Question

tan theta by sec theta minus 1 is equal to tan theta + sec theta + 1 by tan theta + sec Theta - 1

Answer 1

AnswEr :

• To Prove :

$\Large \frac{ \tan( \alpha ) }{ \sec( \alpha ) - 1} = \frac{ \tan( \alpha ) + \sec( \alpha ) + 1}{\tan( \alpha ) + \sec( \alpha ) - 1}$

• Proof :

$\dfrac{ \tan( \alpha ) }{ \sec( \alpha ) - 1} = \dfrac{ \tan( \alpha ) + \sec( \alpha ) + 1}{\tan( \alpha ) + \sec( \alpha ) - 1}$

⋆ I'm taking RHS to Prove this :

$\longrightarrow\dfrac{ \tan( \alpha ) + \sec( \alpha ) + 1}{\tan( \alpha ) + \sec( \alpha ) - 1}$

$\longrightarrow\dfrac{ (\tan( \alpha )) + ( \sec( \alpha ) + 1)}{(\tan( \alpha )) + ( \sec( \alpha ) - 1)}$

$\longrightarrow\dfrac{ (\tan( \alpha )) + ( \sec( \alpha ) + 1)}{(\tan( \alpha )) + ( \sec( \alpha ) - 1)} \times 1$

$\small\longrightarrow\dfrac{ (\tan( \alpha )) + ( \sec( \alpha ) + 1)}{(\tan( \alpha )) + ( \sec( \alpha ) - 1)} \times \dfrac{ (\tan( \alpha )) - ( \sec( \alpha ) - 1)}{(\tan( \alpha )) - ( \sec( \alpha ) - 1)}$

$\longrightarrow \dfrac{( \tan( \alpha )) + ( \sec( \alpha ) + 1)( \tan( \alpha ) ) - (\sec( \alpha ) - 1) }{ ( \tan ( \alpha ) + ( \sec( \alpha ) - 1) ( \tan ( \alpha ) - ( \sec( \alpha ) - 1)}$

⠀

• (a + b)(a - c) = a² - a(c - b) - bc
• (a + b)(a - b) = a² - b²

⠀

$\longrightarrow \small \dfrac{ \tan^{2} ( \alpha ) - \tan( \alpha )( \cancel{\sec( \alpha )} - 1 - \cancel{\sec( \alpha )} - 1) - ( \sec ^{2} ( \alpha ) - 1) }{ \tan^{2} ( \alpha ) - ( \sec( \alpha ) - 1)^{2} }$

$\longrightarrow \dfrac{ \cancel{\tan^{2} ( \alpha )} + 2 \tan( \alpha ) - \cancel{\tan^{2} ( \alpha )} }{\tan^{2}( \alpha ) - ( \sec^{2} ( \alpha ) + 1 - 2 \sec( \alpha ) ) }$

$\longrightarrow \dfrac{ 2 \tan( \alpha )}{\tan^{2}( \alpha ) - \sec^{2} ( \alpha ) - 1 + 2 \sec( \alpha ) }$

• taking Minus Common in denominator

$\longrightarrow \dfrac{ 2 \tan( \alpha )}{ - ( - \tan^{2}( \alpha ) + \sec^{2} ( \alpha ) + 1 - 2 \sec( \alpha ) ) }$

$\longrightarrow \dfrac{ 2 \tan( \alpha )}{ - (( \sec^{2} ( \alpha ) - \tan^{2}( \alpha ) ) + 1 - 2 \sec( \alpha ) ) }$

• sec² θ - tan² θ = 1

$\longrightarrow \dfrac{ 2 \tan( \alpha )}{ - (1+ 1 - 2 \sec( \alpha ) ) }$

$\longrightarrow \dfrac{ 2 \tan( \alpha )}{ - (2 - 2 \sec( \alpha ) ) }$

$\longrightarrow \dfrac{ 2 \tan( \alpha )}{ -2 + 2 \sec( \alpha) }$

$\longrightarrow \dfrac{ 2 \tan( \alpha )}{ 2 \sec( \alpha ) - 2 }$

$\longrightarrow \dfrac{ \cancel2 \tan( \alpha )}{ \cancel2 (\sec( \alpha ) - 1)}$

$\longrightarrow \large{\dfrac{ \tan( \alpha )}{\sec( \alpha ) - 1}} \: \: \: \: \: \: \: \: \small\bf{ \underline{Hence \:Proved}}$