Question

[(sinA+sinB)/(cosA-cosB)]+[(cosA+cosB)/(sinA-sinB)]​

Answer 1

Answer:

[(sinA +sinB) /(cosA - cosB)] +[(cosA +cosB) /sinA-sinB)] =0

Step-by-step explanation:

=(sinA+sinB) (sinA-sinB) +(cosA+cosB)(cosA-cosB)/

(cosA-cosB) (sinA-sinB)

=sin^2 A-sin^2 B + cos^2 A-cos^2B/

(cosA-cosB) (sinA-sinB)

=sin^2 A +cos^2 A-(sin^2 B+cos^2 B) /

(cosA-cosB) (sinA-sinB)

= 1 - 1/

(cosA-cosB) (sinA-sinB)

= 0

NOTE:- 1) sin^2A-sin^2B =(sinA-sinB) (sinA+sinB)

2)cos^2A-cos^2B=(cosA-cosB)(cosA+cosB)

3)sin^2A+cos^2A=1

4)sin^2B+cos^2B=1

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