Find the value of cot − 1 ( − 3 ) .

Let θ = cot − 1 − 3 We know that cot - 1 x ∈ 0 , π for all x ∈ R . ⇒ cot θ = − 3 ⇒ cot θ = - cot π 6 ⇒ cot θ = cot π - π 6 ⇒ cot θ = cot 5 π 6 ⇒ θ = 5 π 6

Find the value of cot − 1 ( − 3 ) .

Let θ = cot − 1 − 3 We know that cot - 1 x ∈ 0 , π for all x ∈ R . ⇒ cot θ = − 3 ⇒ cot θ = - cot π 6 ⇒ cot θ = cot π - π 6 ⇒ cot θ = cot 5 π 6 ⇒ θ = 5 π 6

Find the value of cot - 1 cot 5 π 4 .

Let θ = cot − 1 cot 5 π 4 We know that cot - 1 x ∈ 0 , π for all x ∈ R . ⇒ θ = cot − 1 cot π + π 4 ⇒ θ = cot − 1 cot π 4 ⇒ θ = π 4

Find the value of: tan - 1 tan 3 π 4

Given that t a n - 1 t a n 3 π 4 We know that t a n - 1 t a n x = x if x ∈ - π 2 , π 2 Clearly 3 π 4 ∉ - π 2 , π 2 Therefore, tan - 1 tan π - π 4 = tan - 1 - tan π 4 t a n - 1 - t a n π 4 = t a n - 1 t a n - π 4 = - π 4 ∈ - π 2 , π 2 Hence, t a n - 1 t a n 3 π 4 = - π 4

Find the value of arcsin 1 2 + arccos 1 2 .

Let A = sin - 1 1 2 + cos - 1 1 2 ⇒ A = π 2 Since, sin - 1 x + cos - 1 x = π 2 , x ∈ - 1 , 1 .

Find the value of t a n - 1 tan ⁡ 7 π 6

Given that, t a n - 1 t a n 7 π 6 We know that , tan - 1 t a n x = x ,if x ∈ - π 2 , π 2 ∴ t a n - 1 tan ⁡ 7 π 6 = 7 π 6 But 7 π 6 ∉ - π 2 , π 2 However, = t a n - 1 t a n π + π 6 = t a n - 1 t a n π 6 = π 6 ,since tan π + θ = tan θ Therefore, t a n - 1 t a n 7 π 6 = π 6

Find the value of cos - 1 cos 13 π 6

Given that cos - 1 cos 13 π 6 We know that cos - 1 cos x = x if x ∈ 0 , π , ∴ cos - 1 cos 13 π 6 = 13 π 6 But 13 π 6 ∉ 0 , π which is the principal branch of cos - 1 x . However, cos - 1 cos 13 π 6 = cos - 1 cos 2 π + π 6 = cos - 1 cos π 6 ∵ cos ( 2 π + θ ) = cosθ = π 6 ∈ 0 , π Therefore, cos - 1 cos 13 π 6 = π 6

Find the value of sin - 1 sin 3 π 5 .

Consider the given expression sin - 1 sin 3 π 5 L e t , x = sin - 1 sin 3 π 5 Now, the range of principal value of sin - 1 is - π 2 , π 2 ∴ x = 3 π 5 i s n o t p o s s i b l e So, sin x = sin 3 π 5 sin x = sin π - 3 π 5 ∴ sin x = sin 2 π 5 Since, the above value of x in the range of principal value Hence, sin - 1 sin 3 π 5 = 2 π 5

Find the principal value of tan − 1 3

Let y = tan - 1 ( 3 ) ⇒ tan y = tan π 3 Now, the range of tan - 1 x is - π 2 , π 2 . Since, π 3 ∈ - π 2 , π 2 Hence, the principle value of tan − 1 3 is π 3 .

Find the principal value of sin - 1 - 1 2 .

Let y = sin - 1 - 1 2 = sin - 1 - sin π 6 ∵ sin π 6 = 1 2 = - sin - 1 sin π 6 ∵ sin - 1 ( - x ) = - sin - 1 ( x ) ∀ x ∈ - 1 , 1 = - π 6 ∵ sin - 1 ( sin y ) = y , if - π 2 ≤ y ≤ π 2 We know that the range of sin - 1 x is - π 2 , π 2 , and - π 6 ∈ - π 2 , π 2 Hence, the principal value of sin - 1 - 1 2 is - π 6 .

Prove that: sin - 1 3 5 - sin - 1 8 17 = c o s - 1 84 75

L.H.S. = sin - 1 3 5 - sin - 1 8 17 = sin - 1 3 5 1 - 8 17 2 - 8 17 1 - 3 5 2 = sin - 1 3 5 1 - 64 289 - 8 17 1 - 9 25 = sin - 1 3 5 225 289 - 8 17 16 25 = sin - 1 3 5 × 15 17 - 8 17 × 4 5 = sin - 1 45 - 32 85 = sin - 1 13 85 Let sin - 1 13 85 = A ⇒ A = sin 13 85 , A ∈ 0 , π 2 ∵ sin A > 0 ∴ cos A = 1 - sin 2 A = 1 - 13 85 2 = 7225 - 169 7225 = 84 85 ∵ cos A > 0 ∀ A ∈ 0 , π 2 ⇒ A = cos - 1 84 85 ∴ sin - 1 3 5 - sin - 1 8 17 = cos - 1 84 85 . Hence proved.

Prove that: cos - 1 4 5 + cos - 1 12 13 = cos - 1 33 65

To prove: cos - 1 4 5 + cos - 1 12 13 = cos - 1 33 65 LHS = cos - 1 4 5 + cos - 1 12 13 = cos - 1 4 5 · 12 13 - 1 - 16 25 1 - 144 196 cos - 1 48 65 - 3 5 · 5 13 cos - 1 33 65 Hence, proved

Find the value of sin - 1 sin 3 π 5 .

Consider the given expression sin - 1 sin 3 π 5 L e t , x = sin - 1 sin 3 π 5 Now, the range of principal value of sin - 1 is - π 2 , π 2 ∴ x = 3 π 5 i s n o t p o s s i b l e So, sin x = sin 3 π 5 sin x = sin π - 3 π 5 ∴ sin x = sin 2 π 5 Since, the above value of x in the range of principal value Hence, sin - 1 sin 3 π 5 = 2 π 5

Prove that sin - 1 5 13 + cos - 1 3 5 = tan - 1 63 16

We have to show: sin - 1 5 13 + cos - 1 3 5 = tan - 1 63 16 Let, sin - 1 5 13 = x ⇒ sin x = 5 13 And cos 2 x = 1 - sin 2 x = 1 - 25 169 = 144 169 ⇒ cos x = 144 169 = 12 13 ∴ tan x = sin x cos x = 5 13 12 13 = 5 12 Again, let cos - 1 3 5 = y ⇒ cos y = 3 5 ∴ sin y = 1 - cos 2 y = 1 - 3 5 2 = 1 - 9 25 ⇒ sin y = 16 25 = 4 5 ⇒ tan y = sin y cos y = 4 5 3 5 = 4 3 We know that, tan ( x + y ) = tan x + tan y 1 - tan x tan y ∴ tan ( x + y ) = 5 12 + 4 3 1 - 5 12 4 3 ⇒ tan ( x + y ) = 5 + 16 12 36 - 20 36 ⇒ tan ( x + y ) = 21 × 3 16 ⇒ tan ( x + y ) = 63 16 ⇒ x + y = tan - 1 63 16 ⇒ sin - 1 5 13 + cos - 1 3 5 = tan - 1 63 16 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 .