Find the principal value of tan - 1 ( - 3 ) .

Let y = tan - 1 ( - 3 ) ⇒ tan y = - 3 ⇒ tan y = - tan π 3 ⇒ tan y = tan - π 3 . . . . i We know that the general solution of tan θ = tan α is θ = n π + α , where n ∈ Z . Thus, the general solution of equation i is y = n π - π 3 , n ∈ Z ⇒ y = . . . . , - 4 π 3 , - π 3 , 2 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 the given trigonometric function is - π 3 .

Find the principal value of tan - 1 ( - 3 ) .

Let y = tan - 1 ( - 3 ) ⇒ tan y = - 3 ⇒ tan y = - tan π 3 ⇒ tan y = tan - π 3 . . . . i We know that the general solution of tan θ = tan α is θ = n π + α , where n ∈ Z . Thus, the general solution of equation i is y = n π - π 3 , n ∈ Z ⇒ y = . . . . , - 4 π 3 , - π 3 , 2 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 the given trigonometric function is - π 3 .

Evaluate: sin π 3 - sin - 1 - 1 2

We need to find the value of sin π 3 - sin - 1 - 1 2 . Let y = sin - 1 - 1 2 , where y is the principal value of sin - 1 - 1 2 i.e., y ∈ - π 2 , π 2 . Thus, y = sin - 1 - 1 2 = - sin - 1 1 2 ∵ Sin - 1 ( - x ) = - Sin - 1 ( x ) , for all x ∈ - 1 , 1 = - sin - 1 sin π 6 = - π 6 ∵ sin - 1 ( sin y ) = y , if - π 2 ≤ y ≤ π 2 Thus, sin π 3 - sin - 1 - 1 2 = sin π 3 - - π 6 = sin π 3 + π 6 = sin π 2 = 1 Hence, sin π 3 - sin - 1 - 1 2 = 1

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 .

Find the value of tan - 1 2 cos 2 sin - 1 1 2 .

Given that, tan - 1 2 cos 2 sin - 1 1 2 = tan - 1 2 cos 2 sin - 1 sin π 6 [Since sin π 6 = 1 2 ] = tan - 1 2 cos 2 × π 6 = tan - 1 2 cos π 3 = tan - 1 2 1 2 [Since cos π 3 = 1 2 ] = tan - 1 1 = tan - 1 tan π 4 [Since tan - 1 ( tan y ) = y , if - π / 2 < y < π / 2 ] = π 4 Hence, the value of tan - 1 2 cos 2 sin - 1 1 2 is π 4 .

Show that sin - 1 8 17 + sin - 1 3 5 = tan - 1 77 36

Given sin - 1 8 17 + sin - 1 3 5 = tan - 1 77 36 L . H . S = sin - 1 8 17 + sin - 1 3 5 We know that sin - 1 ( x ) + sin - 1 ( y ) = sin - 1 ( x 1 - y 2 + y 1 - x 2 ) if x 2 + y 2 ≤ 1 , - 1 ≤ x , y ≤ 1 = sin - 1 8 17 1 - 3 5 2 + 3 5 1 - 8 17 2 = sin - 1 8 17 × 25 - 9 25 + 3 5 289 - 64 289 = sin - 1 8 17 × 16 25 + 3 5 225 289 = sin - 1 8 17 × 4 5 + 3 5 × 15 17 = sin - 1 32 85 + 45 85 = sin - 1 77 85 . . . . . . i Let sin - 1 77 85 = θ ⇒ sin θ = 77 85 = P H Now, H 2 = P 2 + B 2 ⇒ B 2 = 85 2 - 77 2 = 7225 - 5929 = 1296 ⇒ B = 36 Thus, tan θ = P B = 77 36 Taking tan - 1 on both sides, we have θ = tan - 1 77 36 . . . . . i i [Since tan - 1 ( tan y ) = y , if - π / 2 < y < π / 2 ] Using i , and i i , we have sin - 1 8 17 + sin - 1 3 5 = tan - 1 77 36 Hence proved

Show that sin - 1 8 17 + sin - 1 3 5 = tan - 1 77 36

Given sin - 1 8 17 + sin - 1 3 5 = tan - 1 77 36 L . H . S = sin - 1 8 17 + sin - 1 3 5 We know that sin - 1 ( x ) + sin - 1 ( y ) = sin - 1 ( x 1 - y 2 + y 1 - x 2 ) if x 2 + y 2 ≤ 1 , - 1 ≤ x , y ≤ 1 = sin - 1 8 17 1 - 3 5 2 + 3 5 1 - 8 17 2 = sin - 1 8 17 × 25 - 9 25 + 3 5 289 - 64 289 = sin - 1 8 17 × 16 25 + 3 5 225 289 = sin - 1 8 17 × 4 5 + 3 5 × 15 17 = sin - 1 32 85 + 45 85 = sin - 1 77 85 . . . . . . i Let sin - 1 77 85 = θ ⇒ sin θ = 77 85 = P H Now, H 2 = P 2 + B 2 ⇒ B 2 = 85 2 - 77 2 = 7225 - 5929 = 1296 ⇒ B = 36 Thus, tan θ = P B = 77 36 Taking tan - 1 on both sides, we have θ = tan - 1 77 36 . . . . . i i [Since tan - 1 ( tan y ) = y , if - π / 2 < y < π / 2 ] Using i , and i i , we have sin - 1 8 17 + sin - 1 3 5 = tan - 1 77 36 Hence proved

Find the value of tan - 1 2 cos 2 sin - 1 1 2 .

Given that, tan - 1 2 cos 2 sin - 1 1 2 = tan - 1 2 cos 2 sin - 1 sin π 6 [Since sin π 6 = 1 2 ] = tan - 1 2 cos 2 × π 6 = tan - 1 2 cos π 3 = tan - 1 2 1 2 [Since cos π 3 = 1 2 ] = tan - 1 1 = tan - 1 tan π 4 [Since tan - 1 ( tan y ) = y , if - π / 2 < y < π / 2 ] = π 4 Hence, the value of tan - 1 2 cos 2 sin - 1 1 2 is π 4 .

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 .

Evaluate: sin π 3 - sin - 1 - 1 2

We need to find the value of sin π 3 - sin - 1 - 1 2 . Let y = sin - 1 - 1 2 , where y is the principal value of sin - 1 - 1 2 i.e., y ∈ - π 2 , π 2 . Thus, y = sin - 1 - 1 2 = - sin - 1 1 2 ∵ Sin - 1 ( - x ) = - Sin - 1 ( x ) , for all x ∈ - 1 , 1 = - sin - 1 sin π 6 = - π 6 ∵ sin - 1 ( sin y ) = y , if - π 2 ≤ y ≤ π 2 Thus, sin π 3 - sin - 1 - 1 2 = sin π 3 - - π 6 = sin π 3 + π 6 = sin π 2 = 1 Hence, sin π 3 - sin - 1 - 1 2 = 1