Grade 10Trigonometry

cosecA +cotA=P then secA-tanA=?

SHUBHAM KUSHWAHA

10 Years agoGrade 10

5 Answers

Akshat Dubey

10 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

10 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

10 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

10 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

10 Years ago

$\text{Using Weierstrass Substitution,}\\ p=\csc A+\cot A=\dfrac{1+\cos A}{\sin A}=\dfrac1{\tan\dfrac A2}\iff\tan\dfrac A2=?\\ \sec A-\tan A=\dfrac{1-\sin A}{\cos A}=\dfrac{1-\tan\dfrac A2}{1+\tan\dfrac A2}=\cdots$