Inverse Trigonometric Functions

Let

Then sin y =

We know that the principal value of is

Therefore the principal value of is

Let

cos y =

We need to find the value of y.

We know that the value of cos is negative for the second quadrant and hence the value lies in [0, π].

cos y = – cos

cos y =

y =

Let

Then

We know that the principal value of is

Therefore the principal value of is .

Let y =

Therefore, sin y =

We know that the principal value of is

And

Therefore the principal value of is .

Domain of lies in the interval [–1, 1].

Therefore domain of lies in the interval [–1, 1].

–1 1

But cannot take negative values,

So, 0 1

–1 1

Hence domain of is [–1, 1].

Domain of lies in the interval [–1, 1].

–1 1.

The domain of sin x lies in the interval

–1.57 1.57

From the above we can see that the domain of sin^–1x + sinx is the intersection of the domains of sin^–1x and sin x.

So domain of sin^–1x + sinx is [–1, 1].

Domain of lies in the interval [–1, 1].

Therefore, Domain of lies in the interval [–1, 1].

–1 1

0 1

1 2

and

Domain of is [– [1,.

Domain of lies in the interval [–1, 1].

–1 1

Therefore, the domain of lies in the interval

–1 1

The domain of sin^–1x + sin^–12x is the intersection of the domains of sin^–1x and sin^–12x.

So, Domain of sin^–1x + sin^–12x is .

Range of sin^–1x is .

Give that sin^–1x + sin^–1y + sin^–1z + sin^–1t = 2π

Each of sin^–1x, sin^–1y, sin^–1z, sin^–1t takes value of .

So,

x = 1, y = 1, z = 1 and t = 1.

Hence,

= x² + y² + z² + t²

= 1 + 1 + 1 + 1

= 4

Range of sin^–1x is .

Given that

Each of , and takes the value of .

x = 1, y = 1, and z = 1.

Hence,

= x² + y² + z²

= 1 + 1 + 1

= 3.

=

We know,

=

= 0

=

=

=

=

=

=

= 1

=

= 0

=

=

= 0

A.

⇒

⇒

⇒

⇒

A. …eq(i)

…eq(ii)

Subtracting (ii) from (i)

⇒

⇒

⇒

⇒

⇒

⇒

⇒

⇒

⇒

⇒

Now, putting the value of in eq(ii)

⇒

⇒

⇒

⇒ and

A.

⇒

⇒

⇒

⇒

(using sin(A-B) = sinAcosB – cosAsinB)

⇒

A. Using a²+b² = (a+b)² - 2ab

⇒

⇒

Substituting sin^-1x with ‘a’

⇒

⇒

⇒

Using quadratic formulae

⇒

⇒

A.

⇒

⇒

⇒

⇒

(using cos(A-B) = cosAcosB + sinAsinB)

⇒

(The value of switches between 1 and -1)

A.

⇒

⇒

⇒

⇒

A.

⇒

⇒

⇒

⇒

A.

⇒

⇒

⇒

A.

⇒

⇒

⇒

⇒

Given:-

Take

LHS

=

We know that, Formula

Thus,

=

=

=

=

= RHS

Hence, Proved.

Given:-

Take

LHS

We know that, Formula

Thus,

=

=

We know that, Formula

Thus,

=

=

We know that, Formula

=

= π

So,

Hence, Proved.

Given:-

Take

LHS

We know that, Formula

Thus,

=

=

=

=

Let,

Therefore,

So,

⇒

Hence, Proved.

Given:-

Take

We know that, Formula

Thus,

=

=

=

=

=

So,

Given:-

Take

LHS

We know that, Formula

Thus,

⇒

⇒

⇒

⇒

⇒ 5x = –1 + 6x²

⇒ 6x² – 5x – 1 = 0

⇒ 6x² – 6x + x – 1 = 0

⇒ 6x(x – 1) + 1(x – 1) = 0

⇒ (6x + 1)(x – 1) = 0

⇒ 6x + 1 = 0 or x – 1 = 0

⇒

Since,

So,

is the root of the given equation

Therefore,

Given:-

Take

LHS

We know that, Formula

Thus,

⇒

⇒

⇒

⇒

⇒ 62x = 8 – 8x²+ 8

⇒ 4x² + 62x – 16 = 0

⇒ 6x² + 31x – 8 = 0

⇒ 4x(x + 8) – 1(x + 8) = 0

⇒ (4x – 1)(x + 8) = 0

⇒ 6x + 1 = 0 or x – 1 = 0

⇒

Since,

So,

is the root of the given equation

Therefore,

Given:-

Take

We know that, Formula

Thus,

⇒

⇒

⇒

⇒

We know that, Formula

Thus,

⇒

⇒

⇒

⇒

⇒ 4x – x³ = 6x – 9x³

⇒ 9x³ – x³ + 4x – 6x = 0

⇒ 8x³ – 2x = 0

⇒ 2x(4x² – 1) = 0

⇒

All satisfies x value

So,

is the root of the given equation

Therefore,

Given:-

Take

We know that, Formula

Thus,

⇒

⇒

⇒

⇒

⇒

Given:-

Take

We know that, Formula

Thus,

⇒

We know that, Formula

Thus,

⇒

⇒

⇒

We know that, Formula

Thus,

⇒

⇒

By rationalisation

⇒

⇒

⇒ (x+1)² = (1+√3)²

⇒ x+1 = ±(1+√3)

⇒ x +1 = 1+√3 or x +1 = –1–√3

⇒ x = √3 or x = –2 – √3

as given, x > 0

Therefore

x = √3

Given:-

Take

We know that, Formula

Thus,

⇒

⇒

⇒

⇒

⇒ 40 – 8x² = 158x

⇒ 8x² + 158x – 40 = 0

⇒ 4x² + 79x – 20 = 0

⇒ 4x² + 80x – x – 20 = 0

⇒ 4x(x + 20) – 1(x + 20) = 0

⇒ (4x – 1)(x + 20) = 0

⇒ 4x – 1 = 0 or x + 20 = 0

⇒

Since,

x > 0

So,

is the root of the given equation

Therefore,

Given:-

Take

We know that, Formula

⇒

⇒

⇒

⇒

⇒

⇒ 5x = 6 – x²

⇒ x² + 5x – 6 = 0

⇒ x² + 6x – x – 6 = 0

⇒ x(x + 6) – 1(x + 6) = 0

⇒ x = –6, 1

as given

0<x< √6

Therefore

x = 1

Given:-

Take

We know that, Formula

⇒

⇒

⇒

⇒

⇒

⇒

⇒

⇒ x² – 8 = – 6

⇒ x² = 2

⇒ x = ±√2

Given:-

Take

We know that, Formula

Thus,

⇒

⇒

⇒

⇒

⇒ 2x² – 8 + 2 = 12

⇒ 2x² = 18

⇒

Since,

x < –√3 or x > √3

So,

x= +3, –3 is the root of the given equation

Therefore,

x= +3, –3

Given:-

Take

We know that, Formula

⇒

⇒

⇒

⇒

⇒

⇒

⇒

⇒ 2x² – 4 = – 5

⇒ 2x² = 1

⇒

Given:-

Take

⇒

We know that, Formula

⇒

So,

.

.

.

Adding all the terms, we get

Given:-

Take

We know that, Formula

=

=

=

We know that, Formula

Therefore,

=

=

=

So,

Given:-

Take

RHS

We know that, Formula

Thus,

=

=

By pathagorous theorem

=

=

We know that, Formula

Thus,

=

=

Now,

LHS

=

=

So,

LHS = RHS

Given:-

Take

LHS

We know that, Formula

Thus,

=

=

By pathagorous theorem

=

=

We know that, Formula

Thus,

=

=

Now,

RHS

=

=

So,

LHS = RHS

Given:-

Take

LHS

=

=

We know that, Formula

Thus,

=

Now,

Assume that

Then,

⇒

And

⇒

Therefore,

⇒ LHS =RHS

⇒

Hence Proved

Given:-

Take

⇒

⇒

We know that, Formula

Thus,

⇒

⇒

⇒

⇒

⇒

⇒

⇒

Given:-

Take

⇒

We know that, Formula

Thus,

⇒

We know that, Formula

Thus,

⇒

⇒

⇒

⇒

⇒

⇒

⇒

⇒ x = ±1

Given:-

Take

We know that, Formula

Thus,

⇒

⇒

⇒

⇒

Now lets take square of both side, we get

⇒

⇒

⇒

⇒

⇒

⇒

Hence Proved

Given:-

Take

⇒

We know that, Formula

Thus,

⇒

⇒

⇒

Squaring both side or removing square root, we get

⇒

⇒

⇒

⇒ (b² – a²)a²b² = x²(b² – a²)

⇒ x² = a²b²

⇒ x = ab

Given:-

Take

We know that, Formula

Thus,

⇒

⇒

⇒

⇒

Squaring both sides, we get

⇒ 3x⁴ = 1 – x² – 3x² + 3x⁴

⇒ 1 – 4x² = 0

⇒

⇒

Given:-

Take

LHS

We know that, Formula

Thus,

=

=

=

=

= RHS

So,

Hence Proved

Given:-

Now, as we know

and

=

=

We know that,

=

=

=

Given:-

Let

Therefore,

⇒

⇒

Now, by Pythagoras theorem

⇒

⇒

⇒

As given, and putting assume value, we get

=

We know that: Formula

=

=

=

= ; by rationalisation

=

=

Hence

Given:-

We know that : Formula

; choose that formula which actually simplifies function

Thus, given function changes to

=

=

As we know

=

Hence,

Given:-

We know that :- Formula (- obtain by Pythagoras theorem)

; Formula of tan in terms of sine, so that it make simplification easier

And

; Formula of tan in terms of cos, so that it make simplification easier

Now given function becomes,

=

=

=

=

Hence,

Given:-

Take

LHS

=

We know that, Formula

Thus,

=

=

=

Again we know that, Formula

Thus,

=

=

=

= RHS

So,

Hence Proved

Given:-

Take

LHS

=

We know that, Formula

Thus,

=

=

=

=

Multiplying and dividing by 2

=

We know that, Formula

=

=

=

= RHS

So,

Now,

=

We know that, Formula

=

Thus,

=

=

=

= RHS

So,

Hence Proved

Given:-

Take

LHS

=

Multiplying and dividing by 2

=

We know that, Formula

=

=

=

= RHS

So,

Hence Proved

Given:-

Take

LHS

=

We know that, Formula

Thus,

=

=

=

We know that, Formula

=

=

=

=

= RHS

So,

Hence Proved

Given:-

Take

LHS

=

We know that, Formula

And,

Thus,

=

=

=

We know that, Formula

=

=

=

=

= RHS

So,

Hence Proved

Given:-

Take

LHS

=

We know that, Formula

Thus,

=

=

=

We know that, Formula

Thus,

=

=

=

We know that, Formula

=

=

=

=

=

= RHS

So,

Hence Proved

Given:-

Take

LHS

=

We know that, Formula

Thus,

=

=

=

We know that, Formula

=

=

=

=

= RHS

So,

Hence Proved

Given:-

Take

LHS

=

We know that, Formula

Thus,

=

=

=

We know that, Formula

=

=

=

=

=

= RHS

So,

Hence Proved

Given:-

Take

LHS

=

We know that, Formula

Thus,

=

=

=

We know that, Formula

=

=

=

= RHS

So,

Hence Proved

Given:-

Take

LHS

=

We know that, Formula

Thus,

=

=

We know that, Formula

=

=

=

=

=

= RHS

So,

Hence Proved

Given:-

Take

⇒

We know that, Formula

And

Thus,

⇒

⇒

⇒

We know that, Formula

Thus,

⇒

On comparing we get,

⇒

Hence Proved

Given:-

Take

LHS

=

We know that, Formula

Thus,

=

We know that, Formula

=

=

=

=

=

= RHS

So,

Hence Proved

Given:-

Take

LHS

=

We know that, Formula

Thus,

=

Again,

Thus,

=

We know that, Formula

=

=

=

=

=

= 1

= RHS

So,

Hence Proved

Given:-

Take

We know that, Formula

Thus,

⇒

⇒

⇒

We know that, Formula

Thus,

⇒

On comparing we get,

⇒

Hence Proved

Given:-

Take

We know that, Formula

Thus,

=

=

Now as given,

For, x ≥1

=

=

= π

= Constant

So,

Given:-

Take

We know that, Formula

Therefore,

cos(

Thus,

=

=

=

=

=

So,

Given:-

Take

We know that, Formula

Therefore,

=cos(

= 0

So,

⇒

Given:-

Take

⇒

We know that, Formula

Thus,

⇒

⇒

⇒

We know that, Formula

⇒

⇒

⇒

We know that, Formula

And

Thus,

⇒

⇒

⇒

We know that, Formula

Thus,

⇒, here

On comparing we get,

⇒

⇒

⇒235x+226 = 226x-235

⇒235x – 226x = 226 + 235

⇒

Given:-

Take

⇒

We know that, Formula

And

Thus,

⇒

⇒

⇒

⇒

⇒

⇒

Given:-

Take

⇒

We know that, Formula

Thus,

⇒

We know that, Formula

Thus,

⇒

⇒

⇒

⇒

⇒

Given:-

Take

We know that, Formula

Thus, here x = sinx

⇒

We know that, Formula

⇒

⇒

⇒

⇒

Thus the solution is

Given:-

Take

⇒

⇒

We know that, Formula

And,

Thus,

⇒

⇒

⇒

⇒

⇒

Given:-

Take

⇒

We know that, Formula

Thus,

⇒

⇒

We know that, Formula

⇒

⇒

⇒

On comparing we get,

⇒

⇒

⇒(2x+3)(x-2)= -(x-1)

⇒2x² – 4x + 3x – 6 = - x + 1

⇒2x² – x – 6 = - x + 1

⇒2x²= 7

⇒

Given:-

Take

LHS

=

We know that, Formula

Thus,

=

=

=

=

=

Dividing numerator and denominator by , we get

=

We know that, Formula

Thus,

=

= RHS

So,

Hence Proved

Given:-

Take

LHS

=

We know that, Formula

Thus,

=

=

=

Formula used:- a² + b² + 2ab = (a+b)²

=

=

As given

a= ax–by and β= ay+bx

Thus,

= RHS

So,

Hence Proved

Given:-

Take

LHS

=

Dividing numerator and denominator of 1^st term and 2^nd term by b³ and y³ respectively.

=

=

We know that, Formula

Thus,

=

=

=

We know that, Formula

Thus,

=

=

=

As given,

ay + bx =β, -ax + by = α

=

We know that, Formula

Thus,

=

=

=

= RHS

So,

Hence Proved

We are given that,

We need to find the value of x².

Take,

Multiply on both sides by tangent.

Since, we know that tan(tan^-1 x) = x.

So,

Or

Now, we need to simplify it in order to find x². So, rationalize the denominator by multiplying and dividing by .

Note the denominator is in the form: (x + y)(x – y), where

(x + y)(x – y) = x² – y²

So,

…(i)

Numerator:

Applying the algebraic identity in the numerator, (x – y)² = x² + y² – 2xy.

We can write as,

Again using the identity, (x + y)(x – y) = x² – y².

…(ii)

Denominator:

Solving the denominator, we get

…(iii)

Substituting values of Numerator and Denominator from (ii) and (iii) in equation (i),

By cross-multiplication,

⇒ x² tan α = 1 – √(1 – x⁴)

⇒ √(1 – x⁴) = 1 – x² tan α

Squaring on both sides,

⇒ [√(1 – x⁴)]² = [1 – x² tan α]²

⇒ 1 – x⁴ = (1)² + (x² tan α)² – 2x² tan α [∵, (x – y)² = x² + y² – 2xy]

⇒ 1 – x⁴ = 1 + x⁴ tan² α – 2x² tan α

⇒ x⁴ tan² α – 2x² tan α + x⁴ + 1 – 1 = 0

⇒ x⁴ tan² α – 2x² tan α + x⁴ = 0

Rearranging,

⇒ x⁴ + x⁴ tan² α – 2x² tan α = 0

⇒ x⁴ (1 + tan² α) – 2x² tan α = 0

⇒ x⁴ (sec² α) – 2x² tan α = 0 [∵, sec² x – tan² x = 1 ⇒ 1 + tan² x = sec² x]

Taking x² common from both terms,

⇒ x² (x² sec² α – 2 tan α) = 0

⇒ x² = 0 or (x² sec² α – 2 tan α) = 0

But x² ≠ 0 as according to the question, we need to find some value of x².

⇒ x² sec² α – 2 tan α = 0

⇒ x² sec² α = 2 tan α

In order to find the value of x², shift sec² α to Right Hand Side (RHS).

Putting ,

⇒ x² = 2 sin α cos α

Using the trigonometric identity, 2 sin x cos x = sin 2x.

⇒ x² = sin 2α

We need to find the value of

Let,

Let us find sin a and cos b.

For sin a,

We know the trigonometric identity, sin² a + cos² a = 1

⇒ sin² a = 1 – cos² a

⇒ sin a = √(1 – cos² a)

Substituting the value of cos a,

We have .

So, we can find tan a.

⇒ tan a = 7 …(i)

For cos b,

We know the trigonometric identity,

sin² b + cos² b = 1

⇒ cos² b = 1 – sin² b

⇒ cos b = √(1 – sin² b)

Substituting the value of sin b,

We have .

So, we can find tan b.

⇒ tan b = 4 …(ii)

We can write as,

Now, we need to solve Right Hand Side (RHS).

We know the trigonometric identity,

Substituting the values of tan a and tan b from (i) and (ii),

So,

We need to find the value of 2 tan^-1 |cosec(tan^-1 x) – tan(cot^-1 x)|.

So, take

2 tan^-1 |cosec(tan^-1 x) – tan(cot^-1 x)|

Using property of inverse trigonometry,

Now, let y = tan^-1 x

So, tan y = x

Substituting the value of tan^-1 x and x in the equation,

Put,

Since, we know the trigonometric identity,

1 – cos 2y = 2 sin² y

Also, sin 2y = 2 sin y cos y

We get,

Since,

Then,

⇒ 2 tan^-1 |cosec(tan^-1 x) – tan(cot^-1 x)| = y

Put y = tan^-1 x as let above.

⇒ 2 tan^-1 |cosec(tan^-1 x) – tan(cot^-1 x)| = tan^-1 x

We are given that,

We need to find the value of

By property of inverse trigonometry,

cos^-1 a + cos^-1 b = cos^-1(ab - √(1 – a²)√(1 – b²))

So,

Simplifying further,

Taking cosine on both sides,

Using the property of inverse trigonometric function,

cos(cos^-1 x) = x

To simplify it further, take square on both sides.

Using algebraic identity,

(x – y)² = x² + y² – 2xy

Simplifying it further,

Shifting all terms at one side,

Using trigonometric identity,

sin² x + cos² x = 1

⇒ sin² x = 1 – cos² x

We get,

We need to find the positive integral solution of the equation:

Using property of inverse trigonometry,

Also,

Taking,

Using the property of inverse trigonometry,

Similarly,

Taking tangent on both sides of the equation,

Using property of inverse trigonometry,

tan(tan^-1 A) = A

Applying this property on both sides of the equation,

Simplifying the equation,

Cross-multiplying in the equation,

⇒ xy + 1 = 3(y – x)

⇒ xy + 1 = 3y – 3x

⇒ xy + 3x = 3y – 1

⇒ x(y + 3) = 3y – 1

We need to find positive integral solutions using the above result.

That is, we need to find solution which is positive as well as in integer form. A positive integer are all natural numbers.

That is, x, y > 0.

So, keep the values of y = 1, 2, 3, 4, … and find x.

Note that, only at y = 2, value is x is positive integer.

Thus, the positive integral solution of the given equation is x = 1, y = 2.

We are given that,

…(i)

We need to find the value of x.

By using the property of inverse trigonometry,

We can find the value of sin^-1 x in the terms of cos^-1 x.

Substituting the value of sin^-1 x in equation (i),

Simplifying it further,

Multiplying cosine on both sides of the equation,

Using property of inverse trigonometry,

cos[cos^-1 x] = x

And we know the value,

Therefore,

We need to find the value of

sin [cot^-1 {tan (cos^-1 x)}] …(i)

We can solve such equation by letting the inner most trigonometric function (here, cos^-1 x) as some variable, and solve systematically following BODMAS rule and other trigonometric identities.

Let cos^-1 x = y

We can re-write the equation (i),

sin [cot^-1 {tan (cos^-1 x)}] = sin [cot^-1 {tan y}]

Using trigonometric identity,

[∵, lies in 1^st Quadrant and sine, cosine, tangent and cot are positive in 1^st Quadrant]

Using property of inverse trigonometry,

cot^-1(cot x) = x

Using trigonometric identity,

Substituting this value of ,

We had let above that cos^-1 x = y.

If,

cos^-1 x = y

⇒ x = cos y

Therefore,

We need to find the number of solutions of the equation,

We shall apply the property of inverse trigonometry, that is,

So,

Taking tangent on both sides of the equation,

Using property of inverse trigonometry,

tan(tan^-1 A) = A

Also,

We get,

Simplifying it,

⇒ 5x = 1 – 6x²

⇒ 6x² + 5x – 1 =0

Since, this is a quadratic equation, it is clear that it will have 2 solutions.

Let us check:

We have,

6x² + 5x – 1 = 0

⇒ 6x² + 6x – x – 1 = 0

⇒ 6x(x + 1) – (x + 1) = 0

⇒ (6x – 1)(x + 1) = 0

⇒ (6x – 1) = 0 or (x + 1) = 0

⇒ 6x = 1 or x = -1

Hence, there are 2 solutions of the given equation.

We are given that,

Take,

We can write as,

Then,

Also, by trigonometric identity

[∵, lies in III Quadrant and tangent is positive in III Quadrant]

Using the property of inverse trigonometry, that is, tan^-1(tan A) = A.

Now, take

We can write as,

Then,

By trigonometric identity,

[∵, lies in II Quadrant and tangent is negative in II Quadrant]

Using the property of inverse trigonometry, that is, tan^-1(tan A) = A.

We have,

⇒ 4α = π and 3β = π

Since, the values of 4α and 3β are same, that is,

4α = 3β = π

Therefore,

4α = 3β

We are given with equation:

√(1 + cos 2x) = √2 sin^-1(sin x) …(i)

Where -π ≤ x ≤ π

We need to find the number of real solutions of the given equation.

Using trigonometric identity,

cos 2x = cos² x – sin² x

⇒ cos 2x = cos² x – (1 – cos² x) [∵, sin² x + cos² x = 1 ⇒ sin² x = 1 – cos² x]

⇒ cos 2x = cos² x – 1 + cos² x

⇒ cos 2x = 2 cos² x – 1

⇒ 1 + cos 2x = 2 cos² x

Substituting the value of (1 + cos 2x) in equation (i),

√(2 cos² x) = √2 sin^-1(sin x)

⇒ √2 |cos x| = √2 sin^-1(sin x)

√2 will get cancelled from each sides,

⇒ |cos x| = sin^-1(sin x)

Take interval :

|cos x| is positive in interval , hence |cos x| = cos x.

And, sin x is also positive in interval , hence sin^-1(sin x) = x.

So, |cos x| = sin^-1(sin x)

⇒ cos x = x

If we draw y = cos x and y = x on the same graph, we will notice that they intersect at one point, thus giving us 1 solution.

∴, There is 1 solution of the given equation in interval .

Take interval :

|cos x| is negative in interval , hence |cos x| = -cos x.

And, sin x is also negative in interval , hence sin^-1(sin (π + x)) = π + x.

So, |cos x| = sin^-1(sin x)

⇒ -cos x = π + x

⇒ cos x = -π – x

If we draw y = cos x and y = -π – x on the same graph, we will notice that they intersect at one point, thus giving us 1 solution.

∴, There is 1 solution of the given equation in interval .

Take interval :

|cos x| is negative in interval , hence |cos x| = -cos x.

And, sin x is positive in interval , hence sin^-1(sin (-π – x)) = -π – x.

So, |cos x| = sin^-1(sin x)

⇒ -cos x = -π – x

⇒ -cos x = -(π + x)

⇒ cos x = π + x

If we draw y = cos x and y = π + x on the same graph, we will notice that they doesn’t intersect at any point, thus giving us no solution.

∴, There is 0 solution of the given equation in interval .

Hence, we get 2 solutions of the given equation in interval [-π, π].

We are given that,

xy = 1, x < 0 and y < 0

We need to find the value of tan^-1 x + tan^-1 y.

Using the property of inverse trigonometry,

We already know the value of xy, that is, xy = 1.

Also, we know that x, y < 0.

Substituting xy = 1 in denominator,

And since (x + y) = negative value = integer = -a (say).

⇒ tan^-1 x + tan^-1 y = tan^-1 -∞ …(i)

Using value of inverse trigonometry,

Substituting the value of tan^-1 -∞ in the equation (i), we get

We are given with

u = cot^-1{√tan θ} – tan^-1{√tan θ}

We need to find the value of .

Let √tan θ = x

Then, u = cot^-1{√tan θ} – tan^-1{√tan θ} can be written as

u = cot^-1 x – tan^-1 x …(i)

We know by the property of inverse trigonometry,

Or,

Substituting the value of cot^-1 x in equation (i), we get

u = (cot^-1 x) – tan^-1 x

Rearranging the equation,

Now, divide by 2 on both sides of the equation.

Taking tangent on both sides, we get

Using property of inverse trigonometry,

tan(tan^-1 x) = x

Recall the value of x. That is, x = √tan θ

We are given with,

…(i)

We need to find the value of

Take Left Hand Side (LHS) of equation (i),

Using the property of inverse trigonometry,

Putting and ,

Equate LHS to RHS.

Taking cosine on both sides,

Using property of inverse trigonometry,

cos(cos^-1 A) = A

Simplifying the equation,

Squaring on both sides,

Using algebraic identity,

(A – B)² = A² + B² – 2AB

Using trigonometric identity,

cos 2θ = cos² θ – sin² θ …(ii)

sin² θ + cos² θ = 1 ⇒ sin² θ = 1 – cos² θ …(iii)

Putting value of sin² θ from equation (iii) in equation (ii), we get

cos 2θ = cos² θ – (1 – cos² θ)

Or, cos 2θ = cos² θ – 1 + cos² θ

Or, cos 2θ = 2 cos² θ – 1

Or, 2 cos² θ = cos 2θ + 1

Replace θ by θ/2.

Substituting the value of in

We are given with,

We need to find the value of α – β.

So,

Using the property of inverse trigonometry,

So,

Simplifying it further,

The term (2x² + 2y² – 2xy) gets cancelled from numerator and denominator.

Using the value of inverse trigonometry,

We are given with,

We need to find .

We just need to find put in f(x).

So,

Simplify the equation,

Using trigonometric identity,

Using trigonometric identity,

cos (-θ) = cos θ

Using property of inverse trigonometry,

cos^-1(cos θ) = θ

We need to find the value of

Using property of inverse trigonometry,

Replacing the values of A by and B by ,

Solving it further,

Thus, none of this match the result.

We are given with,

We need to find the value of 9x² – 12xy cos θ + 4y².

Using property of inverse trigonometry,

Take Left Hand Side (LHS) of:

Replace A by and B by .

Further solving,

We shall equate LHS to RHS,

Taking cosine on both sides,

Using property of inverse trigonometry,

cos(cos^-1 A) = A

So,

By cross-multiplying,

⇒ xy - √(4 – x²) √(9 – y²) = 6 cos θ

Rearranging it,

⇒ xy – 6 cos θ = √(4 – x²) √(9 – y²)

Squaring on both sides,

⇒ [xy – 6 cos θ]² = [√(4 – x²) √(9 – y²)]²

Using algebraic identity,

(a – b)² = a² + b² – 2ab

⇒ (xy)² + (6 cos θ)² – 2(xy)(6 cos θ) = (4 – x²)(9 – y²)

⇒ x²y² + 36 cos² θ – 12xy cos θ = 36 – 9x² – 4y² + x²y²

⇒ x²y² – x²y² + 9x² – 12xy cos θ + 4y² = 36 – 36 cos² θ

⇒ 9x² – 12xy cos θ + 4y² = 36 (1 – cos² θ)

Using trigonometric identity,

sin² θ + cos² θ = 1

⇒ sin² θ = 1 – cos² θ

Substituting the value of (1 – cos² θ), we get

⇒ 9x² – 12xy cos θ + 4y² = 36 sin² θ

We are given with,

tan^-1 3 + tan^-1 x = tan^-1 8

We need to find the value of x.

Using property of inverse trigonometry,

Let us replace A by 3 and B by x.

Since, according to the question

tan^-1 3 + tan^-1 x = tan^-1 8

So,

Taking tangent on both sides,

Using property of inverse trigonometry,

tan(tan^-1 A) = A

Now, in order to find x, we need to solve the linear equation.

By cross-multiplying,

⇒ 3 + x = 8(1 – 3x)

⇒ 3 + x = 8 – 24x

⇒ 24x + x = 8 – 3

⇒ 25x = 5

We need to find the value of .

Using the trigonometric identity,

As the function lies in I Quadrant and so it will be positive.

Using the trigonometric identity,

Using property of inverse trigonometry,

sin^-1(sin A) = A

We need to find the value of:

Let us simplify the trigonometric function.

We can write as:

Similarly,

Since, lies on IV Quadrant and cosine is positive in IV Quadrant.

And since, lies on IV Quadrant and sine is negative in IV Quadrant.

Using property of inverse trigonometry,

sin^-1(sin A) = A and cos^-1(cos A) = A

=0

We need to find the value of:

Let

Take cosine on both sides, we get

Using property of inverse trigonometry,

cos(cos^-1 A) = A

We have the value of cos x, let us find the value of sin x.

By trigonometric identity,

sin² x + cos² x = 1

⇒ sin² x = 1 – cos² x

Putting ,

Now,

Using the trigonometric identity,

sin 2x = 2 sin x cos x

Putting the value of and ,

We are given that,

θ = sin^-1 {sin (-600°)}

We know that,

sin (2π – θ) = sin (4π – θ) = sin (6π – θ) = sin (8π – θ) = … = -sin θ

As, sin (2π – θ), sin (4π – θ), sin (6π – θ), … all lie in IV Quadrant where sine function is negative.

So,

If we replace θ by 600°, then we can write as

sin (4π – 600°) = -sin 600°

Or,

sin (4π – 600°) = sin (-600°)

Or,

sin (720° – 600°) = sin (-600°) …(i)

[∵, 4π = 4 × 180° = 720° < 600°]

Thus, we have

θ = sin^-1 {sin (-600°)}

⇒ θ = sin^-1 {sin (720° – 600°)} [from equation (i)]

⇒ θ = sin^-1 {sin 120°} …(ii)

We know that,

sin (π – θ) = sin (3π – θ) = sin (5π – θ) = … = sin θ

As, sin (π – θ), sin (3π – θ), sin (5π – θ), … all lie in II Quadrant where sine function is positive.

So,

If we replace θ by 120°, then we can write as

sin (π – 120°) = sin 120°

Or,

sin (180° - 120°) = sin 120° …(iii)

[∵, π = 180° < 120°]

Thus, from equation (ii),

θ = sin^-1 {sin 120°}

⇒ θ = sin^-1 {sin (180° - 120°)} [from equation (iii)]

⇒ θ = sin^-1 {sin 60°}

Using property of inverse trigonometry,

sin^-1 (sin A) = A

⇒ θ = 60°

We are given that,

We need to find the value of x.

We know that by trigonometric identity, we can represent sin θ, cos θ and tan θ in terms of tan θ.

Note,

So, in the equation given in the question, let x = tan θ.

Re-writing the equation,

Substituting the values of trigonometric identities,

Using the property of inverse trigonometry, we have

sin^-1 (sin A) = A, cos^-1 (cos A) = A and tan^-1 (tan A) = A

Now, in order to find the value of x, recall

x = tan θ

Substitute the value of θ derived above,

We are given that,

4 cos^-1 x + sin^-1 x = π …(i)

We need to find the value of x.

Using the property of inverse trigonometry,

Replacing θ by x, we get

Substituting the value of sin^-1 x in (i),

4 cos^-1 x + sin^-1 x = π

Taking cosines on both sides,

We are given that,

…(i)

We need to find the value of x.

Using the property of inverse trigonometry,

Replace A by and B by .

Putting this value in equation (i),

Taking tangent on both sides,

Using the property of inverse trigonometry,

tan(tan^-1 A) = A

Cross-multiplying, we get

Simplifying the equation in order to find the value of x,

Let us cancel the denominator from both sides of the equation.

⇒ x(x + 1) + (x – 1)(x – 1) = -7[x(x – 1) – (x + 1)(x – 1)]

⇒ x² + x + (x – 1)² = -7[x² – x – (x + 1)(x – 1)]

Using the algebraic identity,

(a – b) = a² + b² – 2ab

And, (a + b)(a – b) = a² – b²

⇒ x² + x + x² + 1 – 2x = -7[x² – x – (x² – 1)]

⇒ 2x² – x + 1 = -7[x² – x – x² + 1]

⇒ 2x² – x + 1 = -7[1 – x]

⇒ 2x² – x + 1 = -7 + 7x

⇒ 2x² – x – 7x + 1 + 7 = 0

⇒ 2x² – 8x + 8 = 0

⇒ 2(x² – 4x + 4) = 0

⇒ x² – 4x + 4 = 0

We need to solve the quadratic equation to find the value of x.

⇒ x² – 2x – 2x + 4 = 0

⇒ x(x – 2) – 2(x – 2) = 0

⇒ (x – 2)(x – 2) = 0

⇒ x = 2 or x = 2

Hence, x = 2.

We are given that,

cos^-1 x > sin^-1 x

We need to find the range of x.

Using the property of inverse trigonometry,

Or,

So, re-writing the inequality,

cos^-1 x > sin^-1 x

Adding cos^-1 x on both sides of the inequality,

Dividing both sides of the inequality by 2,

Taking cosine on both sides of the inequality,

is the minimum value of x, while the maximum value of cosine function is 1.

We are given that,

∆ABC is a right-angled triangle at C.

Let the sides of the ∆ABC be

AC = b

BC = a

AB = c

By Pythagoras theorem, where C is the right angle,

(AC)² + (BC)² = (AB)²

⇒ b² + a² = c²

Or,

a² + b² = c² …(i)

Using the property of inverse trigonometry,

Replacing A by and B by ,

Substituting the value of a² + b² from equation (i),

=tan^-1

We need to find the value of

Let

Now, take sine on both sides,

Using the property of inverse trigonometry,

sin(sin^-1 A) = A

Let us find the value of cos x.

We know by trigonometric identity, that

sin² x + cos² x = 1

⇒ cos² x = 1 – sin² x

Put the value of sin x,