Question

prove the given question .

Answer 1

HEY Buddy....!! here is ur answer

we have to prove that :

$\frac{ { \tan }^{3} \alpha - { \cot}^{3} \alpha }{ \tan\alpha - \cot\alpha } = { \sec }^{2} \alpha + { \cot}^{2} \alpha \\ \\ On \: taking \: L.H.S. \: \\ \\ = > \frac{ { \tan}^{3} \alpha - { \cot }^{3} \alpha }{ \tan \alpha - \cot\alpha } \\ \\ As \: we \: know \: that :\: {a}^{3} - {b}^{3} = (a - b)( {a}^{2} + ab + {b}^{2} ) \\ \\ = > \frac{( \tan\alpha - \cot\alpha )( { \tan }^{2} \alpha + \ \tan \alpha \cot \alpha + { \cot }^{2} \alpha) }{ (\tan\alpha - \cot\alpha) } \\ \\ = > { \tan}^{2} \alpha + 1 + { \cot }^{2} \alpha \\ \\ = > { \sec }^{2} \alpha - 1 + 1 + { \cot}^{2} \alpha \\ \\ As \: we \: know \: that :\: \ \tan\alpha \cot \alpha = 1 \: and \: { \sec }^{2} \alpha = 1 + { \tan}^{2} \alpha \\ \\ = > { \sec }^{2} \alpha + { \cot }^{2} = R.H.S. \: \\ \\ HENCE\: PROVED \:$

I hope it will be helpful for you....!!

THANK YOU ✌️✌️

MARK IT AS BRAINLIEST