Question

secθ-tanθ/secθ+tanθ=1-2secθ tanθ+tan²θ,Prove it by using trigonometric identities.

Answer 1

HELLO DEAR,

it seems ther is some typing mistake

Your questions is-----------> Secθ-tanθ/secθ+tanθ= 1 - 2secθtanθ + 2tan²θ,
i am taking A instead of theta

secA - tanA/ secA + tanA

= (secA - tanA)/(secA + tanA) * (secA - tanA)/(secA - tanA)

= (secA - tanA)² / sec²A -tan²A

= sec²A + tan²A - 2secAtanA

= (1 + tan²A) + tan²A - 2secAtanA

= 1 + 2tan²A - 2secAtanA.


I HOPE ITS HELP YOU,
THANKS

Answer 2

Hi ,

Here I am using ' x ' instead of theta.

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1 ) ( a + b )( a - b ) = a² - b²

2) sec² x - tan² x = 1

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LHS = ( secx - tanx)/( secx + tanx )

=[(secx-tanx)(secx-tanx)]/[(secx+tanx)(secx-tanx)]

= ( secx - tanx )²/[ sec²x - tan²x ]

= ( secx - tanx )²

= sec²x + tan²x - 2secxtanx

= 1 + tan²x + tan²x - 2secxtanx

= 1 + 2tan² x - 2secxtanx

= RHS

I hope this helps you.

: )