Question

answer please immediately​

Answer 1

Answer:

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