Question

sin7A +Sin3A/cos7B+cos3A=?​

Answer 1

Answer:

sinA+sin3A+sin5A+sin7A/cosA+cos3A+cos5A+cos7A

=(sinA+sin7A)+(sin3A+sin5A)/(cosA+cos7A)+(cos3A+cos5A)

=(2sin4Acos3A+2sin4AcosA)/(2cos4Acos3A+2cos4AcosA)

=2sin4A(cos3A+cosA)/2cos4A(cos3A+cosA)

=sin4A/cos4A

=tan4A

Hence proved