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Standard Formulae - 1

148.If cot in...

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148.If cot inverse x + tan inverse 3= pi/2 , then x = (1) 1/3 (2) 1/4 (3)4 (4) 4

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Q. 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

{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

Q. 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

{c o t}^{- 1} 9 + {c o s e c}^{1} \frac{\sqrt{41}}{4} = \frac{π}{4}

{t a n}^{- 1} \frac{1}{7} + {t a n}^{- 1} \frac{1}{13} = {t a n}^{- 1} \frac{2}{9}

{t a n}^{- 1} 2 + {t a n}^{- 1} 3 = 3 π / 4

Q. Considering only the principal values of inverse functions, the set

A = {x \geq 0 : {tan}^{- 1} (2 x) + {tan}^{- 1} (3 x) = \frac{π}{4}}

⎡ ⎢ ⎣ \begin{matrix} x & 1 & 1 2 & 3 & 4 1 & 1 & 1 \end{matrix} ⎤ ⎥ ⎦

has no inverse, then

x =

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) Find whether

x = 2

satisfies the equation

{t a n}^{- 1} \frac{x + 1}{x - 1} + {t a n}^{- 1} \frac{x - 1}{x} = {t a n}^{- 1} (- 7)

If not, then how should the equation be re-written ?

{t a n}^{- 1} \frac{4}{3} + {t a n}^{- 1} \frac{5}{6} + {t a n}^{- 1} \frac{39}{2} - π = . . . . .

x_{1}, x_{2}, x_{3}, x_{4}

are roots of equation

x^{4} - x^{3} s i n 2 β + x^{2} c o s 2 β - x c o s β - s i n β = 0

, then prove that

\sum_{i = 1}^{4} {t a n}^{- 1} x_{1} = \frac{π}{2} - β

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