Question

Tan^2a-cot^2a = sec^2a (1-cot^2a) prove​

Answer 1

Answer:

See explanation

Step-by-step explanation:

Use the definitions:

$\tan \alpha=\dfrac{\sin \alpha}{\cos \alpha}\\ \\\cot \alpha=\dfrac{\cos \alpha}{\sin \alpha}\\ \\\sec \alpha=\dfrac{1}{\cos \alpha}\\ \\\csc \alpha=\dfrac{1}{\sin \alpha}$

Now,

$\tan^2\alpha -\cot^2\alpha=\dfrac{\sin^2\alpha}{\cos^2\alpha}-\dfrac{\cos^2\alpha}{\sin^2\alpha}=\dfrac{\sin^4\alpha-\cos ^4\alpha}{\sin^2\alpha\cos ^2\alpha }=\\ \\=\dfrac{(\sin^2\alpha-\cos ^2\alpha)(\sin^2\alpha-\cos ^2\alpha)}{\sin^2\alpha\cos ^2\alpha }=\dfrac{(\sin^2\alpha-\cos ^2\alpha)\cdot 1}{\sin^2\alpha\cos ^2\alpha }$

and

$\sec^2\alpha(1-\cot^2\alpha)=\dfrac{1}{\cos^2 \alpha}\left(1-\dfrac{\cos^2\alpha}{\sin^2\alpha}\right)=\dfrac{1}{\cos^2 \alpha}\left(\dfrac{\sin^2\alpha-\cos^2\alpha}{\sin^2\alpha}\right)=\\ \\=\dfrac{\sin^2\alpha-\cos ^2\alpha}{\sin^2\alpha\cos ^2\alpha}$

As you can see, left and right parts simplify to the same expression, so left and right parts are the same.