Question

SHOW THAT, (cos^2A-cos^2B) / (cos^2A .cos^2B)=(tan^2B-tan^2A).

Answer 1

Here, we will only use one simple identity:

$\sec^2 \theta = 1+\tan^2 \theta$


We can solve the question as follows:

$LHS \\ \\ \\ = \frac{\cos^2A-\cos^2B}{cos^2A \cos^2B} \\ \\ \\ = \frac{\cos^2A}{\cos^2A \cos^2B} - \frac{\cos^2B}{\cos^2A \cos^2B} \\ \\ \\ = \frac{1}{\cos^2B} - \frac{1}{\cos^2A} \\ \\ \\ = \sec^2B - \sec^2A \\ \\ \\ = (1+\tan^2B) - (1+tan^2A) \\ \\ \\ = 1 + \tan^B - 1 - \tan^2A \\ \\ \\ = \tan^2B -\tan^2A \\ \\ \\ = RHS$

Hope it helps
Purva
Brainly Community

Answer 2

I HOPE IT'S HELP YOU......