Question 33

If $$\frac{\ \sin^2A-\sin^2B}{\cos^2A\cos^2B}=x$$, then the value of $$x$$ is

Solution

Given, $$x=\frac{\ \sin^2A - \sin^2B}{\cos^2A\cos^2B}$$

$$=$$> $$x=\frac{\ 1-\cos^2A-\left(1-\cos^2B\right)}{\cos^2A\ \cos^2B}$$

$$=$$> $$x=\frac{\ \cos^2B-\cos^2A}{\cos^2A\ \cos^2B}$$

$$=$$> $$x=\frac{\ 1}{\cos^2A}-\frac{\ 1}{\cos^2B}$$

$$=$$> $$x=\sec^2A-\sec^2B$$

$$=$$> $$x=1+\tan^2A-1-\tan^2B$$

$$=$$> $$x=\tan^2A-\tan^2B$$

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