Question

prove that - cos2a + tan2a-1/sin2a=tan2a​

Answer 1

The equality $\frac{cos^2a+tan^2a-1}{sin^2a}=tan^2a$ is proved

Step-by-step explanation:

Given that $\frac{cos^2a+tan^2a-1}{sin^2a}=tan^2a$

To prove the given equality :

Taking LHS

$\frac{cos^2a+tan^2a-1}{sin^2a}$

$=\frac{cos^2a+tan^2a-(cos^2a+sin^2a)}{sin^2a}$ ( by using the identity $cos^2x+sin^2x=1$ here x=a )

$=\frac{cos^2a+tan^2a-cos^2a-sin^2a}{sin^2a}$

$=\frac{tan^2a-sin^2a}{sin^2a}$

$=\frac{tan^2a}{sin^2a}-\frac{sin^2a}{sin^2a}$

$=\frac{tan^2a}{sin^2a}-1$

$=\frac{\frac{sin^2a}{cos^2a}}{sin^2a}-1$ ( we know $tanx=\frac{sinx}{cosx}$ )

$=(\frac{sin^2a}{cos^2a}\times \frac{1}{sin^2a})-1$

$=\frac{1}{cos^2a}-1$

$=sec^2a-1$ ( by using $secx=\frac{1}{cosx}$ )

$=tan^2a=RHS$ ( by using the identity $sec^2x-tan^2x=1$ )

Therefore LHS=RHS

The equality $\frac{cos^2a+tan^2a-1}{sin^2a}=tan^2a$ is proved