Question

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

Answer 1

To Prove :-

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

Proof :-

$\textsf{LHS\: =}$

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

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

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

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

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

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

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

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

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

$\textsf { = \: RHS}$

• $\mathsf {Hence\: Proved}$

$\sf \underline{\red { Need \: To \:Know :-}}$

• $\: \mathsf{ sec\: \phi = \left(\dfrac{1}{cos\: \phi}\right)}$

• $\: \mathsf{tan\: \phi = \left(\dfrac{sin\: \phi}{cos\: \phi}\right)}$

• $\: \mathsf{cos^{2}\: \phi + sin^{2}\: \phi= 1}$

• $\sf \: cos^{2}\: \phi = 1-sin^{2}\: \phi$
• $\mathsf{a^{2}-b^{2}= (a+b)(a-b)}$

Answer 2

$\sf \green{ \: \dfrac{a^m}{a^n} \: = \: a^{m-n}}$

$\sf \green { \: a^m \times a^n = a^{m+n}}$

$\sf \green { \: (a^m)^{n} = a^{mn}}$