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cosecA +cotA=P then secA-tanA=?

cosecA +cotA=P then secA-tanA=?

Grade:10

5 Answers

Akshat Dubey
11 Points
8 years ago
coseca+cota=p
now, cosec2a – cot2a = 1      …(1)
or, (coseca+cota)  (coseca-cota) = 1
or, p(coseca-cota)=1
or, coseca-cota = 1/p   ...(2)
 
add eqn (1) and (2)
coseca + cota + coseca – cota = p+ 1/p
or, 2coseca = (p2+1)/p
or, 2/sina = (p2+1)/p
or, sina = 2p/(p2+1)
 
now, (1) – (2)
coseca + cota – coseca + cota = p – 1/p
or, 2cota = (p2 - 1)/p
or, 2(cosa/sina) = (p2 - 1)/p
or, 2 [cosa / 2p/(p2+1)] = (p2 - 1)/p
or, 2 cosa * (p2 + 1) / 2p = (p2 - 1)/p
or, cosa = (p2 – 1)/ (p2 + 1)
or, seca = (p2 + 1) / (p2 – 1)   …(3)
 
now, tana = sina/cosa = [2p/+1)2(p] / [(p2 – 1)/  + 1)2(p]
or, tana = 2p /(p2 – 1)   ...(4)
 
now, (3) – (4)
seca – tana = [(p2 + 1) / (p2 – 1)] – [2p /(p2 – 1)] 
or, seca – tana = (p2 + 1 – 2p) / (p2 – 1)
 
I know there can be easier and shorter solutions, but this is the one that struck me first!
I hope that’s a correct solution :)
Raghu Vamshi Hemadri
72 Points
8 years ago
 
coseca+cota=p
now, cosec2a – cot2a = 1      …(1)
or, (coseca+cota)  (coseca-cota) = 1
or, p(coseca-cota)=1
or, coseca-cota = 1/p   ...(2)
 
add eqn (1) and (2)
coseca + cota + coseca – cota = p+ 1/p
or, 2coseca = (p2+1)/p
or, 2/sina = (p2+1)/p
or, sina = 2p/(p2+1)
 
now, (1) – (2)
coseca + cota – coseca + cota = p – 1/p
or, 2cota = (p2 - 1)/p
or, 2(cosa/sina) = (p2 - 1)/p
or, 2 [cosa / 2p/(p2+1)] = (p2 - 1)/p
or, 2 cosa * (p2 + 1) / 2p = (p2 - 1)/p
or, cosa = (p2 – 1)/ (p2 + 1)
or, seca = (p2 + 1) / (p2 – 1)   …(3)
 
now, tana = sina/cosa = [2p/+1)2(p] / [(p2 – 1)/  + 1)2(p]
or, tana = 2p /(p2 – 1)   ...(4)
 
now, (3) – (4)
seca – tana = [(p2 + 1) / (p2 – 1)] – [2p /(p2 – 1)] 
or, seca – tana = (p2 + 1 – 2p) / (p2 – 1)
 
Forum Team
94 Points
8 years ago
Dear Raghu,
 
Please do not post exactly same answer which other student has already posted before you. This answer will have no value addition to questioner and will increase one mail at his end unnecessarily. Points earned by this will not be counted for redemption.
 
Thanks
Forum Team
SHAIK HAFEEZUL KAREEM
109 Points
8 years ago
 
 
coseca+cota=p
now, cosec2a – cot2a = 1      …(1)
or, (coseca+cota)  (coseca-cota) = 1
or, p(coseca-cota)=1
or, coseca-cota = 1/p   ...(2)
 
add eqn (1) and (2)
coseca + cota + coseca – cota = p+ 1/p
or, 2coseca = (p2+1)/p
or, 2/sina = (p2+1)/p
or, sina = 2p/(p2+1)
 
now, (1) – (2)
coseca + cota – coseca + cota = p – 1/p
or, 2cota = (p2 - 1)/p
or, 2(cosa/sina) = (p2 - 1)/p
or, 2 [cosa / 2p/(p2+1)] = (p2 - 1)/p
or, 2 cosa * (p2 + 1) / 2p = (p2 - 1)/p
or, cosa = (p2 – 1)/ (p2 + 1)
or, seca = (p2 + 1) / (p2 – 1)   …(3)
 
now, tana = sina/cosa = [2p/+1)2(p] / [(p2 – 1)/  + 1)2(p]
or, tana = 2p /(p2 – 1)   ...(4)
 
now, (3) – (4)
seca – tana = [(p2 + 1) / (p2 – 1)] – [2p /(p2 – 1)] 
or, seca – tana = (p2 + 1 – 2p) / (p2 – 1)
 
I know there can be easier and shorter solutions, but this is the one that struck me first!
I hope that’s a correct solution :)
Lab Bhattacharjee
121 Points
8 years ago
\text{Using Weierstrass Substitution,}\\ p=\csc A+\cot A=\dfrac{1+\cos A}{\sin A}=\dfrac1{\tan\dfrac A2}\iff\tan\dfrac A2=?\\ \sec A-\tan A=\dfrac{1-\sin A}{\cos A}=\dfrac{1-\tan\dfrac A2}{1+\tan\dfrac A2}=\cdots

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