Asked by Rahul
Prove that 1/sin theta-tan theta and - 1/cos theta = 1/cos theta and + 1/sec theta+tan theta
Answers
Answered by
Reiny
The way you typed it
1/sinØ-tanØ + - 1/cosØ = 1/cosØ + 1/secØ + tanØ
the statement is false and thus cannot be proven true
Please ret-type it using brackets.
e.g.
did you mean
1/(sinØ-tanØ) + (- 1/cosØ) = 1/cosØ + 1/(secØ + tanØ) ?
1/sinØ-tanØ + - 1/cosØ = 1/cosØ + 1/secØ + tanØ
the statement is false and thus cannot be proven true
Please ret-type it using brackets.
e.g.
did you mean
1/(sinØ-tanØ) + (- 1/cosØ) = 1/cosØ + 1/(secØ + tanØ) ?
Answered by
Rahul
i mean to say
1/sinØ-tanØ - 1/cosØ = 1/cosØ + 1/secØ + tanØ
1/sinØ-tanØ - 1/cosØ = 1/cosØ + 1/secØ + tanØ
Answered by
Reiny
In that case the statement is false
try Ø=30°
Left Side = 1/sin30 - tan30 - 1/cos30
= 2-1/√3 - 2/√3 = appr. .2679
Right Side = 1/cos30 + 1/sec30 + tan30 = appr. 2.598
You will need brackets to tell us how the equation is.
You must put brackets around the denominator
e.g. 1/sinØ+ tanØ ≠ 1/(sinØ+tanØ)
try Ø=30°
Left Side = 1/sin30 - tan30 - 1/cos30
= 2-1/√3 - 2/√3 = appr. .2679
Right Side = 1/cos30 + 1/sec30 + tan30 = appr. 2.598
You will need brackets to tell us how the equation is.
You must put brackets around the denominator
e.g. 1/sinØ+ tanØ ≠ 1/(sinØ+tanØ)
Answered by
Rahul
the answer is = 2/cosØ , L.H.S = R.H.D
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