Solve the equation 2 cos - 1 x = sin - 1 2 x 1 - x 2

Given that 2 cos - 1 x = sin - 1 2 x 1 - x 2 . We know range of sin - 1 2 x 1 - x 2 i s [ - π 2 , π 2 ] Range of 2 cos - 1 x i s [ 0 , π ] Common range is [ 0 , π 2 ] 0 ≤ 2 cos - 1 x ≤ π 2 0 ≤ cos - 1 x ≤ π 4 1 2 ≤ x ≤ 1 As we know cos is a decreasing function. So inequality will get reversed. Hence 1 2 ≤ x ≤ 1 . This is Required Solution.

Solve the equation 2 cos - 1 x = sin - 1 2 x 1 - x 2

Given that 2 cos - 1 x = sin - 1 2 x 1 - x 2 . We know range of sin - 1 2 x 1 - x 2 i s [ - π 2 , π 2 ] Range of 2 cos - 1 x i s [ 0 , π ] Common range is [ 0 , π 2 ] 0 ≤ 2 cos - 1 x ≤ π 2 0 ≤ cos - 1 x ≤ π 4 1 2 ≤ x ≤ 1 As we know cos is a decreasing function. So inequality will get reversed. Hence 1 2 ≤ x ≤ 1 . This is Required Solution.

Solve sin - 1 x 1 + x 2 - sin 1 1 + x 2 = sin - 1 1 + x 1 + x 2 find the value for x

sin - 1 x 1 + x 2 - sin - 1 1 1 + x 2 = sin - 1 1 + x 1 + x 2 sin - 1 x 1 + x 2 - sin - 1 1 1 + x 2 = sin - 1 x 1 + x 2 1 - 1 1 + x 2 - 1 1 + x 2 1 - x 2 1 + x 2 ⇒ sin - 1 x 2 1 + x 2 - 1 1 + x 2 = sin - 1 x 2 - 1 x 2 + 1 As given ⇒ sin - 1 x 2 - 1 x 2 + 1 = sin - 1 x + 1 x 2 + 1 ⇒ x 2 - 1 x 2 + 1 = x + 1 x 2 + 1 ⇒ x 2 - 2 x - 2 = 0 ⇒ ( x - 2 ) ( x + 1 ) = 0 ⇒ x = 2 , - 1

Find the sum ∑ k = 1 n tan - 1 2 k 2 + k 2 + k 4 .

Given, ∑ k = 1 n tan - 1 2 k 2 + k 2 + k 4 T k = tan - 1 2 k 2 + k 2 + k 4 ⇒ T k = tan - 1 2 k 1 + k 4 + 2 k 2 + 1 - k 2 ⇒ T k = tan - 1 2 k 1 + k 2 + 1 - k 2 ⇒ T k = tan - 1 2 k 1 + k 2 + k + 1 k 2 - k + 1 ⇒ T k = tan - 1 k 2 + k + 1 - k 2 - k + 1 1 + k 2 + k + 1 k 2 - k + 1 ⇒ T k = tan - 1 k 2 + k + 1 - tan - 1 k 2 - k + 1 ∴ S = ∑ tan - 1 k 2 + k + 1 - tan - 1 k 2 - k + 1 k = 1 n ⇒ S = tan - 1 3 - tan - 1 1 + tan - 1 7 - tan - 1 3 + tan - 1 13 - tan - 1 7 + . . . . . . . . + tan - 1 n 2 + n + 1 - tan - 1 n 2 - n + 1 ⇒ S = - tan - 1 1 + tan - 1 n 2 + n + 1 ⇒ S = tan - 1 n 2 + n + 1 - π 4 .

Show that for x > 0 , tan - 1 x 1 + 1 . 2 x 2 + tan - 1 x 1 + 2 . 3 . x 2 + … + tan - 1 x 1 + n n + 1 x 2 = tan - 1 n x 1 + n + 1 x 2

Given, L . H . S . = tan - 1 x 1 + 1 . 2 x 2 + tan - 1 x 1 + 2 . 3 x 2 + . . . . . . . . tan - 1 x 1 + n n + 1 x 2 , x > 0 Using tan - 1 x - tan - 1 y = tan - 1 x - y 1 + x y , we have tan - 1 x 1 + 1 . 2 x 2 = tan - 1 2 x - x 1 + x . 2 x = tan - 1 2 x - tan - 1 x tan - 1 x 1 + 2 . 3 x 2 = tan - 1 3 x - 2 x 1 + 2 x . 3 x = tan - 1 3 x - tan - 1 2 x . . . . . . . . tan - 1 x 1 + n n + 1 x 2 = tan - 1 n + 1 x - n x 1 + n x . n + 1 x = tan - 1 n + 1 x - tan - 1 n x ∴ tan - 1 2 x - tan - 1 x + tan - 1 3 x - tan - 1 2 x + . . . . . . . . + tan - 1 n + 1 x - tan - 1 n x ⇒ tan - 1 n + 1 x - tan - 1 x ⇒ tan - 1 n + 1 x - x 1 + n + 1 x . x ⇒ tan - 1 n x + x - x 1 + n + 1 x 2 ⇒ tan - 1 n x 1 + n + 1 x 2 = R . H . S . .

Sum to infinite terms the series cot - 1 1 2 + 3 4 + cot - 1 2 2 + 3 4 + cot - 1 3 2 + 3 4 + …

Given, cot - 1 1 2 + 3 4 + cot - 1 2 2 + 3 4 + cot - 1 3 + 3 4 + . . . . i Let T n = cot - 1 n 2 + 3 4 ⇒ tan - 1 1 n 2 + 3 4 ∵ cot - 1 x = tan - 1 1 x ⇒ tan - 1 1 1 + n 2 - 1 4 ⇒ tan - 1 1 1 + n + 1 2 n - 1 2 ⇒ tan - 1 n + 1 2 - n - 1 2 1 + n + 1 2 n - 1 2 ⇒ tan - 1 n + 1 2 - tan - 1 n - 1 2 Therefore, equation i can be written as S n = tan - 1 3 2 - tan - 1 1 2 + tan - 1 5 2 - tan - 1 3 2 + . . . . . . + tan - 1 n + 1 2 - tan - 1 n - 1 2 ⇒ S n = tan - 1 n + 1 2 - tan - 1 1 2 Now, S ∞ = lim n → ∞ S n ⇒ lim n → ∞ tan - 1 n + 1 2 - tan - 1 1 2 ⇒ π 2 - tan - 1 1 2 ∵ tan - 1 x + cot - 1 x = π 2 ⇒ cot - 1 1 2 ⇒ tan - 1 2 .

Solve for x the equation tan - 1 x 2 + cot - 1 x 2 = 5 π 2 8 .

Given, tan - 1 x 2 + cot - 1 x 2 = 5 π 2 8 ⇒ tan - 1 x 2 + π 2 - tan - 1 x 2 = 5 π 2 8 ⇒ tan - 1 x 2 + π 2 4 + tan - 1 x 2 - πtan - 1 x - 5 π 2 8 = 0 ⇒ 2 tan - 1 x 2 - πtan - 1 x - 3 π 2 8 = 0 Solving above equation, we get tan - 1 x = - - π ± - π 2 - 4 . 2 . - 3 π 2 8 2 . 2 ⇒ tan - 1 x = π ± π 2 + 3 π 2 4 ⇒ tan - 1 x = π ± 4 π 2 4 ⇒ tan - 1 x = π ± 2 π 4 ⇒ tan - 1 x = 3 π 4 , - π 4 ⇒ x = tan 3 π 4 , tan - π 4 ⇒ x = - 1 .

Show that the greatest and least values of sin - 1 x 3 + cos - 1 x 3 are 7 π 3 8 and π 2 32 respectively.

Given ⇒ f ( x ) = ( S i n - 1 x ) 3 + ( C o s - 1 x ) 3 ⇒ f ( x ) = ( S i n - 1 x + C o s - 1 x ) { ( S i n - 1 x ) 2 + ( C o s - 1 x ) 2 - S i n - 1 x . C o s - 1 x } ⇒ f ( x ) = π 2 { ( π 2 ) 2 - 3 . S i n - 1 x . C o s - 1 x } ⇒ f ( x ) = π 2 { π 2 4 - 3 . S i n - 1 x . ( π 2 - a ) } where a = S i n - 1 x ⇒ f ( x ) = π 2 { 3 a 2 - 3 a π 2 + π 2 4 } ⇒ f ( x ) = π 8 { 12 a 2 - 6 πa + π 2 } ⇒ f ( x ) = 12 π 8 { a 2 - 1 2 πa + π 2 12 } ⇒ f ( x ) = 12 π 8 { ( a - π 4 ) 2 + π 2 48 } Therefore, the minimum value at x = 1 2 will be π 2 32 and the maximum value at x = - 1 will be 7 π 2 8 .

If sin - 1 x a + sin - 1 y b = sin - 1 c 2 a b , then prove that b 2 x 2 + 2 x y a 2 b 2 - c 4 + a 2 y 2 = c 4 .

Given that, sin - 1 x a + sin - 1 y b = sin - 1 c 2 a b and To prove: b 2 x 2 + 2 x y a 2 b 2 - c 4 + a 2 y 2 = c 4 Consider, sin - 1 x a + sin - 1 y b = sin - 1 c 2 a b ⇒ sin - 1 x a = sin - 1 c 2 a b - sin - 1 y b ⇒ x a = sin sin - 1 c 2 a b - sin - 1 y b ⇒ x a = sin sin - 1 c 2 a b cos sin - 1 y b - cos sin - 1 c a b sin sin - 1 y b [Using sin A - B = sin A cos B - cos A sin B ] ⇒ x a = c 2 a b cos cos - 1 1 - y 2 b 2 - cos cos - 1 1 - c 4 a 2 b 2 · y b [Using sin - 1 sin x = x ; x ∈ - π 2 , π 2 and sin - 1 x = cos - 1 1 - x 2 ] ⇒ x a = c 2 a b · 1 - y 2 b 2 - 1 - c 4 a 2 b 2 · y b ⇒ x a + y b 1 - c 4 a 2 b 2 = c 2 a b 1 - y 2 b 2 ⇒ x a + y b 1 - c 4 a 2 b 2 2 = c 2 a b 1 - y 2 b 2 2 ⇒ x 2 a 2 + y 2 b 2 1 - c 4 a 2 b 2 + 2 x y a b 1 - c 4 a 2 b 2 = c 4 a 2 b 2 1 - y 2 b 2 ⇒ x 2 a 2 + y 2 b 2 - y 2 b 2 · c 4 a 2 b 2 + 2 x y a b 1 - c 4 a 2 b 2 = c 4 a 2 b 2 - c 4 a 2 b 2 · y 2 b 2 ⇒ x 2 a 2 + y 2 b 2 + 2 x y a b 1 - c 4 a 2 b 2 = c 4 a 2 b 2 ⇒ x 2 a 2 + y 2 b 2 + 2 x y a 2 b 2 - c 4 a 2 b 2 = c 4 a 2 b 2 ⇒ b 2 x 2 + a 2 y 2 + 2 x y a 2 b 2 - c 4 = c 4 [Multiplied both sides with a 2 b 2 ] Hence, proved.

If sin - 1 x + sin - 1 y + sin - 1 z = π , then prove that x 1 - x 2 + y 1 - y 2 + z 1 - z 2 = 2 x y z

We are given, sin - 1 x + sin - 1 y + sin - 1 z = π Let, sin − 1 x = A ⇒ sin A = x sin - 1 y = B ⇒ sin B = y sin − 1 z = C ⇒ sin C = z . Here, A , B , C ∈ [ − π / 2 , π / 2 ] ∴ A + B + C = π . . . . . ( 1 ) Now, L H S = x 1 - x 2 + y 1 - y 2 + z 1 - z 2 = 2 x y z = sin A . 1 − sin 2 A + sin B 1 − sin 2 B + sin C . 1 − sin 2 C = sin A . cos A + sin B cos B + sin C . cos C [Multiply & Divide by 2 ] = 2 2 [ sin A cos A + sin B cos B ] + sin C . cos C = [ 2 sin A cos A + 2 sin B cos B ] 2 + sin C . cos C ∵ 2 sin θ cos θ = sin 2 θ = ( sin A + sin 2 B ) 2 + sin C . cos C = 2 sin 2 A + 2 B 2 cos 2 A − 2 B 2 2 + sin C . cos C = sin ( A + B ) cos ( A − B ) + sin C . cos C = sin ( π − C ) cos ( A − B ) + sin C . cos ( π − ( A + B ) ) = sin C [ cos ( A − B ) − cos ( A + B ) ] = sin C [ − 2 sin ( − 2 B 2 ) sin ( 3 A 2 ) ] = 2 sin A sin B sin C = 2 x y z = R H S .

sin - 1 ⁡ sin ⁡ 5 > x 2 - 4 x holds if

x ∈ 2 - 9 - 2 π , 2 + 9 - 2 π

E M B I B E

TRIGONOMETRY FOR JEE MAIN AND ADVANCED>Chapter 13 - Inverse Circular Functions>EXERCISE>Q 28

EASY

Mathematics

IMPORTANT

Earn 100

Solve the equation $2 \cos^{- 1} x = \sin^{- 1} 2 x \sqrt{1 - x^{2}}$

Important Questions on Inverse Circular Functions

HARD

Mathematics

IMPORTANT

TRIGONOMETRY FOR JEE MAIN AND ADVANCED>Chapter 13 - Inverse Circular Functions>EXERCISE>Q 29

Solve $\sin^{- 1} (\frac{x}{\sqrt{1 + x^{2}}}) - \sin (\frac{1}{\sqrt{1 + x^{2}}}) = \sin^{- 1} (\frac{1 + x}{1 + x^{2}})$

find the value for x

MEDIUM

Mathematics

IMPORTANT

TRIGONOMETRY FOR JEE MAIN AND ADVANCED>Chapter 13 - Inverse Circular Functions>EXERCISE>Q 30

If

\tan^{- 1} x, \tan^{- 1} y, \tan^{- 1} z

be in

A . P .

, find the algebraic relation between

x, y

and

z

. If

x, y, z

be in

A . P .

, prove that

x = y = z

.

MEDIUM

Mathematics

IMPORTANT

TRIGONOMETRY FOR JEE MAIN AND ADVANCED>Chapter 13 - Inverse Circular Functions>EXERCISE>Q 31

Find the sum

\sum_{k = 1}^{n} \tan^{- 1} \frac{2 k}{2 + k^{2} + k^{4}}

.

EASY

Mathematics

IMPORTANT

TRIGONOMETRY FOR JEE MAIN AND ADVANCED>Chapter 13 - Inverse Circular Functions>EXERCISE>Q 32

Show that for $x > 0$ ,

$\tan^{- 1} \frac{x}{1 + 1.2 x^{2}} + \tan^{- 1} \frac{x}{1 + 2.3 . x^{2}} + \dots + \tan^{- 1} \frac{x}{1 + n (n + 1) x^{2}} = \tan^{- 1} \frac{n x}{1 + (n + 1) x^{2}}$