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If cos⁻¹ α + cos⁻¹ β + cos⁻¹ γ = 3π, then α(β + γ) + β(γ + α) + γ(α + β) is equal to
We are given: cos⁻¹ α + cos⁻¹ β + cos⁻¹ γ = 3π Step 1: Using the property of inverse cosine The principal value range of cos⁻¹ x is [0, π]. For the sum of three inverse cosine terms to equal 3π, each term must be equal to π. This means: cos⁻¹ α = π, cos⁻¹ β = π, cos⁻¹ γ = π Step 2: Find the values oRead more
We are given:
cos⁻¹ α + cos⁻¹ β + cos⁻¹ γ = 3π
Step 1: Using the property of inverse cosine
The principal value range of cos⁻¹ x is [0, π].
For the sum of three inverse cosine terms to equal 3π, each term must be equal to π. This means:
cos⁻¹ α = π, cos⁻¹ β = π, cos⁻¹ γ = π
Step 2: Find the values of α, β, and γ
When cos⁻¹ α = π, then cos π = -1, so:
α = -1, β = -1, γ = -1
Step 3: Substitute into the expression
We are given to calculate:
α(β + γ) + β(γ + α) + γ(α + β)
Replace α = -1, β = -1, and γ = -1:
(-1)((-1) + (-1)) + (-1)((-1) + (-1)) + (-1)((-1) + (-1))
Simplify each term:
(-1)(-2) + (-1)(-2) + (-1)(-2) = 2 + 2 + 2 = 6
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The value of tan⁻¹(tan 5π/6) + cos⁻¹(cos 13π/6) is
We are given: tan⁻¹(tan 5π/6) + cos⁻¹(cos 13π/6) Step 1: Simplify tan⁻¹(tan 5π/6) The range of tan⁻¹ x is (-π/2, π/2). For any angle θ, tan⁻¹(tan θ) gives the principal value of θ, which must lie in this range. The angle 5π/6 lies outside this range. To bring it into the principal range, we use theRead more
We are given:
tan⁻¹(tan 5π/6) + cos⁻¹(cos 13π/6)
Step 1: Simplify tan⁻¹(tan 5π/6)
The range of tan⁻¹ x is (-π/2, π/2).
For any angle θ, tan⁻¹(tan θ) gives the principal value of θ, which must lie in this range.
The angle 5π/6 lies outside this range. To bring it into the principal range, we use the periodicity of tan and adjust it:
5π/6 – π = -π/6
Thus:
tan⁻¹(tan 5π/6) = -π/6
Step 2: Simplify cos⁻¹(cos 13π/6)
The range of cos⁻¹ x is [0, π].
For any angle θ, cos⁻¹(cos θ) gives the principal value of θ, which must lie in this range.
The angle 13π/6 is outside this range. To bring it into the range, subtract 2π:
13π/6 – 2π = π/6
Thus:
cos⁻¹(cos 13π/6) = π/6
Step 3: Add the two results
Now, add the simplified terms:
tan⁻¹(tan 5π/6) + cos⁻¹(cos 13π/6) = -π/6 + π/6 = 0
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The domain of the function sin⁻¹ (3x -1) is
Let us now solve this function step by step to find out the domain of sin⁻¹(3x - 1). Step 1: Domain of sin⁻¹(y) We know that sin⁻¹(y) is defined only if -1 ≤ y ≤ 1. In the case of sin⁻¹(3x - 1), we must have: -1 ≤ 3x - 1 ≤ 1 Step 2: Solve the inequality 1. Add 1 to all sides: 0 ≤ 3x ≤ 2 2. DivideRead more
Let us now solve this function step by step to find out the domain of sin⁻¹(3x – 1).
Step 1: Domain of sin⁻¹(y)
We know that sin⁻¹(y) is defined only if -1 ≤ y ≤ 1.
In the case of sin⁻¹(3x – 1), we must have:
-1 ≤ 3x – 1 ≤ 1
Step 2: Solve the inequality
1. Add 1 to all sides:
0 ≤ 3x ≤ 2
2. Divide through by 3:
0 ≤ x ≤ 2/3
Step 3: Domain
The domain of sin⁻¹(3x – 1) is:
[0, 2/3]
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sin[π/3 + sin⁻¹(1/2)] is equal to
To solve sin[π/3 + sin⁻¹(1/2)], let’s break it into steps: Step 1: Simplify sin⁻¹(1/2) The angle whose sine is 1/2 in the range of sin⁻¹ (i.e., [-π/2, π/2]) is: sin⁻¹(1/2) = π/6 Step 2: Substitute into the given expression Now substitute sin⁻¹(1/2) = π/6 into the expression: sin[π/3 + sin⁻¹(1/2)] =Read more
To solve sin[π/3 + sin⁻¹(1/2)], let’s break it into steps:
Step 1: Simplify sin⁻¹(1/2)
The angle whose sine is 1/2 in the range of sin⁻¹ (i.e., [-π/2, π/2]) is:
sin⁻¹(1/2) = π/6
Step 2: Substitute into the given expression
Now substitute sin⁻¹(1/2) = π/6 into the expression:
sin[π/3 + sin⁻¹(1/2)] = sin(π/3 + π/6)
Step 3: Simplify π/3 + π/6
Find a common denominator:
π/3 + π/6 = 2π/6 + π/6 = 3π/6 = π/2
Step 4: Simplify sin(π/2)
From the unit circle, sin(π/2) = 1.
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Which of the following is the principal value branch of cosec⁻¹x?
To determine the principal value branch of cosec⁻¹x, let’s analyze the properties: Step 1: Definition of the principal value branch The principal value branch of cosec⁻¹x is defined such that: 1. It includes all possible values of the inverse cosecant function. 2. It avoids discontinuities or undefiRead more
To determine the principal value branch of cosec⁻¹x, let’s analyze the properties:
Step 1: Definition of the principal value branch
The principal value branch of cosec⁻¹x is defined such that:
1. It includes all possible values of the inverse cosecant function.
2. It avoids discontinuities or undefined values (like when cosec x = 0).
Step 2: Range of cosec⁻¹x
For cosec⁻¹x, the principal value is taken from the range:
[-π/2, π/2] – {0}
This excludes 0 because cosec x is undefined at sin x = 0.
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