verify the identities : 1- cosθ =tan²θ / 1+secθ + tan²θ?

Answer 1

RHS
tan ² θ / [ sec θ (sec θ + 1) ]
(sec² θ - 1) / [ sec θ (sec θ + 1) ]
(sec θ - 1) (sec θ + 1) / [ sec θ (sec θ + 1) ]
(sec θ - 1) / sec θ
1 - 1 / sec θ
1 - cos θ

LHS
1 - cos θ

LHS = RHS

Answer 2

Prove the identity.

1 - cosθ = tan²θ / (1 + secθ + tan²θ)
_____________

Right Hand Side = tan²θ / (1 + secθ + tan²θ)

= tan²θ / [(1 + tan²θ) + secθ]

= (sec²θ - 1) / (sec²θ + secθ)

= [(secθ - 1)(secθ + 1)] / [secθ(secθ + 1)]

= (secθ - 1) / secθ = 1 - 1/secθ

= 1 - cosθ = Left Hand Side

Answer 3

1- cosθ ≡ tan²θ / 1+secθ + tan²θ

tan²θ / 1+secθ + tan²θ
(sin²θ/ cos²θ ) / 1 + 1/cosθ + sin²θ/cos²θ (multiply up an down by cos²θ)

sin²θ / cos²θ + cosθ + sin²θ
sin²θ / cos²θ + cosθ + ( 1 - cos²θ)
(1 -cos²θ) / cosθ + 1
(1- cosθ)(1+cosθ) / (1+cosθ)
(1-cosθ)

1- cosθ ≡ 1- cosθ

Answer 4

RHS:

(tanA)^2/1+secA + (tanA)^2 ::

(sinA)^2/ (cosA)^2
= -------------------------------------------
1 + (1/cosA) + (sinA^2/cosA^2)

= (sinA)^2/ (cosA)^2
-----------------------------------
|[(cosA)^2+cosA+(sinA)^2]/(cosA)^2
{ taking LCM in denominator}

= (sinA)^2
--------------------------------
(cosA)^2+cosA+(sinA)^2

= 1^2 - (cosA)^2
-------------------------------------
cosA + 1
{ sinA^2 + cosA^2 = 1
sinA^2= 1- cosA^2 }

= (1+cosA) (1-cosA)
------------------------------------
1+cosA
{ (a^2 - b^2) = (a+b) (a-b) }

= 1 - cosA

= LHS