Question

(sec Φ - tan Φ)^2 = (1 - sinΦ) ÷(1 + sinΦ). prove it

Answer 1

$\underline{\underline{\Huge{\textbf{Question:}}}}$

$\left(sec\: \phi - tan\: \phi\right)^{2} = \left(\dfrac{1-sin\: \phi}{1+sin\: \phi}\right)$



$\underline{\underline{\Huge{\textbf{Solution:}}}}$



$\underline{\huge{\textsf{Step-1:}}}$
Consider LHS

$\textbf{LHS\: =}$

$\left(sec\: \phi - tan\: \phi\right)^{2}$



$\underline{\huge{\textsf{Step-2:}}}$
Express $sec\: \phi$ and $tan\: \phi$ in terms of $cos\: \phi$

We have,
$\bigstar \: \mathbf{ sec\: \phi = \left(\dfrac{1}{cos\: \phi}\right)}$

$\bigstar \: \mathbf{tan\: \phi = \left(\dfrac{sin\: \phi}{cos\: \phi}\right)}$

On Substituting in LHS

$\implies \left(\dfrac{1}{cos\: \phi} - \dfrac{sin\: \phi}{cos\: \phi}\right)^{2}$ ———[1]



$\underline{\huge{\textsf{Step-3:}}}$
Simplify [1]

$\implies \left(\dfrac{1-sin\: \phi}{cos\: \phi}\right)^{2}$

$\implies \left(\dfrac{\left(1-sin\: \phi\right)^{2}}{cos^{2} \: \phi} \right)$ ———[2]



$\underline{\huge{\textsf{Step-4:}}}$
Express $cos^{2}\: \phi$ in terms of $sin\: \phi$

$\bigstar \: \mathbf{cos^{2}\: \phi + sin^{2}\: \phi= 1}$

$\implies\: cos^{2}\: \phi = 1-sin^{2}\: \phi$

Substitute in [2]

$\implies \left(\dfrac{\left(1-sin\: \phi\right)^{2}}{\left(1-sin^{2} \: \phi\right)} \right)$ ———[3]



$\underline{\huge{\textsf{Step-4:}}}$
Using the Identity $\mathbf{a^{2}-b^{2}= (a+b)(a-b)}$

$\implies 1-sin^{2} \: \phi = \left(1+sin\: \phi\right)\left(1-sin\: \phi\right)$

Substitute in [3]

$\implies \left(\dfrac{\left(1-sin\: \phi\right)^{2}}{\left(1+sin\: \phi\right)\left(1-sin\: \phi\right)} \right)$

$\implies \left(\dfrac{\left(1-sin\: \phi\right)^{ \cancel{2}}}{\left(1+sin\: \phi\right) \cancel{\left(1-sin\: \phi\right)}} \right)$

$\implies \left(\dfrac{\left(1-sin\: \phi\right)}{\left(1+sin\: \phi\right)} \right)$

$\textbf{ = \: RHS}$

$\mathbf{Hence\: Proved}$