For any a- b- x- y-0- prove that-23 tan -13 a b 2-a 3 b 3-3 a 2 b-23 tan -13 x y 2-x 3 y 3-3 x 2 y-tan -12 - 2- 2where -a x-b y- -b x-a y

Let #160;a=btanm #160;and #160;x=ytann Then, 23tan 13ab2 a3b3 3a2b+23tan 13xy2 x3y3 3x2y=23tan 13b3tanm b3tan3mb3 3b3tan2m+23tan 13y3tann y3tan3ny3 3y3ta ...

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Standard XII

Mathematics

First Fundamental Theorem of Calculus

For any a, b,...

Question

For any a, b, x, y > 0, prove that:
$\frac{2}{3} \tan^{- 1} (\frac{3 a b^{2} - a^{3}}{b^{3} - 3 a^{2} b}) + \frac{2}{3} \tan^{- 1} (\frac{3 x y^{2} - x^{3}}{y^{3} - 3 x^{2} y}) = \tan^{- 1} \frac{2 αβ}{α^{2} - β^{2}}$
where α = − ax + by, β = bx + ay

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Solution

$Let a = b \tan m and x = y \tan n$
Then,

$\frac{2}{3} \tan^{- 1} (\frac{3 a b^{2} - a^{3}}{b^{3} - 3 a^{2} b}) + \frac{2}{3} \tan^{- 1} (\frac{3 x y^{2} - x^{3}}{y^{3} - 3 x^{2} y}) = \frac{2}{3} \tan^{- 1} (\frac{3 b^{3} \tan m - b^{3} \tan^{3} m}{b^{3} - 3 b^{3} \tan^{2} m}) + \frac{2}{3} \tan^{- 1} (\frac{3 y^{3} \tan n - y^{3} \tan^{3} n}{y^{3} - 3 y^{3} \tan^{2} n}) = \frac{2}{3} \tan^{- 1} (\frac{3 \tan m - \tan^{3} m}{1 - 3 \tan^{2} m}) + \frac{2}{3} \tan^{- 1} (\frac{3 \tan n - \tan^{3} n}{1 - \tan^{2} n}) = \frac{2}{3} \tan^{- 1} (\tan 3 m) + \frac{2}{3} \tan^{- 1} (\tan 3 n) [∵ \tan 3 x = \frac{3 \tan x - \tan^{3} x}{1 - 3 \tan^{2} x}] = \frac{2}{3} (3 m) + \frac{2}{3} (3 n) = 2 m + 2 n = 2 (\tan^{- 1} \frac{a}{b} + \tan^{- 1} \frac{x}{y}) [∵ a = b \tan m, x = y \tan n] = 2 \tan^{- 1} (\frac{\frac{a}{b} + \frac{x}{y}}{1 - \frac{a}{b} \frac{x}{y}}) = 2 \tan^{- 1} (\frac{a y + b x}{b y - a x}) = \tan^{- 1} \{\frac{2 \frac{a y + b x}{b y - a x}}{1 - {(\frac{a y + b x}{b y - a x})}^{2}}\} = \tan^{- 1} \{\frac{2 (a y + b x) (b y - a x)}{{(b y - a x)}^{2} - {(a y + b x)}^{2}}\} = \tan^{- 1} \{\frac{2 α β}{α^{2} - β^{2}}\} [∵ β = a y + b x and α = b y - a x]$

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Similar questions

Q. Prove that:

\tan^{- 1} \frac{2 a b}{a^{2} - b^{2}} + \tan^{- 1} \frac{2 x y}{x^{2} - y^{2}} = \tan^{- 1} \frac{2 αβ}{α^{2} - β^{2}},

where α = ax − by and β = ay + bx.

Q. Solve for x and y:

a x + b y - a + b = 0, b x - a y - a - b = 0 .

\int_{0}^{π / 2} \frac{1}{2 + \cos x} d x

equals

(a)

\frac{1}{3} \tan^{- 1} (\frac{1}{\sqrt{3}})

(b)

\frac{2}{\sqrt{3}} \tan^{- 1} (\frac{1}{\sqrt{3}})

(c)

\sqrt{3} \tan^{- 1} (\sqrt{3})

(d)

2 \sqrt{3} \tan^{- 1} \sqrt{3}

Q. Match the following provided that

a

and

b

any two rational numbers different from zero and

x, y

are any two rational numbers.

$(1)$	$a^{x} \times a^{y}$	$(a)$	$a^{x - y}$
$(2)$	$a^{x} \div a^{y}$	$(b)$	$a^{x y}$
$(3)$	$(a^{x})^{y}$	$(c)$	$a^{x + y}$
$(4)$	$(a b)^{x}$	$(d)$	$\frac{a^{x}}{b^{x}}$
$(5)$	${(\frac{a}{b})}^{x}$	$(e)$	$a^{x} \times b^{x}$

Q. Solve for

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a x + b y = a - b

b x - a y = a + b

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For any a, b, x, y > 0, prove that: 23tan-13ab2-a3b3-3a2b+23tan-13xy2-x3y3-3x2y=tan-12αβα2-β2 where α = − ax + by, β = bx + ay

For any a, b, x, y > 0, prove that:
$\frac{2}{3} \tan^{- 1} (\frac{3 a b^{2} - a^{3}}{b^{3} - 3 a^{2} b}) + \frac{2}{3} \tan^{- 1} (\frac{3 x y^{2} - x^{3}}{y^{3} - 3 x^{2} y}) = \tan^{- 1} \frac{2 αβ}{α^{2} - β^{2}}$
where α = − ax + by, β = bx + ay