Question

In a ΔABC, `(sin  "C"/2)/(cos(("A" - "B")/2))` = ______ 

विकल्प

• `("b + c")/"a"`

• `"c"/("a + b")`

• `"c"/("a - b")`

• `"a"/("b - c")`

Answer 1

In a ΔABC, `(sin  "C"/2)/(cos(("A" - "B")/2))` = `underline("c"/("a + b"))`

Explanation:

`(sin  "C"/2)/(cos(("A" - "B")/2))`

= `(sin(("A" + "B")/2) sin  "C"/2)/(sin(("A" + "B")/2) cos(("A" - "B")/2))`

= `(sin(pi/2 - "C"/2) sin  "C"/2)/(sin(("A" + "B")/2) cos(("A" - "B")/2))` .............[∵ A + B + C = π]

= `(2cos  "C"/2 sin  "C"/2)/(2sin(("A" + "B")/2)cos(("A" - "B")/2))`

= `sin"C"/(sin"A" + sin"B")`

= `"c"/("a + b")` ............[by sine rule]