if sec theta+tan theta =p ,prove that sin theta = (p square- 1) / (p square + 1)

= Taking R.H.S i.e, (p2-1) / (p2+1)

= Putting value of p in given equation

= [(sec0+tan0)2-1] / [(sec0+tan0)2+1]

= (sec20+tan20+2sec0tan0-1) / (sec20+tan20+2sec0tan0+1)

= (sec20-1)+tan20+2sec0tan0 / (1+tan20)+sec20+2sec0tan0

= tan20+tan20+2sec0tan0 / sec20+sec20+2sec0tan0 [ As sec20-1=tan20 1+tan20=sec20 ]

= 2tan20+2sec0tan0 / 2sec20+2sec0tan0

= 2tan0+(tan0+sec0) / 2sec0+(sec0+tan0) [ Taking 2tan0 2sec0 as common ]

= tan0 / sec0 [ 2 and tan0+sec0 gets cancelled above ]

= sin0/cos0 / 1/cos0 [ As tan0=sin0/cos0 and sec0=1/cos0 ]

= sin0/1

= sin0

= L.H.S

Hence Proved...........

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it is there in RD SHARMA
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The answer:

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Hope this helps !

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Hope this helps 👍

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Hi thanks....ppl
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Hope it will help cheers!!!

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This may be helpful

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Solution

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yes
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sec theta + tan theta =p
sec theta - tan theta =1/p
2 sec theta =pSq+1/p
2 tan theta =pSq-1/p
2 tan theta/2 sec theta =pSq-1/pSq+1
hence ,proved
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Ans

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Hi.Your school is my centre of examination.
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Hi; your school is my centre of examination.
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Hope it helps....

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