Question

a.

$\cot^{2} - \cos ^2a = \cot ^{2} a \times \cos ^{2} a$
​

Answer 1

Step-by-step explanation:

We need to prove that,

$\cot^2 a-\cos^2 a=\cot^2 a\times \cos^2 a$

Taking LHS,

$\cot^2 a-\cos^2 a$

We know that,

$\cot a=\dfrac{\cos a}{\sin a}$

​$=(\dfrac{\cos a}{\sin a})^2 -\cos^2 a\\\\=\cos^2 a(\dfrac{1}{\sin^2 a}-1)\\\\=\cos^2 a(\dfrac{1-\sin^2a}{\sin^2 a})\\\\=\cos^2 a\times \dfrac{\cos^2 a}{\sin^2 a}\\\\=\dfrac{\cos ^2 a}{\sin^2 a}\times \cos^2a\\\\\because \ \dfrac{\cos ^2 a}{\sin^2 a}=\cot^2 a\\\\=\cot^2 a \times \cos^2a\\\\=RHS$

So, LHS = RHS

Hence, proved.