Prove that- tan-114-tan-129-12cos-135-

LHS=tan^ 114+tan^ 129=tan^ 1(1/4+2/91 1/4×2/9)=tan^ 1(17/3617/18)=tan^ 112 Let x2=tan^ 112 tanx2=12 cos x=1 tan^2 θ/21+tan^2 θ/2=1 1/41+1/4= ...

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Domain and Range of Basic Inverse Trigonometric Functions

Prove that: ...

Question

Prove that:
${tan}^{- 1} \frac{1}{4} + {tan}^{- 1} \frac{2}{9} = \frac{1}{2} {cos}^{- 1} \frac{3}{5}$ .

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{tan}^{- 1} \frac{1}{4} + {tan}^{- 1} \frac{2}{9} = \frac{1}{2} {sin}^{- 1} \frac{4}{5}

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}

{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}{21} + {t a n}^{- 1} \frac{1}{13} + {t a n}^{- 1} (- \frac{1}{8}) = 0

{t a n}^{- 1} \frac{2}{3} = \frac{1}{2} {t a n}^{- 1} \frac{12}{5}

{t a n}^{- 1} \frac{1}{4} + {t a n}^{- 1} \frac{2}{9} = \frac{1}{2} {c o s}^{- 1} \frac{3}{5}

{tan}^{- 1} (\frac{1}{\sqrt{3}} tan \frac{x}{2}) = \frac{1}{2} {cos}^{- 1} (\frac{1 + 2 cos x}{2 + cos x})

t a n^{- 1} \frac{1}{3} + t a n^{- 1} \frac{1}{5} = \frac{1}{2} c o s^{- 1} \frac{33}{65}

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