Question

Sec a(1 - sin a)(sec a+ tan a)= 1

Answer 1

Step-by-step explanation:

$sec \: a(1 - sina)( sec \: a \: + tan \: a ) = 1$

$(sec \: a \: - sec \: a \: sin \: a)(sec \: a \: + tan \:a)$

$(sec \: a \: - \frac{1}{cos \: a} \times sin \: a)(sec \: a + tan \: a)$

$(sec \: a \: - \frac{sin \: a}{cos \: a} )(sec \: a \: + tan \: a)$

$(sec \: a \: - tan \: a)(sec \: a \: + tan \: a)$

$sec {}^{2} a \: - tan {}^{2} a$

$1$

$= rhs \:$

HENCE. ,

PROVED

Answer 2

Answer:

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

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

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

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

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

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

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

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

$\underline\textsf{Now,}$

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

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

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

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