Question

prove that sec a - tan a is equal to 1 upon sec a + tan a​

Answer 1

Given LHS :

• $\sf{sec\:A-\:tan\:A}$

Given RHS :

• $\sf{\dfrac{1}{sec\:A\:+\:tan\:A}}$

To prove :

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

Proof :

Taking RHS :

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

Multiply numerator and denominator by sec A - Tan A.

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

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

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

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

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

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

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

Hence proved.

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