show that (1/secA-tanA)
Answers
Answered by
1
Step-by-step explanation:
Given LHS = \frac{1}{secA-tanA}
secA−tanA
1
On rationalizing we get,
\frac{1}{secA-tanA} * \frac{secA+tanA}{secA+tanA}
secA−tanA
1
∗
secA+tanA
secA+tanA
\frac{secA+tanA}{(secA-tanA)(secA+tanA)}
(secA−tanA)(secA+tanA)
secA+tanA
\frac{secA+tanA}{sec^2A-tan^2A}
sec
2
A−tan
2
A
secA+tanA
\frac{secA+tanA}{1}
1
secA+tanA
secA+tanA.
LHS = RHS.
Hope this helps!
Answered by
3
Answer:Answer:
LHS=1/secA+tanA
1/secA+tanA*secA-tanA/secA-tanA
secA-tanA/sec^2A-tan^2A
secA-tanA/1
SecA-tanA
LHS=RHS
Step-by-step explanation:
Step-by-step explanation:
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