Question

cos squared theta minus sin square theta equal to 1 minus 10 squared theta by 1 + 10 squared theta prove the identity

Answer 1

Answer and Explanation:

To prove : The identity $\cos^2\theta-\sin^2\theta=\frac{1-\tan^2\theta}{1+\tan^2\theta}$

Solution :

Taking RHS,

$\frac{1-\tan^2\theta}{1+\tan^2\theta}$

Write, $\tan\theta=\frac{\sin\theta}{\cos\theta}$

$\frac{1-(\frac{\sin\theta}{\cos\theta})^2}{1+(\frac{\sin\theta}{\cos\theta})^2}}$

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

$\frac{\frac{\cos^2\theta-\sin^2\theta}{\cos^2\theta}}{\frac{\cos^2\theta+\sin^2\theta}{\cos^2\theta}}$

$\frac{\cos^2\theta-\sin^2\theta}{\cos^2\theta+\sin^2\theta}$

$\frac{\cos^2\theta-\sin^2\theta}{1}$

$\cos^2\theta-\sin^2\theta$

$=LHS$

So, RHS=LHS

Answer 2

Step-by-step explanation:

The identity \cos^2\theta-\sin^2\theta=\frac{1-\tan^2\theta}{1+\tan^2\theta}cos

2

θ−sin

2

θ=

1+tan

2

θ

1−tan

2

θ

Solution :

Taking RHS,

\frac{1-\tan^2\theta}{1+\tan^2\theta}

1+tan

2

θ

1−tan

2

θ

Write, \tan\theta=\frac{\sin\theta}{\cos\theta}tanθ=

cosθ

sinθ