If c o s − 1 √ p + c o s − 1 √ 1 − p + c o s − 1 √ 1 − q = 3 π 4 , then the value of q is Options: A . &nbsp 1 B . &nbsp 1 √ 2 C . &nbsp 1 3 D . &nbsp 1 2

Answer: Option D : D Let α = c o s − 1 √ p : β = c o s − 1 √ 1 − p and γ = c o s − 1 √ 1 − q or c o s α = √ p : c o s β = √ 1 − p and c o s γ = √ 1 − q . Therefore s i n α = √ 1 − p , s i n β = √ p and s i n γ = √ q . The given equation may be written as α + β + γ = 3 π 4 o r α + β = 3 π 4 − γ or c o s ( α + β ) = c o s ( 3 π 4 − γ ) ⇒ c o s α c o s β − s i n α s i n β = c o s { π − ( π 4 + γ ) } = − c o s ( π 4 + γ ) ⇒ √ p √ 1 − p − √ 1 − p √ p = − ( 1 √ 2 √ 1 − q − 1 √ 2 . √ q ) ⇒ 0 = √ 1 − q − √ q ⇒ 1 − q = q ⇒ q = 1 2 .

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Discussion Forum : Inverse Trigonometric Functions

Question -

If $c o s^{- 1} \sqrt{p} + c o s^{- 1} \sqrt{1 - p} + c o s^{- 1} \sqrt{1 - q} = \frac{3 π}{4}$ , then the value of q is

Options:

A .

1

B .

\frac{1}{\sqrt{2}}

C .

\frac{1}{3}

D .

\frac{1}{2}

Answer: Option D
:
D
Let

α = c o s^{- 1} \sqrt{p} : β = c o s^{- 1} \sqrt{1 - p}

and

γ = c o s^{- 1} \sqrt{1 - q}

c o s α = \sqrt{p} : c o s β = \sqrt{1 - p}

and

c o s γ = \sqrt{1 - q} .

Therefore

s i n α = \sqrt{1 - p}, s i n β = \sqrt{p}

and

s i n γ = \sqrt{q} .

The given equation may be written as

α + β + γ = \frac{3 π}{4} o r α + β = \frac{3 π}{4} - γ

c o s (α + β) = c o s (\frac{3 π}{4} - γ)

\Rightarrow c o s α c o s β - s i n α s i n β

c o s {π - (\frac{π}{4} + γ)} = - c o s (\frac{π}{4} + γ)

\Rightarrow \sqrt{p} \sqrt{1 - p} - \sqrt{1 - p} \sqrt{p} = - (\frac{1}{\sqrt{2}} \sqrt{1 - q} - \frac{1}{\sqrt{2}} . \sqrt{q})

\Rightarrow 0 = \sqrt{1 - q} - \sqrt{q} \Rightarrow 1 - q = q \Rightarrow q = \frac{1}{2}

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Discussion Forum : Inverse Trigonometric Functions

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