Question

Any JEE ASPIRANT who can solve this..???? ​

Answer 1

Step-by-step explanation:

1+tan²ø=sec²ø

1=sec²ø-tan²ø

Answer 2

Answer:

0

Concept

Sec²x - tan²x = 1

log(1) = 0 at any base.

Solution

We have,

To find the value of $log_{22 \frac{1 \degree}{2} }( \sec ^{2} x - \tan^{2} x )$

Putting 1 in place of sec²x - tan²x

$log_{22 \frac{1 \degree}{2} }( 1 ) \\ \\ = \: 0$

Additional Concept

• Logarithm of any number with the same base gives 1.

$log_{x} {x}\:=\:1$

• sin²x + cos²x = 1.

• cosec²x - cot²x = 1.

• sin2x = 2sinx × cosx

• cos2x = cos²x - sin²x