smallest value, either positive or negative of

angle θ is called principal value of the inverse trigonometric

function.

True.

Graph of any inverse function can be obtained by interchanging

x and y axis in the graph of corresponding function. If (p, q) are

two points on f(x) then (q, p) will be on f^-1(x).

False

As, tan is an increasing function, so applying tan on both sides

we get,

As,

So,

⇒ n > π

⇒ n > 3.14

As, n is a natural number, so least value of n is 4.

True

Principal value of sin^-1 x is

Principal value of cos^-1 x is [0, π]

We have,

As, , so,

As, , so,

As, , so,