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Trigonometric Ratios of Compound Angles

Prove that ta...

Question

Prove that tan1 is greater than tan–¹π/4

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Solution

1 rad = 57.3 degree
tan 1 = tan(57.3)=1.557
tan–¹π/4=tan–¹(45)=1
so tan1>tan–¹π/4

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{tan}^{- 1} (\frac{cos x}{1 + sin x}) = \frac{π}{4} - \frac{x}{2}, x ϵ (- \frac{π}{2}, \frac{π}{2})

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}

{tan}^{- 1} (1) + {tan}^{- 1} (2) + {tan}^{- 1} (3) = π

{tan}^{- 1} \frac{x - 1}{x - 2} + {tan}^{- 1} \frac{x + 1}{x + 2} = \frac{π}{4}

Q. Prove that :

tan [\frac{π}{4} + \frac{1}{2} {cos}^{- 1} \frac{a}{b}] + tan [\frac{π}{4} - \frac{1}{2} {cos}^{- 1} \frac{a}{b}] = \frac{2 b}{a}

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