Show that:cos−1[1+cosα cosβ cosα +cosβ ] =2tan−1(tan2α tan2β )

cos−1[1+cosα cosβ cosα +cosβ ] =2tan−1(tan2α tan2β ) RHS =2tan−1[tan2αtan2β]=cos−1[1+tan2α/2tan2β/21−tan2α /2tan2β/2]=cos−1⎣⎡1+cos2α/2cos2β/2sin2α/2sin2β/2cos2α/2cos2β/21−sin2α/2sin2β/2⎦⎤=cos−1[cos2α/2cos2β/2+sin2α/2sin2β/2cos2α/2cos2β/2−sin2α/2sin2β/2]=cos−1[2cos2α/2cos2β/2+2sin2α/2sin2β/22cos2α/2cos2β/2−sin2α/2sin2β/2]=cos−1[(1+cosα)(1+cosβ)+(1−cosα)(1−cosβ)(1+cosα)(1+cosβ)−(1−cosα)(1−cosβ)]=cos−1[1−cosαcosβcosα+cosβ]=LHS

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Question

Show that:
$cos^{- 1} [\frac{cos α + cos β}{1 + cos α cos β}] = 2 tan^{- 1} (tan \frac{α}{2} tan \frac{β}{2})$

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Show that:

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

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Text solutionVerified

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

RHS

= 2 t a n_{- 1} [t an \frac{α}{2} t an \frac{β}{2}]

= co s_{- 1} [\frac{1 - t a n ^{2} α /2 t a n ^{2} β /2}{1 + t a n ^{2} α /2 t a n ^{2} β /2}]

= co s^{- 1} ⎣ ⎡ \frac{\frac{1 - s i n ^{2} α /2 s i n ^{2} β /2}{co s ^{2} α /2 co s ^{2} β /2}}{1 + \frac{s i n ^{2} α /2 s i n ^{2} β /2}{co s ^{2} α /2 co s ^{2} β /2}} ⎦ ⎤

= co s^{- 1} [\frac{co s ^{2} α /2 co s ^{2} β /2 - s i n ^{2} α /2 s i n ^{2} β /2}{co s ^{2} α /2 co s ^{2} β /2 + s i n ^{2} α /2 s i n ^{2} β /2}]

= co s^{- 1} [\frac{2 co s ^{2} α /2 co s ^{2} β /2 - s i n ^{2} α /2 s i n ^{2} β /2}{2 co s ^{2} α /2 co s ^{2} β /2 + 2 s i n ^{2} α /2 s i n ^{2} β /2}]

= co s^{- 1} [\frac{( 1 + cos α ) ( 1 + cos β ) - ( 1 - cos α ) ( 1 - cos β )}{( 1 + cos α ) ( 1 + cos β ) + ( 1 - cos α ) ( 1 - cos β )}]

= co s^{- 1} [\frac{cos α + cos β}{1 - cos α cos β}]

= L H S

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Question Text	Show that: $cos^{- 1} [\frac{cos α + cos β}{1 + cos α cos β}] = 2 tan^{- 1} (tan \frac{α}{2} tan \frac{β}{2})$
Answer Type	Text solution:1
Upvotes	150

Show that:cos−1[1+cosα cosβ cosα +cosβ ​ ] =2tan−1(tan2α ​ tan2β ​ )

Text solutionVerified

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Show that:
$cos^{- 1} [\frac{cos α + cos β}{1 + cos α cos β}] = 2 tan^{- 1} (tan \frac{α}{2} tan \frac{β}{2})$