Question

sec A(1-sin A)(sec A+tan A)=1​

Answer 1

Step-by-step explanation:

L.H.S.=secA(1-sinA)(secA+tanA)

=(SecA-secAsinA)(secA+tanA)

=(secA-tanA)(secA+tanA)

since secAsinA= sinA/cosA=tanA

= sec^2A-tan^2A

=1

R.H.S=L.H.S

Hence proved

Answer 2

Answer:

$\huge\underline\textsf{Question:- }$

$\boxed{\textsf{ Sec A(1-sin A)(sec A +tan A) =1}}$

$\huge\underline\textsf{Explantion:- }$

$\large\implies\tt \frac{1}{ \cos A } (1 - \sin A )( \frac{1}{ \cos A } + \frac{ \sin A }{ \cos A } )$

$\large\implies\tt \frac{(1 - \sin A )}{ \cos A } ( \frac{1}{ \cos A } + \frac{ \sin A}{ \cos A })$

$\large\implies\tt \frac{(1 - \sin A) (1 + \sin A )}{ \cos A \times \cos A }$

$\large \underline\textsf{identity Used:-3 }$

$\red{\boxed{\bf(a + b)(a - b) = a {}^{2} - b {}^{2} }}$

$\underline\textsf{Now,}$

$\large\implies\tt \frac{1 {}^{2} - \sin {}^{2} A }{ \cos {}^{2} A }$

$\large\implies{\overbrace{\boxed{\tt \frac{1 - \sin {}^{2} A }{ \cos {}^{2} A} }}}$

$\large\leadsto {\boxed{\tt= 1 = RHS}}$

$\large\underline{\underline{\texttt{\purple{\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:Hence proved.}}}}$