Q1: Find the value of cos-1 (1/2) + 2 sin-1 (1/2).
Ans: First, solve for cos-1 (1/2):
Let us take, y = cos-1 (1/2)
This can be written as:
cos y = (1/2)
cos y = cos (π /3).
Thus, the range of principal value of cos-1 is [0, π ]
Therefore, the principal value of cos-1 (1/2) is π/3.
Now, solve for sin-1 (1/2):
Let y = sin-1 (1/2)
sin y = 1/2
sin y = sin ( π/6)
Thus, the range of principal value of sin-1 is [(-π)/2, π/2 ]
Hence, the principal value of sin-1 (1/2) is π/6.
Now we have cos-1 (1/2) = π/3 & sin-1 (1/2) = π/6
Now, substitute the obtained values in the given formula, we get:
= cos-1 (1/2) + 2sin-1 (1/2)
= π /3 + 2( π/6)
= π/3 + π/3
= ( π+π )/3
= 2π /3
Thus, the value of cos-1 (1/2) + 2 sin-1 (1/2) is 2π /3.
Q2: The value of tan-1 √3 – sec-1(–2) is equal to:
Ans: Now, solve the first part of the expression: tan-1 √3
Let us take y = tan-1√3
This can be written as:
tan y = √3
Now, use the trigonometry table to find the radian value
tan y = tan (π/3)
Thus, the range of principal value of tan-1 is (−π/2, π/2)
Therefore, the principal value of tan-1√3 is π/3.
Now, solve the second part of the expression: sec-1(–2)
Now, assume that y = sec-1 (–2)
sec y = -2
sec y = sec (2π/3)
We know that the principal value range of sec-1 is [0,π] – {π/2}
Therefore, the principal value of sec-1 (–2) = 2π/3
Now we have:
tan-1(√3) = π/3
sec-1 (–2) = 2π/3
Now, substitute the values in the given expression:
= tan-1 √3 – sec-1 (−2)
= π/3 − (2π/3)
= π/3 − 2π/3
= (π − 2π)/3
= – π/3
Q3: Determine the principal value of cos-1( -1/2).
Ans: Let us assume that, y = cos-1( -1/2)
We can write this as:
cos y = - 1/2
cos y = cos (2π/3).
Thus, the Range of the principal value of cos-1 is [0, π ].
Therefore, the principal value of cos-1( -1/2) is 2π /3.
Q4: Prove that sin-1 (3/5) – sin-1 (8/17) = cos-1 (84/85).
Ans: Let sin-1 (3/5) = a and sin-1 (8/17) = b
Thus, we can write sin a = 3/5 and sin b = 8/17
Now, find the value of cos a and cos b
To find cos a:
Cos a = √[1 – sin2 a]
= √[1 – (3/5)2 ]
= √[1 – (9/25)]
= √[(25-9)/25]
= 4/5
Thus, the value of cos a = 4/5
To find cos b:
Cos b = √[1 – sin2 b]
= √[1 – (8/17)2 ]
= √[1 – (64/289)]
= √[(289-64)/289]
= 15/17
Thus, the value of cos b = 15/17
We know that cos (a- b) = cos a cos b + sin a sin b
Now, substitute the values for cos a, cos b, sin a and sin b in the formula, we get:
cos (a – b) = (4/5) x (15/17) + (3/5) x (8/17)
cos (a – b) = (60 + 24)/(17x 5)
cos (a – b) = 84/85
(a – b) = cos-1 (84/85)
Substituting the values of a and b sin-1 (3/5)- sin-1 (8/7) = cos-1 (84/85)
Hence proved.
Q5: Find the value of cot (tan-1 α + cot-1 α).
Ans: Given that: cot (tan-1 𝛂 + cot-1 𝛂)
= cot (𝝅/𝟐) (since, tan-1 x + cot-1 x = 𝜋/2)
= cot (180°/2) ( we know that cot 90° = 0)
= cot (90°)
= 0
Therefore, the value of cot (tan-1 α + cot-1 α) is 0.
