In a triangle ABC, right angled at B. If cosec C = 2 then find sec A?
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A+90°+C=180°
=>A+C=90°
sinB
secAsinC−tanAtanC
=(secAsinC−
cosA
sinA
×
cosC
sinC
)÷sin90°
=(
cosA
sinC
−
cosA
sinA
×
cosC
sinC
)
=(
cosA
sin(90°−A)
−
cosA
sinA
×
cos(90°−A)
sin(90°−A)
)
=
cosA
cosA
−
cosA
sinA
×
sinA
cosA
=1−1
=0
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