SOLUTION: a) prove that (1+tan A + sec A)(1+cot A-cosecA)=2
b) Prove that Sin 20degree.sin40degree.sin 60degree.Sin80degree = 3/16
c) sin 5A+2sin7A+sin9A/Cos3A+2cos5A+Cos7A = Sin2A+Cos2A.T
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Question 735516: a) prove that (1+tan A + sec A)(1+cot A-cosecA)=2
b) Prove that Sin 20degree.sin40degree.sin 60degree.Sin80degree = 3/16
c) sin 5A+2sin7A+sin9A/Cos3A+2cos5A+Cos7A = Sin2A+Cos2A.Tan5A
d) Show that Sin112+Cos12 / Cos12-Sin12 = Cot35
e) Prove that Sin 7x+Sinx / Cos5x-Cos3x = Sin2x-Cos2x.cotx
Answer by mananth(16946) (Show Source): You can put this solution on YOUR website!
a) prove that (1+tan A + sec A)(1+cot A-cosecA)=2
LHS =>
= 2
LHS = RHS
b) Prove that Sin 20degree.sin40degree.sin 60degree.Sin80degree = 3/16
sin(A+B) = sin A cos B + cos A sin B
sin(40) as sin(60-20)
sin(80) as sin(60+20)
sin(20)•sin(60-20)•[√(3)/2]•sin(60+20)
- Simplify sin(60-20)•sin(60+20)
sin(60-20) = sin60•cos20 - sin20•cos60
sin(60+20)= sin60•cos20 + sin20•cos60
multiply
= sin²60•cos²20 - sin²20•cos²60
cos²A = 1 - sin²A
sin²60•cos²20 - sin²20•cos²60
= sin²60•(1-sin²20) - sin²20•(1-sin²60)
= sin²60- sin²20•sin²60 - sin²20 + sin²20•sin²60
= sin²60 - sin²20
= 3/4 - sin²20
We get
= sin20•[√(3)/2]•(3/4 - sin²20)
= [√(3)/8]•(3sin20 - 4sin³20)
we know that
sin(3A) = 3sin(A) - 4sin³(A)
(3sin20 - 4sin³20) = sin60
we have
= [√(3)/8]•sin60
= [√(3)/8]•[√(3)/2)
= 3/16
LHS = RHS
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