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Sum of Trigonometric Ratios in Terms of Their Product

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Question

Show that $2 {tan}^{- 1} {tan \frac{α}{2} \times tan {\frac{π}{4} - \frac{β}{2}}} = {tan}^{- 1} {\frac{sin α cos β}{cos α + sin β}}$

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Solution

Consider the L.H.S.

$2 {tan}^{- 1} {tan \frac{a}{2} . tan {\frac{π}{4} - \frac{β}{2}}}$

We know that,

$tan (A - B) = \frac{tanA - tanB}{1 + tan A tan B}$

Then,

$2 {tan}^{- 1} ⎧ ⎪ ⎪ ⎪ ⎨ ⎪ ⎪ ⎪ ⎩ tan \frac{α}{2} ⎛ ⎜ ⎜ ⎜ ⎝ \frac{tan \frac{π}{4} - tan \frac{β}{2}}{1 + tan \frac{π}{4} tan \frac{β}{2}} ⎞ ⎟ ⎟ ⎟ ⎠ ⎫ ⎪ ⎪ ⎪ ⎬ ⎪ ⎪ ⎪ ⎭$

$= 2 {tan}^{- 1} ⎧ ⎪ ⎪ ⎪ ⎨ ⎪ ⎪ ⎪ ⎩ tan \frac{α}{2} ⎛ ⎜ ⎜ ⎜ ⎝ \frac{1 - tan \frac{β}{2}}{1 + tan \frac{β}{2}} ⎞ ⎟ ⎟ ⎟ ⎠ ⎫ ⎪ ⎪ ⎪ ⎬ ⎪ ⎪ ⎪ ⎭$

$= 2 {tan}^{- 1} ⎧ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎨ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎩ \frac{sin \frac{α}{2}}{cos \frac{α}{2}} \frac{⎛ ⎜ ⎜ ⎜ ⎝ 1 - \frac{sin \frac{β}{2}}{cos \frac{β}{2}} ⎞ ⎟ ⎟ ⎟ ⎠}{⎛ ⎜ ⎜ ⎜ ⎝ 1 + \frac{sin \frac{β}{2}}{cos \frac{β}{2}} ⎞ ⎟ ⎟ ⎟ ⎠} ⎫ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎬ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎭$

$= 2 {tan}^{- 1} ⎧ ⎪ ⎪ ⎪ ⎪ ⎨ ⎪ ⎪ ⎪ ⎪ ⎩ \frac{sin \frac{α}{2}}{cos \frac{α}{2}} \frac{(cos \frac{β}{2} - sin \frac{β}{2})}{(cos \frac{β}{2} + sin \frac{β}{2})} ⎫ ⎪ ⎪ ⎪ ⎪ ⎬ ⎪ ⎪ ⎪ ⎪ ⎭$

Let $x = ⎧ ⎪ ⎪ ⎪ ⎪ ⎨ ⎪ ⎪ ⎪ ⎪ ⎩ \frac{sin \frac{α}{2}}{cos \frac{α}{2}} \frac{(cos \frac{β}{2} - sin \frac{β}{2})}{(cos \frac{β}{2} + sin \frac{β}{2})} ⎫ ⎪ ⎪ ⎪ ⎪ ⎬ ⎪ ⎪ ⎪ ⎪ ⎭$

Using formula,

$2 {tan}^{- 1} x = {tan}^{- 1} \frac{2 x}{1 - x^{2}}$

Therefore,

$= {tan}^{- 1} \frac{2 ⎧ ⎪ ⎪ ⎪ ⎪ ⎨ ⎪ ⎪ ⎪ ⎪ ⎩ \frac{sin \frac{α}{2}}{cos \frac{α}{2}} \frac{(cos \frac{β}{2} - sin \frac{β}{2})}{(cos \frac{β}{2} + sin \frac{β}{2})} ⎫ ⎪ ⎪ ⎪ ⎪ ⎬ ⎪ ⎪ ⎪ ⎪ ⎭}{1 - {⎧ ⎪ ⎪ ⎪ ⎪ ⎨ ⎪ ⎪ ⎪ ⎪ ⎩ \frac{sin \frac{α}{2}}{cos \frac{α}{2}} \frac{(cos \frac{β}{2} - sin \frac{β}{2})}{(cos \frac{β}{2} + sin \frac{β}{2})} ⎫ ⎪ ⎪ ⎪ ⎪ ⎬ ⎪ ⎪ ⎪ ⎪ ⎭}^{2}}$

$= {tan}^{- 1} \frac{2 ⎧ ⎪ ⎪ ⎪ ⎪ ⎨ ⎪ ⎪ ⎪ ⎪ ⎩ \frac{sin \frac{α}{2}}{cos \frac{α}{2}} \frac{(cos \frac{β}{2} - sin \frac{β}{2})}{(cos \frac{β}{2} + sin \frac{β}{2})} ⎫ ⎪ ⎪ ⎪ ⎪ ⎬ ⎪ ⎪ ⎪ ⎪ ⎭}{\frac{{cos}^{2} \frac{α}{2} {(cos \frac{β}{2} + sin \frac{β}{2})}^{2} - {sin}^{2} \frac{α}{2} {(cos \frac{β}{2} - sin \frac{β}{2})}^{2}}{{cos}^{2} \frac{α}{2} {(cos \frac{β}{2} + sin \frac{β}{2})}^{2}}}$

$= {tan}^{- 1} \frac{2 sin \frac{α}{2} (cos \frac{β}{2} - sin \frac{β}{2}) cos \frac{α}{2} (cos \frac{β}{2} + sin \frac{β}{2})}{{cos}^{2} \frac{α}{2} ({cos}^{2} \frac{β}{2} + {sin}^{2} \frac{β}{2} + 2 cos \frac{β}{2} sin \frac{β}{2}) - {sin}^{2} \frac{α}{2} ({cos}^{2} \frac{β}{2} + {sin}^{2} \frac{β}{2} - 2 cos \frac{β}{2} sin \frac{β}{2})}$

$= {tan}^{- 1} \frac{2 sin \frac{α}{2} cos \frac{α}{2} ({cos}^{2} \frac{β}{2} - {sin}^{2} \frac{β}{2})}{{cos}^{2} \frac{α}{2} (1 + sin β) - {sin}^{2} \frac{α}{2} (1 - sin β)}$

Since, $2 sin \frac{A}{2} cos \frac{A}{2} = sin A$

${cos}^{2} \frac{A}{2} - {sin}^{2} \frac{A}{2} = cos A$

Therefore,

$= {tan}^{- 1} \frac{sin α cos β}{{cos}^{2} \frac{α}{2} + {cos}^{2} \frac{α}{2} sin β - {sin}^{2} \frac{α}{2} + {sin}^{2} \frac{α}{2} sin β}$

$= {tan}^{- 1} \frac{sin α cos β}{({cos}^{2} \frac{α}{2} - {sin}^{2} \frac{α}{2}) + sin β ({sin}^{2} \frac{α}{2} + {cos}^{2} \frac{α}{2})}$

$= {tan}^{- 1} \frac{sin α cos β}{cos α + sin β (1)}$

$= {tan}^{- 1} \frac{sin α cos β}{cos α + sin β}$

Hence proved.

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Similar questions

2 {tan}^{- 1} [tan \frac{α}{2} tan (\frac{π}{4} - \frac{β}{2})] = {tan}^{- 1} \frac{sin α c o s β}{cos α + sin β}

2 {tan}^{- 1} {tan \frac{α}{2} tan (\frac{π}{4} - \frac{β}{2})}

= {tan}^{- 1} (\frac{sin α cos β}{cos α + sin β})

2 {tan}^{- 1} [tan \frac{^{a}}{2} tan (\frac{π}{4} - \frac{β}{2})] = {tan}^{- 1} \frac{sin α cos β}{sin α + cos β}

s i n α + s i n β = a

c o s α + c o s β = b,

s i n (α + β) = \frac{2 a b}{a^{2} + b^{2}}

c o s (α + β) = \frac{b^{2} - a^{2}}{b^{2} + a^{2}}

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

.
(a) Prove that

2 {t a n}^{2} {t a n \frac{α}{2} t a n (\frac{π}{4} - \frac{β}{2})} = {t a n}^{- 1} [\frac{s i n α c o s β}{s i n β + c o s α}]

{c o s}^{- 1} (\frac{2 + 3 c o s x}{3 + 2 c o s x}) = 2 {t a n}^{- 1} (\frac{1}{\sqrt{5}} t a n \frac{x}{2})

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Sum of Trigonometric Ratios in Terms of Their Product

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