Question

$\mathfrak{If\:sec\theta+tan\theta=p\\then\:find\:cosec\theta}$

Answer 1

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Answer 2

Given :

sec θ + tan θ = p ..................( 1 )

Multiplying both sides by ( sec θ - tan θ ) we get :

➡( sec θ - tan θ )( sec θ + tan θ ) = p ( sec θ - tan θ )

Use the identity of ( a + b )( a - b ) = a² - b²

➡sec² θ - tan² θ = p ( sec θ - tan θ )

We know that sec²θ - tan²θ = 1

➡1 = p ( sec θ - tan θ )

sec θ - tan θ = 1 / p .................. ( 2 )

Adding ( 1 ) and ( 2 ) we get :

➡2 sec θ = p + 1 / p

➡sec θ = 1 / cos θ

➡2 / cos θ = p + 1 / p

➡2 / cos θ = ( p² + 1 ) / p

➡cos θ / 2 = p / ( p² + 1 )

➡cos θ = 2 p / ( p² + 1 )

Square both sides :

➡cos²θ = 4 p² / ( p² + 1 )²

Multiply both sides by -1

➡ - cos²θ = - 4 p² / ( p² + 1 )²

Add 1 both sides :

➡1 - cos²θ = - 4 p² / ( p² + 1 )²+ 1

We know that 1 - cos²θ = sin²θ

➡sin²θ = [ - 4 p² + ( p² + 1 )² ] / ( p² + 1 )²

➡sin² θ = [ - 4 p² + p⁴ + 2 p² + 1 ] / ( p² + 1 )²

➡sin² θ = [ p⁴ - 2 p² + 1 ] / ( p² + 1 )²

➡sin² θ = ( p² - 1 )² / ( p² + 1 )²

By invertendo :

➡1 / sin²θ = ( p² + 1 )² / ( p² - 1 )²

Take square root both sides :

➡1 / sin θ = ( p² + 1 ) / ( p² - 1 )

We know that 1 / sin θ = cosec θ

➡cosec θ = ( p² + 1 ) / ( p² - 1 )

$\mathfrak{\huge{\textsf{ANSWER}}}$

$\mathsf{\boxed{\boxed{cosec\theta=\frac{p^2+1}{p^2-1}}}}$