Prove that 1/(secA - tanA) - 1/cosA = 1/cosA - 1/(secA - tanA) .
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Answered by
7
Answer:
hii
your answer is here !
Explanation:
Consider 1/secA-tanA - 1/cosA
Multiplying by secA + tanA in the numerator and denominator of
first term,
we get secA + tanA/(secA +tanA)(secA - tanA) - 1/cosA
= secA + tanA - secA (Since sec²A - tan²A = 1)
= tanA
Adding and subtracting secA , we get
secA + tanA - secA
= 1/cosA - (secA - tanA)
Now multiplying and dividing (secA - tanA) by (secA + tanA), we
get 1/cosA - (sec²A - tan²A)/(secA + tanA)
= 1/ cosA - 1/secA + tanA
= R.H.S
Hence, Proved.
Hope, it helps !
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Answer:
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