Question

tan2a-tan2b=cos2b-cos2a/cos2acos2b​

Answer 1

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Answer 2

Proved below.

Step-by-step explanation:

Given:

$tan^2a-tan^2b=\frac{cos^2b-cos^2a}{cos^2bcos^2a}$

LHS = $tan^2a-tan^2b$        

       = $\frac{sin^2a}{cos^2a} -\frac{sin^b}{cos^2b}$   [∵ $tan^2a=\frac{sin^2a}{cos^2a}, tan^2b= \frac{sin^2b}{cos^2b}$]

       = $\frac{sin^2a \times cos^2b - sin^2b\times cos^2a}{cos^2a \times cos^2b}$

      = $\frac{(1-cos^2a)\times cos^2b - (1 - cos^2b) \times cos^2a}{cos^2a \times cos^2b}$                [ ∵ $sin^2a+cos^a=1$ ]

      = $\frac{cos^2b - cos^2a \times cos^2b - cos^2a + cos^2a \times cos^2b}{cos^2a \times cos^2b}$

       = $\frac{cos^2b - cos^2a}{cos^2a \timts cos^2b}$

      = RHS.

Hence proved.