Question

1/sec theta-1 - 1/sec theta+1 = 2cot
square theta

Answer 1

Answer:

1/sec teta-1-1/sec teta+1 is not equal to 2cot²teta

Answer 2

$\huge \red{ \mathfrak{solution}}$

To prove :

$\frac{1}{sec \alpha - 1} - \frac{1}{sec \alpha + 1} = 2 {cot}^{2} \alpha$

LHS :

$\implies \frac{1}{sec \alpha - 1} - \frac{1}{sec \alpha + 1} \\ \\ \implies \frac{ \sec( \alpha ) + 1 - \sec( \alpha ) + 1 }{ {sec}^{2} \alpha - 1} \\ \\ \implies \: \frac{2}{ {tan }^{2} \alpha } \\ \\ \implies \: 2 {cot }^{2} \alpha$

LHS=RHS

$\boxed{ \red{ \bold{proved}}}$

Formula used :

◆

$\purple{\boxed{ \bold{(x + y)(x - y) = {x}^{2} - {y}^{2} }}}$

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$\pink{ \boxed{{sec}^{2} \alpha - {tan}^{2} \alpha = 1} }\\ \\ \pink{ \boxed{ \leadsto \: sec ^{2} \alpha - 1 = {tan}^{2} \alpha }}$

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$\green{\boxed{\frac{1}{ \tan( \alpha ) } = \cot( \alpha ) }}$