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

You visited us 1 times! Enjoying our articles? Unlock Full Access!

Byju's Answer

Standard X

Mathematics

Trigonometric Identities

tan -1x/a+√a2...

Question

$t a n^{- 1} {\frac{x}{a + \sqrt{a^{2} - x^{2}}}}$

Open in App

Solution

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

Suggest Corrections

Similar questions

Q. The value of

\int \frac{\sqrt{x^{2} - a^{2}}}{x} d x

will be

\frac{d}{d x} [{tan}^{- 1} (\frac{x}{\sqrt{a^{2} - x^{2}}})] =

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}}\}

Q. Write each of the following in the simplest form:
(i)

\sin^{- 1} \{x \sqrt{1 - x} - \sqrt{x} \sqrt{1 - x^{2}}\}

(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}}\}

(xi)

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

Q. If

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

- a < x < a

.
find the value of p.

Join BYJU'S Learning Program

tan−1{xa+√a2−x2}

$t a n^{- 1} {\frac{x}{a + \sqrt{a^{2} - x^{2}}}}$