2 tan -1-tan-2tan-4-2- is equal to

The correct option is D We have ,2 tan^ 1[tanα/2tan(π/4 β/2 ) ] =tan^ 12tanπ/2tan(π/4 β/2 )/1 tan^2α/2tan^2(π/4 β/2 ) =tan^ 12sin(α/2)cos(α/2) sin(π/4 β/2 )co ...

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Byju's Answer

Standard XII

Mathematics

Sum of Trigonometric Ratios in Terms of Their Product

2 tan -1[tanα...

Question

$2 t a n^{- 1} [t a n \frac{α}{2} t a n (\frac{π}{4} - \frac{β}{2})]$ is equal to

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Solution

The correct option is D

$W e h a v e, 2 t a n^{- 1} [t a n \frac{α}{2} t a n (\frac{π}{4} - \frac{β}{2})]$
$= t a n^{- 1} \frac{2 t a n \frac{π}{2} t a n (\frac{π}{4} - \frac{β}{2})}{1 - t a n^{2} \frac{α}{2} t a n^{2} (\frac{π}{4} - \frac{β}{2})}$
$= t a n^{- 1} \frac{2 s i n (\frac{α}{2}) c o s (\frac{α}{2}) s i n (\frac{π}{4} - \frac{β}{2}) c o s (\frac{π}{4} - \frac{β}{2})}{c o s^{2} (\frac{α}{2}) c o s^{2} (\frac{π}{4} - \frac{β}{2}) - s i n^{2} (\frac{α}{2}) s i n^{2} (\frac{π}{4} - \frac{β}{2})}$
$= t a n^{- 1} \frac{s i n α . s i n (\frac{π}{2} - β)}{2 [c o s (\frac{α}{2} - \frac{β}{2} + \frac{π}{4}) c o s (\frac{α}{2} + \frac{β}{2} - \frac{π}{4})]}$
$= t a n^{- 1} \frac{s i n α c o s β}{c o s α + c o s (\frac{π}{2} - β)} = t a n^{- 1} \frac{s i n α c o s β}{c o s α + s i n β}$

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Similar questions

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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}

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