Question

prove that cot²A-cos²A=cot²A cos²A​

Answer 1

Required Answer:-

Given to prove:

• $\sf \cot^{2} (x) - \cos^{2} (x) = \cot^{2} (x) \cos^{2} (x)$

Proof:

Taking LHS,

$\sf \cot^{2} (x) - \cos^{2} (x)$

$\sf = \frac{1}{ { \tan}^{2}(x) } - \frac{1}{ { \sec}^{2}(x)}$

$\sf = \frac{ \sec^{2} (x) - { \tan}^{2}(x)}{ { \tan}^{2}(x) \sec^{2} (x) }$

We know that,

$\sf \sec^{2} (x) - \tan^{2} (x) = 1$

So, we get,

$\sf = \frac{ 1}{ { \tan}^{2}(x) \sec^{2} (x) }$

$\sf = \cot^{2} (x) \cos^{2} (x)$

Taking RHS,

$\sf = \cot^{2} (x) \cos^{2} (x)$

Hence, LHS = RHS

So,

$\sf \blue{ \cot^{2} (x) - \cos^{2} (x) = \cot^{2} (x) \cos^{2} (x)}$

Hence Proved.

Note:

• tan(x) is the reciprocal of cot(x)
• sec(x) is the reciprocal of cos(x)

Formula Used:

• sec²(x) - tan²(x) = 1