Prove that cot²A-cos²A=cot²A cos²A

1.	Prove that cot²A-cos²A=cot²A cos²A
Answer» Required Answer:- GIVEN to prove: $\sf \cot^{2} (x) - \cos^{2} (x) = \cot^{2} (x) \cos^{2} (x)$ Proof: Taking LHS, $\sf \cot^{2} (x) - \cos^{2} (x)$ $\sf = \frac{1}{ { \tan}^{2}(x) } - \frac{1}{ { \<klux>SEC</klux>}^{2}(x)}$ $\sf = \frac{ \sec^{2} (x) - { \tan}^{2}(x)}{ { \tan}^{2}(x) \sec^{2} (x) }$ 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

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