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Monotonicity in an Interval

19. How to ad...

Question

19. How to adjust the tan(inverse)x + tan (inverse) y when xy>1 and x,y>0 and also when xy>1 and x,y<0

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{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

[2 {c o s}^{- 1} c o t (2 {t a n}^{- 1} x)] = 0

A B C

A = t a n^{- 1} 2, B = t a n^{- 1} 3

x > 0, y > 0

x y > 1

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

{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} (\frac{1}{2} t a n 2 A) + {t a n}^{- 1} (c o t A) + {t a n}^{- 1} ({c o t}^{3} A)

0

π / 4 < A < π / 2

= π

0 < A < π / 4

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

(\frac{x + y}{1 - x y})

(\frac{x + y}{1 - x y}) - n

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