Question

secθ+tanθ=p, then obtain the values of secθ, tanθ and sinθ in terms of p.Prove it by using trigonometric identities.

Answer 1

GIVEN ;-

  ⇒  secθ + tanθ = p-----------( 1 )

let theta = A for simplicity in solution

⇒ This sum can be solved through an identity, so -


⇒ Remember the identity - sec²A  –  tan ²A  = 1

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THEREFORE USING THIS IDENTITY , WE CAN SOLVE THIS SUM
         
⇒ ( sec A  +  tan  A ) ( sec A  –  tan A )  =  1
                       

⇒ ( sec A  –  tan A  )  ×  p                      = 1

           
⇒( sec A   – tan A   )                              = 1/p-----------( 2 )


NOW, From-------( 1 ) & --------( 2 )

Sec A + tan A = p
sec A - tan A = 1/p
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2sec A = (p² + 1)/p

⇒sec A = (p² + 1)/2p [ put in ( 1 )]

⇒tan A = p - (p² + 1)/2p

⇒tan A = (2p² - p² - 1)/2p

⇒tan A = (p² - 1)/2p


now, sin A = tan A * cos A


⇒sin a = tanA * 1/secA


⇒sin A = (p² - 1)/2p * 2p/(p² + 1)

⇒sin A = (p² - 1)/(p² + 1)


I HOPE ITS HELP YOU DEAR,
THANKS

Answer 2

Hi ,

Here I am using ' A ' instead of theta.


secA + tanA = p ---( 1 )

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we know the trigonometric identity

sec²A - tan²A = 1

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( secA + tanA )( secA - tanA ) = 1

p( secA - tanA ) = 1 [ from ( 1 ) ]

secA - tanA = 1/p ---( 2 )

add ( 1 ) and ( 2 ) , equations ,

2secA = p + 1/p

secA = ( p² + 1 )/2p ----( 3 )

Subtract ( 2 ) from ( 1 ) , we get

2tanA = p - 1/p

2tanA = ( p² - 1 )/p

tanA = ( p² - 1 )/2p ----( 4 )

Do ( 4 ) ÷ ( 3 ) , we get

tanA/secA = [ (p² - 1 )/2p ]/[(p²+1)/2p ]

[(sinA/cosA)/(1/cosA) = ( p² - 1 )/( p² + 1 )

sinA = ( p² - 1 )/( p² + 1 )

I hope this helps you


: )