Question

(1+tan²x)/1+sin²x+cos²x = 2sinx+cosx​

Answer 1

What is (1-sin²x) (1+tan²x) equal to?

We know that,

sin²x + cos²x = 1 ( eq1 )

1-sin²x = sin²x + cos²x - sin²x = cos²x

1-sin²x = cos²x

And,

sec²x - tan²x = 1 (eq2)

1+tan²x = sec²x - tan²x + tan²x = sec²x

1+tan²x = sec²x

NOW,

(1-sin²x) (1+tan²x) = cos²x sec²x

[Since , cos²x = 1/sec²x]

(1/sec²x )(sec²x ) =1

Clearly,

(1-sin²x) (1+tan²x) = 1

OR

Fact:- sin²x+cos²x=1; tanx=sinx/cosx

(1-sin²x) (1+tan²x)=(cos²x) (1+sin²x/cos²x)

=(cos²x) ((cos²x+sin²x)/(cos²x))

=(cos²x) (1/(cos²x)) =(cos²x)/(cos²x) =1

Thus, (1-sin²x) (1+tan²x)=1

$bts ❤ exo$