Evaluate the following: c o s ( s i n - 1 3 5 ) .

Given that cos sin - 1 3 5 . We know that sin - 1 x y = cos - 1 y 2 - x 2 y , x y ≥ 0 , y ≠ 0 ⇒ sin - 1 3 5 = cos - 1 25 - 9 5 ⇒ sin - 1 3 5 = cos - 1 4 5 ⇒ cos sin - 1 3 5 = cos cos - 1 4 5 = 4 5

Evaluate the following: c o s ( s i n - 1 3 5 ) .

Given that cos sin - 1 3 5 . We know that sin - 1 x y = cos - 1 y 2 - x 2 y , x y ≥ 0 , y ≠ 0 ⇒ sin - 1 3 5 = cos - 1 25 - 9 5 ⇒ sin - 1 3 5 = cos - 1 4 5 ⇒ cos sin - 1 3 5 = cos cos - 1 4 5 = 4 5

Evaluate the following: s i n ( c o s - 1 4 5 ) .

Given that sin cos - 1 4 5 . We know that cos - 1 x = sin - 1 1 - x 2 , 0 ≤ x ≤ 1 . ⇒ cos - 1 4 5 = sin - 1 1 - 4 5 2 = sin - 1 9 5 = sin - 1 3 5 ∴ sin sin - 1 3 5 = 3 5

Evaluate the following: c o s sin - 1 - 3 5 .

Given that cos sin - 1 - 3 5 . We know that sin - 1 - x = - sin - 1 x , x ∈ - 1 , 1 = cos - sin - 1 3 5 We know that sin - 1 x = cos - 1 1 - x 2 , 1 ≥ x ≥ 0 = cos - cos - 1 1 - 3 5 2 = cos - cos - 1 4 5 = cos cos - 1 4 5 ∵ cos - α = cos α = 4 5

Evaluate the following: t a n ( c o s - 1 8 17 ) .

Given that tan cos - 1 8 17 . We know that cos - 1 x = tan - 1 1 - x 2 x , 1 ≥ x > 0 . Here x = 8 17 ⇒ 1 - x 2 x = 1 - 8 17 2 8 17 . = 225 8 = 15 8 ⇒ tan cos - 1 8 17 = tan tan - 1 15 8 = 15 8

Evaluate the following: c o s e c cos - 1 - 12 13 .

Given that, cosec cos - 1 - 12 13 . cosec π - cos - 1 12 13 ∵ cos - 1 - x = π - cos - 1 x , x ∈ - 1 , 1 We know that, cos - 1 x = sin - 1 1 - x 2 , x ≥ 0 . cos - 1 12 13 = sin - 1 1 - 12 13 2 = sin - 1 25 13 = sin - 1 5 13 = cosec - 1 13 5 ∴ cosec π - cosec - 1 13 5 = cosec cosec - 1 13 5 = 13 5

Evaluate the following: tan 2 tan - 1 1 5 - π 4

It is given that tan 2 tan - 1 1 5 - π 4 . Using 2 tan - 1 x = tan - 1 2 x 1 - x 2 , - 1 < x < 1 = tan tan - 1 2 5 1 - 1 25 - tan - 1 1 = tan tan - 1 5 12 - tan - 1 ( 1 ) Using tan - 1 x - tan - 1 y = tan - 1 x - y 1 + x y , x > 0 , y > 0 , x y < 1 = tan tan - 1 5 12 - 1 1 + 5 12 = tan tan - 1 - 7 17 = - 7 17

Evaluate the following: t a n 1 2 c o s - 1 5 3 .

Given that, tan 1 2 cos - 1 5 3 Let α = 1 2 cos - 1 5 3 ⇒ 2 α = cos - 1 5 3 ⇒ cos 2 α = 5 3 We know that cos 2 α = 1 - tan 2 α 1 + tan 2 α ⇒ 1 - tan 2 α 1 + tan 2 α = 5 3 By using"componendo dividendo rule" ⇒ 1 - tan 2 α + 1 + tan 2 α 1 - tan 2 α - 1 - tan 2 α = 5 + 3 5 - 3 ⇒ 2 - 2 tan 2 α = 5 + 3 5 - 3 ⇒ 1 tan 2 α = 3 + 5 3 - 5 ⇒ tan 2 α = 3 - 5 3 + 5 ⇒ tan 2 α = 3 - 5 2 3 + 5 3 - 5 ⇒ tan 2 α = 3 - 5 2 4 ⇒ tan α = ± 3 - 5 2 , We take only positive value because α = 1 2 cos - 1 5 3 , which is in the first quadrant. ∴ The required answer is 3 - 5 2 .

Evaluate the following: s i n 1 2 c o s - 1 4 5 .

Given that, sin 1 2 cos - 1 4 5 Let t = cos - 1 4 5 ⇒ cos t = 4 5 We know that cos 2 α = 1 - 2 sin 2 α ⇒ 2 sin 2 α = 1 - cos 2 α ⇒ sin α = ± 1 - cos 2 α 2 Here sin t 2 = ± 1 - cos t 2 ⇒ sin t 2 = ± 1 - 4 5 2 ⇒ sin t 2 = ± 1 10 As cos t = 4 5 , cos - 1 4 5 has value between 0 , π 2 . So, we take only positive value. ∴ sin 1 2 cos - 1 4 5 = 1 10

Evaluate the following: c o s s i n - 1 3 5 + s i n - 1 5 13 .

Given that, cos sin - 1 3 5 + sin - 1 5 13 We know that sin - 1 x + sin - 1 y = sin - 1 x 1 - y 2 + y 1 - x 2 , x 2 + y 2 ≤ 1 or x 2 + y 2 > 1 , x y < 0 Here x = 3 5 , y = 5 13 ∴ sin - 1 3 5 + sin - 1 5 13 = sin - 1 3 5 1 - 5 13 2 + 5 13 1 - 3 5 2 = sin - 1 3 5 × 12 13 + 5 13 × 4 5 = sin - 1 36 65 + 20 65 = sin - 1 56 65 We know that sin - 1 x = cos - 1 ± 1 - x 2 As sin - 1 56 65 is in the first quadrant, it will be positive. = cos - 1 1 - 56 65 2 = cos - 1 33 65 ∴ cos cos - 1 33 65 = 33 65