Question

cos square theta + sec square theta divided by cos squared theta minus sec squared theta prove that 1 + tan squared
theta divided by 1 minus tan squared theta
​

Answer 1

Given : (cos²θ  + sec²θ )/((cos²θ  - sec²θ )  = (1 + tan²θ) / (1 -  tan²θ)

To Find : To be Proved

Solution:

let say cos θ = 1  => Sinθ = 0  tanθ  = 0  , secθ =  1

LHS = (cos²θ  + sec²θ )/((cos²θ  - sec²θ )  = 2/0

RHS = (1 + tan²θ) / (1 -  tan²θ)  = 1/1 = 1

LHS  ≠ RHS

Hence Given is not correct

Correct Question should be

(cos²θ  + sin²θ )/((cos²θ  - sin²θ )  = (1 + tan²θ) / (1 -  tan²θ)

sinθ =  cos θ tanθ

LHS =  (cos²θ  + sin²θ )/((cos²θ  - sin²θ )  

= (cos²θ  +cos²θtan²θ )/((cos²θ  -cos²θtan²θ )  

= cos²θ(1 + tan²θ)/cos²θ(1 -  tan²θ)

= (1 + tan²θ) / (1 -  tan²θ)

= RHS

QED

(cos²θ  + sin²θ )/((cos²θ  - sin²θ )  = (1 + tan²θ) / (1 -  tan²θ)

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