If r2-x2-y2-z2 and tan -1y z-x r-tan -1x z-y r-2-tan -1- then

The correct option is D ϕ=xyzrtan^ 1yzxr+tan^ 1xzyr =tan^ 1yzxr+xzyr1 yzxr·xzyr =tan^ 1zxyr(x^2+y^2)r^2 z^2r^2 =tan^ 1zrxy = tan^ 1zrxy=^ 1xyzr = π2 tan^ 1xyzr ...

You visited us 65 times! Enjoying our articles? Unlock Full Access!

Byju's Answer

Standard XII

Mathematics

Logarithmic Differentiation

If r2=x2+y2+z...

Question

If $r^{2} = x^{2} + y^{2} + z^{2}$ and ${tan}^{- 1} \frac{y z}{x r} + {tan}^{- 1} \frac{x z}{y r} = \frac{π}{2} - {tan}^{- 1} ϕ$ then

ϕ = \frac{x + y}{z r}

No worries! We‘ve got your back. Try BYJU‘S free classes today!

ϕ = \frac{y z}{x r} + \frac{x z}{y r}

No worries! We‘ve got your back. Try BYJU‘S free classes today!

ϕ = \frac{z r}{x y}

No worries! We‘ve got your back. Try BYJU‘S free classes today!

ϕ = \frac{x y}{z r}

Right on! Give the BNAT exam to get a 100% scholarship for BYJUS courses

Open in App

Solution

The correct option is D $ϕ = \frac{x y}{z r}$
${tan}^{- 1} \frac{y z}{x r} + {tan}^{- 1} \frac{x z}{y r} = {tan}^{- 1} \frac{\frac{y z}{x r} + \frac{x z}{y r}}{1 - \frac{y z}{x r} \cdot \frac{x z}{y r}}$
$= {tan}^{- 1} \frac{\frac{z}{x y r} (x^{2} + y^{2})}{\frac{r^{2} - z^{2}}{r^{2}}} = {tan}^{- 1} \frac{z r}{x y}$
$= {tan}^{- 1} \frac{z r}{x y} = {cot}^{- 1} \frac{x y}{z r}$
$= \frac{π}{2} - {tan}^{- 1} \frac{x y}{z r}$
$∴ ϕ = \frac{x y}{z r}$

Suggest Corrections

Similar questions

Q. If

x^{2} + y^{2} + z^{2} = r^{2}

, then

{tan}^{- 1} (\frac{x y}{z r}) + {tan}^{- 1} (\frac{y z}{x r}) + {tan}^{- 1} (\frac{x z}{y r}) =

Q. Let

x, y, z

and

r

be positive real numbers such that

x^{2} + y^{2} + z^{2} = r^{2} .

Then the value of

{tan}^{- 1} (\frac{x y}{z r}) + {tan}^{- 1} (\frac{y z}{x r}) + tan (\frac{z x}{y r})

\sum t a n^{- 1} (\frac{y z}{x \sqrt{x^{2} + y^{2} + z^{2}}}) = ?

Q. If

r^{2} = x^{2} + y^{2} + z^{2}

, then prove that

{tan}^{- 1} (\frac{y z}{r x}) + {tan}^{- 1} (\frac{z x}{r y}) + {tan}^{- 1} (\frac{x y}{r z}) = \frac{π}{2}

Inverse circular functions,Principal values of

{s i n}^{- 1} x, {c o s}^{- 1} x, {t a n}^{- 1} x

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

x y < 1

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

x y > 1

(a) If

{c o s}^{- 1} p + {c o s}^{- 1} q + {c o s}^{- 1} r = π

, then prove that

p^{2} + q^{2} + r^{2} + 2 p q r = 1

(b) If

{s i n}^{- 1} x + {s i n}^{- 1} y + {s i n}^{- 1} z = π

, then prove that

x^{4} + y^{4} + z^{4} + 4 x^{2} y^{2} z^{2} = 2 (x^{2} y^{2} + y^{2} z^{2} + z^{2} x^{2})

{t a n}^{- 1} x + {t a n}^{- 1} y + {t a n}^{- 1} z = π

π / 2

show that

x + y + z = x y z

x y + y z + z x = 1

Join BYJU'S Learning Program

If r2=x2+y2+z2 and tan−1yzxr+tan−1xzyr=π2−tan−1ϕ then

The correct option is D ϕ=xyzrtan−1yzxr+tan−1xzyr=tan−1yzxr+xzyr1−yzxr⋅xzyr =tan−1zxyr(x2+y2)r2−z2r2=tan−1zrxy =tan−1zrxy=cot−1xyzr =π2−tan−1xyzr ∴ϕ=xyzr

If $r^{2} = x^{2} + y^{2} + z^{2}$ and ${tan}^{- 1} \frac{y z}{x r} + {tan}^{- 1} \frac{x z}{y r} = \frac{π}{2} - {tan}^{- 1} ϕ$ then