The value of sin -1tancos -1 x is equal to-Bihar CEE 1974-A- xB- -2C- None of theseD- 1

The correct option is A xLet cos^ 1x=θ⇒ x=cos θ⇒ secθ=1/x ⇒ tanθ=√(sec^2θ 1)=√(1/2 1)=1/x√(1 x^2) Now sin cot^ 1tanθ=sin cot^ 1(1/x√(1 x^2) ) Again, putting x= ...

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Standard XII

Mathematics

Integration Using Substitution

The value of ...

Question

The value of $s i n c o t^{- 1} t a n c o s^{- 1} x$ is equal to
[Bihar CEE 1974]

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None of these

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Solution

The correct option is A
x

Let $c o s^{- 1} x = θ \Rightarrow x = c o s θ \Rightarrow s e c θ = \frac{1}{x}$
$\Rightarrow t a n θ = \sqrt{s e c^{2} θ - 1} = \sqrt{\frac{1}{2} - 1} = \frac{1}{x} \sqrt{1 - x^{2}}$
Now $s i n c o t^{- 1} t a n θ = s i n c o t^{- 1} (\frac{1}{x} \sqrt{1 - x^{2}})$
Again, putting $x = s i n θ$
$s i n c o t^{- 1} (\frac{1}{x} \sqrt{1 - x^{2}}) = s i n c o t^{- 1} (\frac{\sqrt{1 - s i n^{2} θ}}{s i n θ})$
$= s i n c o t^{- 1} (c o t θ) = s i n θ = x .$

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Similar questions

Q. sin

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(b)

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The value of sincot−1 tancos−1x is equal to [Bihar CEE 1974]

The correct option is A xLet cos−1x=θ⇒x=cos θ⇒secθ=1x ⇒tanθ=√sec2θ−1=√12−1=1x√1−x2 Now sin cot−1tanθ=sin cot−1(1x√1−x2) Again, putting x=sinθ sin cot−1(1x√1−x2)=sin cot−1(√1−sin2θsinθ) =sin cot−1(cotθ)=sinθ=x.

The value of $s i n c o t^{- 1} t a n c o s^{- 1} x$ is equal to
[Bihar CEE 1974]