Question

pls solve this question. We have to prove LHS=RHS ​

Answer 1

Required Answer:-

Given to prove:

$\rm \mapsto \dfrac{ \tan^{3} (x) - 1}{ \tan(x) - 1} = { \sec}^{2} (x) + \tan(x)$

Proof:

Taking LHS,

$\rm\dfrac{ \tan^{3} (x) - 1}{ \tan(x) - 1}$

$\rm = \dfrac{ \tan(x)^{3} - {1}^{3} }{ \tan(x) - 1}$

As we know that,

➡ a³ - b³ = (a - b)(a² + ab + b²), So,

$\rm = \dfrac{( \tan(x) - 1)( { \tan}^{2}(x) + \tan(x) + 1) }{ \tan(x) - 1}$

$\rm =( {\tan}^{2}(x) + \tan(x) + 1)$

$\rm ={\tan}^{2}(x) + 1+ \tan(x)$

We know that,

$\rm \mapsto { \sec}^{2} \alpha = \tan^{2} \alpha + 1$

Therefore,

$\rm{\tan}^{2}(x) + 1+ \tan(x)$

$\rm = {\sec}^{2}(x) + \tan(x)$

= RHS (Hence Proved)

Note:

• sec²(x) - tan²(x) = 1, So sec²(x) = 1 + tan²(x)

Answer 2

Answer:

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