Question

If A,B,C are interior angles of a triangle ABC, then prove that:i) sec(A+C/2) = cosec(B/2) ii) tan(B+C/2) = cot (A/2)

Answer 1

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1) A + B + C = 180

A + C = 180 - B

(divide both by 2)

A + C/2 = 90 - B/2

Sec(A+C/2) = Sec(90 - B/2)

sec(A+C/2) = Cosec(B/2)
........cosec(90-theta) = sec (theta).

Sec(A+C/2) = Cosec(B/2) ..............(Proved)
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2) A+B+C = 180

B+C = 180-A

(divide both by 2)

B+C/2 = 90 - A/2

Tan(B+C/2) = Tan(90-A/2).
.....Tan(90-theta) = cot (theta)

Tan(B+C/2) = Cot (A/2). .............(Proved)

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THANKS..........

Answer 2

1) A + B + C = 180

A + C = 180 - B

(divide both by 2)

A + C/2 = 90 - B/2

Sec(A+C/2) = Sec(90 - B/2)

sec(A+C/2) = Cosec(B/2)

........cosec(90-theta) = sec (theta).

Sec(A+C/2) = Cosec(B/2) ..............(Proved)

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2) A+B+C = 180

B+C = 180-A

(divide both by 2)

B+C/2 = 90 - A/2

Tan(B+C/2) = Tan(90-A/2).

.....Tan(90-theta) = cot (theta)

Tan(B+C/2) = Cot (A/2). .............(Proved)

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☺ HOPE IT HELPS YOU ☺

THANKS..........