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Help me plz.....m stuck in this question.....kindly explain 37 and 38

Help me plz.....m stuck in this question.....kindly explain 37 and 38

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Grade:11

2 Answers

Arun
25750 Points
4 years ago
 
Dear student
 
Please ask only one question in one thread,
38.)
 
tan 3x – tan 2x = 1 + tan 2x tan3x
 
hence tan 3x – tan2x /(1 + tan 2x tan3x) = 1
 
tan (3x – 2x) = 1
 
tan x = 1
hence x = pi/4 and 5 pi/4
 
but 2x = pi/2 and 5pi/2
 
hence no solution
Ravi Sankar Mukhrejee
18 Points
4 years ago
37.2sec 2α=tan β+cot β
=>2(1/cos 2α)=(sin β/cos β)+(cos β/sin β)[∵sec x=1/cos x;tan x=sin x/cos x;cot x=cos x/sin x]
=>2(1/cos 2α)=(sin²β+cos²β)/sin β•cos β
=>(1/cos 2α)=1/2sin β•cos β[∵sinx+cos²x=1]
=>cos 2α=2sin β•cos β
=>cos 2α=sin 2β[∵2sin x•cos x=sin 2x]
=>cos 2α=cos(π/2-2β)[∵sin x=cos(90°-x)]
=>2α=π/2-2β
=>2α+2β=π/2
=>2(α+β)=π/2
∴α+β=π/4

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