Tan- -2- 1-cos - - 1-cos - -

LHS =tanα β #160; 2 #160; √( 1 cos( α β #160; ) #160; #160; 1+cos( α β #160; ) #160; #160; #160; ) =√( 2sin ^ 2 ( α β #16 ...

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Sin2A and Cos2A in Terms of tanA

tanα -β2=±√ 1...

Question

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

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Solution

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c o s (α + β) + s i n (α - β) = 0

2010 t a n β + 1 = 0

t a n α

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

\frac{t a n α + t a n β}{c o t α + c o t β} + [c o s (α - β) s e c (α + β) + 1]^{- 1} .

cos θ = \frac{cos α - cos β}{1 - cos α cos β}

tan \frac{θ}{2}

= \pm tan \frac{α}{2}

cot \frac{β}{2}

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