Cos -13 - 5- -1x-tan -11 - 2- then x-

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Standard Formulae - 1

cos -13 / 5+ ...

Question

cos^-1(3/5) + sec^-1(x) = tan^-1(1/2) , then x=?

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{cosec}^{- 1} (3 / 2) + {cos}^{- 1} (2 / 3) - 2 {cot}^{- 1} (1 / 7) - {cot}^{- 1} 7

{cot}^{- 1} 7

{sin}^{- 1} x + {cos}^{- 1} x = π / 2, {tan}^{- 1} x + {cot}^{- 1} x = π / 2, {cosec}^{- 1} x = {sin}^{- 1} (1 / x),

{cot}^{- 1} (1 / x) = {tan}^{- 1} (x), {cot}^{- 1} (x) = {tan}^{- 1} (1 / x)

{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}^{- 1} \frac{1}{3} = \frac{π}{4}

{t a n}^{- 1} \frac{1}{2} + {t a n}^{- 1} \frac{1}{5} + {t a n}^{- 1} \frac{1}{8} = \frac{π}{4}

{t a n}^{- 1} \frac{3}{4} + {t a n}^{- 1} \frac{3}{5} - {t a n}^{- 1} \frac{8}{19} = \frac{π}{4}

{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 s (2 {s i n}^{- 1} x) = 1 / 9

{c o s}^{- 1} (3 / 5) - {s i n}^{- 1} (4 / 5) = {c o s}^{- 1} x

s i n ({s i n}^{- 1} \frac{1}{5} + {c o s}^{- 1} x) = 1

1 / 5

{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 {t a n}^{- 1} \frac{1}{5} + {s e c}^{- 1} \frac{5 \sqrt{2}}{7} + 2 {t a n}^{- 1} \frac{1}{8} = \frac{π}{4}

{c o s}^{- 1} \frac{12}{13} + 2 {c o s}^{- 1} \sqrt{(\frac{64}{65})} + {c o s}^{- 1} \sqrt{(\frac{49}{50})} = {c o s}^{- 1} \frac{1}{\sqrt{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

{s i n}^{- 1} (1 - x) - 2 {s i n}^{- 1} x = π / 2

{s i n}^{- 1} x + {s i n}^{- 1} (1 - x) = {c o s}^{- 1} x

0, 1 / 2

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