Question

Sin 11A Sin A+Sin 7A Sin 3ACos 11A Sin A+Cos 7A Sin 3A = tan 8A​

Answer 1

Answer:

Please check the attached images for

Step-by-step explanation:

Answer 2

Hence Proved.

Step-by-step explanation:

Given,

Proved $\frac{sin11A.sinA+sin7A.sin3A}{cos11A.sinA+cos7A.sin3A} =tan8A$

LHS:

$\frac{sin11A.sinA+sin7A.sin3A}{cos11A.sinA+cos7A.sin3A}$

$=\frac{2sin11A.sinA+2sin7A.sin3A}{2cos11A.sinA+2cos7A.sin3A}$ (Multiplying by $2$ on numerator and denominator)

$=\frac{cos10A-cos12A+cos4A-cos10A}{sin12A-sin10A+sin10A-sin4A}$ (Using trigonometry formulas)

$=\frac{cos4A-cos12A}{sin12A-sin4A}$

$=\frac{-2sin8A.sin(-4A)}{2cos8A.sin4A}$ (Using trigonometry formulas)

$=\frac{sin8A.sin4A}{cos8A.sin4A}$ (∵ $sin(-A)=-sinA$)

$=\frac{sin8A}{cos8A}$

$=tan8A$

$=RHS$

Hence Proved.