Question

Sin11A sinA sin7A sin3A/ cos11A sin A cos 7A sinA= tan 8A

Answer 1

hello,
i think it is addition between sinA sin7A  and sinA and cos7A
sin11AsinA+sin7Asin3A/cos11AsinA+cos7Asin3A
=2sin11AsinA+2sin7Asin3A/2cos11AsinA+2cos7Asin3A
=[{cos(11A-A)-cos(11A+A)}+{cos(7A-3A)-cos(7A+3A)}]/
[{sin(11A+A)-sin(11A-A)}+{sin(7A+3A)-sin(7A-3A)}]
=(cos10A-cos12A+cos4A-cos10A)/(sin12A-sin10A+sin10A-sin4A)
=(cos4A-cos12A)/(sin12A-sin4A)
=[2sin(4A+12A)/2sin(12A-4A)/2]/[2cos(12A+4A)/2sin(12A-4A)/2]
=sin8Asin4A/cos8Asin4A
=sin8A/cos8A
=tan8A (Proved)

Answer 2

Step-by-step explanation:

sin11AsinA+sin7Asin3A/cos11AsinA+cos7Asin3A

=2sin11AsinA+2sin7Asin3A/2cos11AsinA+2cos7Asin3A

=[{cos(11A-A)-cos(11A+A)}+{cos(7A-3A)-cos(7A+3A)}]/

[{sin(11A+A)-sin(11A-A)}+{sin(7A+3A)-sin(7A-3A)}]

=(cos10A-cos12A+cos4A-cos10A)/(sin12A-sin10A+sin10A-sin4A)

=(cos4A-cos12A)/(sin12A-sin4A)

=[2sin(4A+12A)/2sin(12A-4A)/2]/[2cos(12A+4A)/2sin(12A-4A)/2]

=sin8Asin4A/cos8Asin4A

=sin8A/cos8A

=tan8A (Proved)