Question

prove it using trigonometric identities ​

Answer 1

To prove :

$\sf \: \dfrac{ \bigg(1 + \tan ^{2} ( A) \: \bigg) \cot (A) }{ \csc^{2} (A) } \: = \: \tan ( A)$

Identity used :

$\sf \bullet \: 1 + \tan ^{2} ( A) = \sec^{2} ( A)$

$\sf \bullet \: \csc( A) = \dfrac{1}{ \sin( A) \: }$

$\sf \bullet \: \sec( A) = \dfrac{1}{ \cos( A) \: }$

$\sf \bullet \: \cot( A) = \dfrac{ \cos(A) }{ \sin(A) }$

$\sf \bullet \: \tan( A) = \dfrac{ \sin(A) }{ \cos(A) }$

Solution :

$\sf \implies LHS = \dfrac{ \bigg(1 + \tan ^{2} ( A) \: \bigg) \cot(A) }{ \csc^{2} (A) }$

$\sf \implies LHS = \dfrac{ \sec ^{2} ( A) \: \cot (A) }{ \csc^{2} (A) }$

$\sf \implies LHS = \dfrac{ \bigg(\: \dfrac{1}{ \cos ^{2} ( A) \: }\bigg)\: \bigg(\: \dfrac{\cos( A) \: }{ \sin( A) \: }\bigg)\: }{ \bigg(\: \dfrac{1}{ \sin ^{2} ( A) \: }\bigg)\: }$

$\sf \implies LHS = \bigg(\: \dfrac{ \: \sin ^{2} ( A) }{ \cos ^{2} ( A) \: }\bigg)\: \bigg(\: \dfrac{\cos( A) \: }{ \sin( A) \: }\bigg)$

$\sf \implies LHS = \bigg(\: \dfrac{ \: \sin ( A) }{ \cos ( A) \: }\bigg)\:$

$\sf \implies LHS = \: \tan ( A)$

Now, RHS = tan(A)

This gives, LHS = RHS

Hence proved.