y = tan^ 1√(a x/a+x) Let x = a cos θ θ = cos^ 1 ( x/a ) √(a x/a+x) = √(a a cos θ/a+a cos θ) = √(1 cos θ/1 + cos θ) = √(sin^2 θ/2 + cos^2 θ/2 cos^2 θ/2 ...

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Quantitative Aptitude

Functions

tan -1√a-x/a+...

Question

$t a n^{- 1} \sqrt{\frac{a - x}{a + x}}$

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Solution

$y = t a n^{- 1} \sqrt{\frac{a - x}{a + x}}$
$L e t x = a c o s θ θ = c o s^{- 1} (\frac{x}{a})$
$\sqrt{\frac{a - x}{a + x}} = \sqrt{\frac{a - a c o s θ}{a + a c o s θ}} = \sqrt{\frac{1 - c o s θ}{1 + c o s θ}} = \sqrt{\frac{s i n^{2} \frac{θ}{2} + c o s^{2} \frac{θ}{2} - c o s^{2} \frac{θ}{2} + s i n^{2} \frac{θ}{2}}{c o s^{2} \frac{θ}{2} + s i n^{2} \frac{θ}{2} + c o s^{2} \frac{θ}{2} - s i n^{2} \frac{θ}{2}}}$
$\sqrt{\frac{a - x}{a + x}} = \sqrt{\frac{2 s i n^{2} \frac{θ}{2}}{2 c o s^{2} \frac{θ}{2}}} = t a n \frac{θ}{2}$
$∴ y = t a n^{- 1} [t a n \frac{θ}{2}] = \frac{θ}{2}$
$∴ y = \frac{1}{2} c o s^{- 1} (\frac{x}{a})$

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Similar questions

t a n^{- 1} \sqrt{\frac{a - x}{a + x}}

Q. The value of

{lim}_{x \to 0} \frac{t a n (a + x) - t a n (a - x)}{t a n^{- 1} (a + x) - t a n^{- 1} (a - x)}

is equal to

Q. If

{tan}^{- 1} (c o s e c {tan}^{- 1} x - tan {cot}^{- 1} x) = a {tan}^{- 1} x (x \neq 0)

.
Find the value of a

\int \tan^{- 1} \sqrt{x} d x i s e q u a l t o (a) (x + 1) \tan^{- 1} \sqrt{x} - \sqrt{x} + C (b) x \tan^{- 1} \sqrt{x} - \sqrt{x} + C (c) \sqrt{x} - x \tan^{- 1} \sqrt{x} + C (d) \sqrt{x} - (x + 1) \tan^{- 1} \sqrt{x} + C

Q. Write each of the following in the simplest form:

(i)

\cot^{- 1} \frac{a}{\sqrt{x^{2} - a^{2}}}, |x| > a

(ii)

\tan^{- 1} \{x + \sqrt{1 + x^{2}}\}, x \in R

(iii)

\tan^{- 1} \{\sqrt{1 + x^{2}} - x\}, x \in R

(iv)

\tan^{- 1} \{\frac{\sqrt{1 + x^{2}} - 1}{x}\}, x \neq 0

(v)

\tan^{- 1} \{\frac{\sqrt{1 + x^{2}} + 1}{x}\}, x \neq 0

(vi)

\tan^{- 1} \sqrt{\frac{a - x}{a + x}}, - a < x < a

(vii)

\tan^{- 1} \{\frac{x}{a + \sqrt{a^{2} - x^{2}}}\}, - a < x < a

(viii)

\sin^{- 1} \{\frac{x + \sqrt{1 - x^{2}}}{\sqrt{2}}\}, - 1 < x < 1

(ix)

\sin^{- 1} \{\frac{\sqrt{1 + x} + \sqrt{1 - x}}{2}\}, 0 < x < 1

(x)

\sin \{2 \tan^{- 1} \sqrt{\frac{1 - x}{1 + x}}\}

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Functions

Other Quantitative Aptitude

tan−1√a−xa+x

y=tan−1√a−xa+x Let x=a cos θ θ=cos−1(xa) √a−xa+x=√a−a cosθa+a cosθ=√1−cosθ1+cosθ=√sin2θ2+cos2θ2−cos2θ2+sin2θ2cos2θ2+sin2θ2+cos2θ2−sin2θ2 √a−xa+x=√2 sin2θ22 cos2θ2=tan θ2 ∴ y=tan−1[tan θ2]=θ2 ∴ y=12cos−1(xa)

$t a n^{- 1} \sqrt{\frac{a - x}{a + x}}$