prove that sec a - tan a is equal to 1 upon sec a + tan a
Answers
Taking RHS :
Multiply numerator and denominator by sec A - Tan A.
Answer:
\underline{\boxed{\orange{\sf Given \: LHS :\: }}}
GivenLHS:
\sf{sec\:A-\:tan\:A}secA−tanA
\underline{\boxed{\purple{\sf Given \: LHS:}}}
GivenLHS:
\sf{\dfrac{1}{sec\:A\:+\:tan\:A}}
secA+tanA
1
\underline{\boxed{\orange{\sf To \: prove: }}}
Toprove:
\sf{sec\:A-\:tan\:A\:=\:\dfrac{1}{sec\:A\:+\:tan\:A}}secA−tanA=
secA+tanA
1
\underline{\boxed{\purple{\sf Proof: }}}
Proof:
Taking RHS :
\sf{\dfrac{1}{sec\:A+tan\:A}}
secA+tanA
1
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}}
secA+tanA
1
×
secA−tanA
secA−tanA
\sf{\dfrac{1\:\times\:sec\:A\:-\:tan\:A}{(sec\:A\:+\:cos\:A)\:\times\:(sec\:A\:-\:Tan\:A)}}
(secA+cosA)×(secA−TanA)
1×secA−tanA
\sf{\dfrac{sec\:A\:-\:tan\:A}{sec^2\:A\:-\:tan^2\:A}}
sec
2
A−tan
2
A
secA−tanA
\bold{\big[\because\:(a+b)(a-b)=a^2-b^2\big]}[∵(a+b)(a−b)=a
2
−b
2
]
\sf{\dfrac{sec\:A\:-\:tan\:A}{1}}
1
secA−tanA
\bold{\big[\because\:sec^2\:A\:-\:tan^2\:A\:=\:1\big]}[∵sec
2
A−tan
2
A=1]
\sf{sec\:A\:-\:tan\:A}secA−tanA
\underline{\boxed{\orange{\sf Hence \: Proved: }}}
HenceProved:
\large{\boxed{\bold{sec\:A-\:tan\:A\:=\:\dfrac{1}{sec\:A\:+\:tan\:A}}}}
secA−tanA=
secA+tanA
1