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Answer 1 · 2018-02-06T10:24:53

# (d+sqrt(a^2-d^2))/(d-sqrt(a^2-d^2))#, OR #(d-sqrt(a^2-d^2))/(d+sqrt(a^2-d^2))#.

#cosA=d/a#.

#:. sin^2A=1-cos^2A#,

#=1-d^2/a^2#,

#=(a^2-d^2)/a^2#.

#rArr sinA=+-sqrt(a^2-d^2)/a#.

#sinA=+sqrt(a^2-d^2)/arArr cosA/sinA=d/sqrt(a^2-d^2)#.

By componendo-dividendo, then,

#(cosA+sinA)/(cosA-sinA)=(d+sqrt(a^2-d^2))/(d-sqrt(a^2-d^2))#.

If, #sinA=sqrt(a^2-d^2)/(-a)," then "cosA/sinA=d/(-sqrt(a^2-d^2)#.

#:. (cosA+sinA)/(cosA-sinA)=(d-sqrt(a^2-d^2))/(d+sqrt(a^2-d^2))#.

Answer 2 · 2018-02-06T10:25:02

When #A# is in #I or II# quadrant
Considering #sinA=+sqrt(1-cos^2A)#

#(cosA+sinA)/(cosA-sinA)#

#=(cosA+sqrt(1-cos^2A))/(cosA-sqrt(1-cos^2A))#

#=(d/a+sqrt(1-d^2/a^2))/(d/a-sqrt(1-d^2/a^2))#

#=(d+sqrt(a^2-d^2))/(d-sqrt(a^2-d^2))#

Again When #A# is in #III or IV# quadrant

considering #sinA=-sqrt(1-cos^2A)#

#(cosA+sinA)/(cosA-sinA)#

#=(cosA-sqrt(1-cos^2A))/(cosA+sqrt(1 -cos^2A))#

#=(d/a-sqrt(1-d^2/a^2))/(d/a+sqrt(1-d^2/a^2))#

#=(d-sqrt(a^2-d^2))/(d+sqrt(a^2-d^2))#