Question

sin11AsinA+sin7Asin3A/cos11AsinA+cos7Asin3A = tan8A​

Answer 1

$\huge\pink{Answer}$

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Answer 2

Solution=

L.H.S= 2SinA.Sin11A+2Sin3A.Sin7A

-------------------------------------------

2SinA.Cos11A+2Sin3A.Cos7A

/ we know

/2sinA.sinA

cos(a-b)-cos(a+b)

2sinA.CosB

=sin(a+b)+sin(a-b).........Hint

cos10A-cos12A+cos4A-cos10A

= -----------------------------------------

sin12A-sin10A+sin10A-sin4A

cos4A - cos 12 A.

= ------------------------

sin2A - sin 4 A

2sin8A.sin4A.

= --------------------

2 co8A.sin4A

sin8A.

= ------- = tan8A

cos8A

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