If x-tan -1 tan 5- 4 and y-tan -1 tan- 3 - then show that- 4x-3y-

x=tan ^ 1 ( tan 5π #160; 4 #160; #160; ) #160; y=tan ^ 1 ( tanπ #160; 3 #160; #160; ) #160; =tan ^ 1 ( tanπ #160; 4 ...

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Domain and Range of Basic Inverse Trigonometric Functions

If x=tan -1...

Question

If $x = {tan}^{- 1} (tan \frac{5 π}{4})$ and $y = {tan}^{- 1} (tan \frac{π}{3})$ , then show that: $4 x = 3 y$ .

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y = {tan}^{- 1} (\frac{2 x}{1 - x^{2}}) + {tan}^{- 1} (\frac{3 x - x^{3}}{1 - 3 x^{2}}) - {tan}^{1} (\frac{4 x - 4 x^{3}}{1 - 6 x^{2} + x^{4}})

\frac{d y}{d x} = \frac{1}{1 + x^{2}}

{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

{t a n}^{- 1} (x + \frac{2}{x}) - {t a n}^{- 1} \frac{4}{x} - {t a n}^{- 1} (x - \frac{2}{x}) = 0

x = . . . . . . . . . .

x = . . . . . . . . (x ϵ R)

{s i n}^{- 1} x + {s i n}^{- 1}

y = 2 π / 3

{c o s}^{- 1} x - {c o s}^{- 1}

y = π / 3

x = . . . . . . . .

y = . . . . . . .

{t a n}^{- 1} y = 4 {t a n}^{- 1} x, (| x | < t a n \frac{π}{8})

t a n (π / 8)

x^{4} - 6 x^{2} + 1 = 0

{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

t a n (\frac{π}{4} + \frac{1}{2} {c o s}^{- 1} \frac{a}{b}) + t a n (\frac{π}{4} - \frac{1}{2} {c o s}^{- 1} \frac{a}{b}) = \frac{2 b}{a}

1 + 4 x^{2} = 4 x sec A

sec A + tan A = \frac{1}{2 x}

	Column - I		Column - II
a	x = $c o s e c^{2}$ $(c o t^{- 1} 3)$ - $s e c^{2}$ $(t a n^{- 1} 2)$	p	x = 2
b	$t a n^{- 1}$ x + $t a n^{- 1}$ $\frac{1}{y}$ = $(t a n^{- 1} 3)$ and $y^{2}$ + y - 56 = 0	q	x = 5
c	$c o s^{- 1}$ x = $t a n^{- 1}$ y and $y^{2}$ = 3	r	$x = \frac{1}{2}$
d	$s i n^{- 1}$ $(t a n \frac{π}{4})$ - $s i n^{- 1}$ $\sqrt{\frac{3}{y}}$ = $\frac{π}{6}$ and $x^{2}$ = y	s	x = - $\frac{1}{2}$

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