Question

prove that sec a - tan a is equal to 1 upon sec a + tan a​
Please ans. This

Answer 1

$\underline{\boxed{\purple{\sf Given \: LHS:}}}$

$\sf{\dfrac{1}{sec\:A\:+\:tan\:A}}$

$\underline{\boxed{\orange{\sf To \: prove: }}}$

$\sf{sec\:A-\:tan\:A\:=\:\dfrac{1}{sec\:A\:+\:tan\:A}}$

$\underline{\boxed{\purple{\sf Proof: }}}$

Taking RHS :

$\sf{\dfrac{1}{sec\:A+tan\:A}}$

Multiply numerator and denominator by sec A - Tan A.

$\sf{\dfrac{1}{sec\:A+\:tan\:A}\:\times\:\dfrac{sec\:A\:-\:tan\:A}{sec\:A\:-\:tan\:A}}$

$\sf{\dfrac{1\:\times\:sec\:A\:-\:tan\:A}{(sec\:A\:+\:cos\:A)\:\times\:(sec\:A\:-\:Tan\:A)}}$

$\sf{\dfrac{sec\:A\:-\:tan\:A}{sec^2\:A\:-\:tan^2\:A}}$

$\bold{\big[\because\:(a+b)(a-b)=a^2-b^2\big]}$

$\sf{\dfrac{sec\:A\:-\:tan\:A}{1}}$

$\bold{\big[\because\:sec^2\:A\:-\:tan^2\:A\:=\:1\big]}$

$\sf{sec\:A\:-\:tan\:A}$

$\underline{\boxed{\orange{\sf Hence \: Proved: }}}$

$\large{\boxed{\bold{sec\:A-\:tan\:A\:=\:\dfrac{1}{sec\:A\:+\:tan\:A}}}}$

Answer 2

$\huge\underline\frak{\fbox\red{★ Solution : }}$

$\sf{\underline{\overline{\pink{Given\: LHS :}}}}$

$\sf{sec\:A-\:tan\:A}$

$\sf{\underline{\overline{\purple{Given\: RHS :}}}}$

$\sf{\dfrac{1}{sec\:A\:+\:tan\:A}}$

$\sf{\underline{\overline{\orange{To\: prove :}}}}$

$\sf{sec\:A-\:tan\:A\:=\:\dfrac{1}{sec\:A\:+\:tan\:A}}$

$\sf{\underline{\overline{\blue{Proof :}}}}$

Taking RHS :

$\sf{\dfrac{1}{sec\:A+tan\:A}}$

Multiply numerator and denominator by $\bold{sec\: A - Tan\: A.}$

$\sf{\dfrac{1}{sec\:A+\:tan\:A}\:\times\:\dfrac{sec\:A\:-\:tan\:A}{sec\:A\:-\:tan\:A}}$

$\sf{\dfrac{1\:\times\:sec\:A\:-\:tan\:A}{(sec\:A\:+\:cos\:A)\:\times\:(sec\:A\:-\:Tan\:A)}}$

$\sf{\dfrac{sec\:A\:-\:tan\:A}{sec^2\:A\:-\:tan^2\:A}}$

$\bold{\big[\because\:(a+b)(a-b)=a^2-b^2\big]}$

$\sf{\dfrac{sec\:A\:-\:tan\:A}{1}}$

$\bold{\big[\because\:sec^2\:A\:-\:tan^2\:A\:=\:1\big]}$

$\sf{sec\:A\:-\:tan\:A}$

Hence proved.

$\large{\boxed{\bold{sec\:A-\:tan\:A\:=\:\dfrac{1}{sec\:A\:+\:tan\:A}}}}$