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Prove that tan 2 Π/16+ tan 2 2Π/16+ tan 2 3Π/16+… tan 2 7Π/16=35

Prove that tan2 Π/16+ tan2 2Π/16+ tan2 3Π/16+… tan2 7Π/16=35

Grade:11

2 Answers

anvesh
45 Points
5 years ago

an²(π/16) + tan²(7π/16) 
= [tan(π/16) + tan(7π/16)]² − 2 tan(π/16)tan(7π/16)
= [tan(π/16) + tan(7π/16)]² − 2 
 
Similarly,
tan²(3π/16) + tan²(5π/16)  = [tan(3π/16) + tan(5π/16)]² − 2 
tan²(2π/16) + tan²(6π/16)  = [tan(2π/16) + tan(6π/16)]² − 2      

Now,  
[tan(π/16) + tan(7π/16)]² 
= [sin(7π/16 + π/16) / ((cos(π/16) cos(7π/16))]² 
=  1/((sin(7π/16) cos(7π/16))²
=  4/sin²(π/8)


Similarly,
[tan(2π/16) + tan(6π/16)]² = 4/sin²(π/4)
[tan(3π/16) + tan(5π/16)]² = 4/sin²(6π/16)

Therefore,
tan²(π/16) + tan²(2π/16) + tan²(3π/16) + tan²(4π/16) + tan²(5π/16) + tan²(6π/16) + tan²(7π/16)
= 4/sin²(π/8) − 2 + 4/sin²(π/4) − 2 + 4/sin²(6π/16) − 2 + tan²π/4
= 4/sin²(π/8) + 4/sin²(6π/16) + 3
= 4[1/sin²(π/8) + 1/sin²(3π/8)] + 3
= 4[sin²(π/8) + cos²(π/8] / [sin²(π/8) cos²(π/8)] + 3
= 16/sin²(π/4)  + 3
35

an²(π/16) + tan²(7π/16) 
= [tan(π/16) + tan(7π/16)]² − 2 tan(π/16)tan(7π/16)
= [tan(π/16) + tan(7π/16)]² − 2 
 
Similarly,
tan²(3π/16) + tan²(5π/16)  = [tan(3π/16) + tan(5π/16)]² − 2 
tan²(2π/16) + tan²(6π/16)  = [tan(2π/16) + tan(6π/16)]² − 2      

Now,  
[tan(π/16) + tan(7π/16)]² 
= [sin(7π/16 + π/16) / ((cos(π/16) cos(7π/16))]² 
=  1/((sin(7π/16) cos(7π/16))²
=  4/sin²(π/8)


Similarly,
[tan(2π/16) + tan(6π/16)]² = 4/sin²(π/4)
[tan(3π/16) + tan(5π/16)]² = 4/sin²(6π/16)

Therefore,
tan²(π/16) + tan²(2π/16) + tan²(3π/16) + tan²(4π/16) + tan²(5π/16) + tan²(6π/16) + tan²(7π/16)
= 4/sin²(π/8) − 2 + 4/sin²(π/4) − 2 + 4/sin²(6π/16) − 2 + tan²π/4
= 4/sin²(π/8) + 4/sin²(6π/16) + 3
= 4[1/sin²(π/8) + 1/sin²(3π/8)] + 3
= 4[sin²(π/8) + cos²(π/8] / [sin²(π/8) cos²(π/8)] + 3
= 16/sin²(π/4)  + 3
35

an²(π/16) + tan²(7π/16) 
= [tan(π/16) + tan(7π/16)]² − 2 tan(π/16)tan(7π/16)
= [tan(π/16) + tan(7π/16)]² − 2 
 
Similarly,
tan²(3π/16) + tan²(5π/16)  = [tan(3π/16) + tan(5π/16)]² − 2 
tan²(2π/16) + tan²(6π/16)  = [tan(2π/16) + tan(6π/16)]² − 2      

Now,  
[tan(π/16) + tan(7π/16)]² 
= [sin(7π/16 + π/16) / ((cos(π/16) cos(7π/16))]² 
=  1/((sin(7π/16) cos(7π/16))²
=  4/sin²(π/8)


Similarly,
[tan(2π/16) + tan(6π/16)]² = 4/sin²(π/4)
[tan(3π/16) + tan(5π/16)]² = 4/sin²(6π/16)

Therefore,
tan²(π/16) + tan²(2π/16) + tan²(3π/16) + tan²(4π/16) + tan²(5π/16) + tan²(6π/16) + tan²(7π/16)
= 4/sin²(π/8) − 2 + 4/sin²(π/4) − 2 + 4/sin²(6π/16) − 2 + tan²π/4
= 4/sin²(π/8) + 4/sin²(6π/16) + 3
= 4[1/sin²(π/8) + 1/sin²(3π/8)] + 3
= 4[sin²(π/8) + cos²(π/8] / [sin²(π/8) cos²(π/8)] + 3
= 16/sin²(π/4)  + 3
35an²(π/16) + tan²(7π/16) 

= [tan(π/16) + tan(7π/16)]² − 2 tan(π/16)tan(7π/16)
= [tan(π/16) + tan(7π/16)]² − 2 
 
Similarly,
tan²(3π/16) + tan²(5π/16)  = [tan(3π/16) + tan(5π/16)]² − 2 
tan²(2π/16) + tan²(6π/16)  = [tan(2π/16) + tan(6π/16)]² − 2      

Now,  
[tan(π/16) + tan(7π/16)]² 
= [sin(7π/16 + π/16) / ((cos(π/16) cos(7π/16))]² 
=  1/((sin(7π/16) cos(7π/16))²
=  4/sin²(π/8)


Similarly,
[tan(2π/16) + tan(6π/16)]² = 4/sin²(π/4)
[tan(3π/16) + tan(5π/16)]² = 4/sin²(6π/16)

Therefore,
tan²(π/16) + tan²(2π/16) + tan²(3π/16) + tan²(4π/16) + tan²(5π/16) + tan²(6π/16) + tan²(7π/16)
= 4/sin²(π/8) − 2 + 4/sin²(π/4) − 2 + 4/sin²(6π/16) − 2 + tan²π/4
= 4/sin²(π/8) + 4/sin²(6π/16) + 3
= 4[1/sin²(π/8) + 1/sin²(3π/8)] + 3
= 4[sin²(π/8) + cos²(π/8] / [sin²(π/8) cos²(π/8)] + 3
= 16/sin²(π/4)  + 3
35an²(π/16) + tan²(7π/16) 

= [tan(π/16) + tan(7π/16)]² − 2 tan(π/16)tan(7π/16)
= [tan(π/16) + tan(7π/16)]² − 2 
 
Similarly,
tan²(3π/16) + tan²(5π/16)  = [tan(3π/16) + tan(5π/16)]² − 2 
tan²(2π/16) + tan²(6π/16)  = [tan(2π/16) + tan(6π/16)]² − 2      

Now,  
[tan(π/16) + tan(7π/16)]² 
= [sin(7π/16 + π/16) / ((cos(π/16) cos(7π/16))]² 
=  1/((sin(7π/16) cos(7π/16))²
=  4/sin²(π/8)


Similarly,
[tan(2π/16) + tan(6π/16)]² = 4/sin²(π/4)
[tan(3π/16) + tan(5π/16)]² = 4/sin²(6π/16)

Therefore,
tan²(π/16) + tan²(2π/16) + tan²(3π/16) + tan²(4π/16) + tan²(5π/16) + tan²(6π/16) + tan²(7π/16)
= 4/sin²(π/8) − 2 + 4/sin²(π/4) − 2 + 4/sin²(6π/16) − 2 + tan²π/4
= 4/sin²(π/8) + 4/sin²(6π/16) + 3
= 4[1/sin²(π/8) + 1/sin²(3π/8)] + 3
= 4[sin²(π/8) + cos²(π/8] / [sin²(π/8) cos²(π/8)] + 3
= 16/sin²(π/4)  + 3
35an²(π/16) + tan²(7π/16) 

= [tan(π/16) + tan(7π/16)]² − 2 tan(π/16)tan(7π/16)
= [tan(π/16) + tan(7π/16)]² − 2 
 
Similarly,
tan²(3π/16) + tan²(5π/16)  = [tan(3π/16) + tan(5π/16)]² − 2 
tan²(2π/16) + tan²(6π/16)  = [tan(2π/16) + tan(6π/16)]² − 2      

Now,  
[tan(π/16) + tan(7π/16)]² 
= [sin(7π/16 + π/16) / ((cos(π/16) cos(7π/16))]² 
=  1/((sin(7π/16) cos(7π/16))²
=  4/sin²(π/8)


Similarly,
[tan(2π/16) + tan(6π/16)]² = 4/sin²(π/4)
[tan(3π/16) + tan(5π/16)]² = 4/sin²(6π/16)

Therefore,
tan²(π/16) + tan²(2π/16) + tan²(3π/16) + tan²(4π/16) + tan²(5π/16) + tan²(6π/16) + tan²(7π/16)
= 4/sin²(π/8) − 2 + 4/sin²(π/4) − 2 + 4/sin²(6π/16) − 2 + tan²π/4
= 4/sin²(π/8) + 4/sin²(6π/16) + 3
= 4[1/sin²(π/8) + 1/sin²(3π/8)] + 3
= 4[sin²(π/8) + cos²(π/8] / [sin²(π/8) cos²(π/8)] + 3
= 16/sin²(π/4)  + 3
35

an²(π/16) + tan²(7π/16) 
= [tan(π/16) + tan(7π/16)]² − 2 tan(π/16)tan(7π/16)
= [tan(π/16) + tan(7π/16)]² − 2 
 
Similarly,
tan²(3π/16) + tan²(5π/16)  = [tan(3π/16) + tan(5π/16)]² − 2 
tan²(2π/16) + tan²(6π/16)  = [tan(2π/16) + tan(6π/16)]² − 2      

Now,  
[tan(π/16) + tan(7π/16)]² 
= [sin(7π/16 + π/16) / ((cos(π/16) cos(7π/16))]² 
=  1/((sin(7π/16) cos(7π/16))²
=  4/sin²(π/8)


Similarly,
[tan(2π/16) + tan(6π/16)]² = 4/sin²(π/4)
[tan(3π/16) + tan(5π/16)]² = 4/sin²(6π/16)

Therefore,
tan²(π/16) + tan²(2π/16) + tan²(3π/16) + tan²(4π/16) + tan²(5π/16) + tan²(6π/16) + tan²(7π/16)
= 4/sin²(π/8) − 2 + 4/sin²(π/4) − 2 + 4/sin²(6π/16) − 2 + tan²π/4
= 4/sin²(π/8) + 4/sin²(6π/16) + 3
= 4[1/sin²(π/8) + 1/sin²(3π/8)] + 3
= 4[sin²(π/8) + cos²(π/8] / [sin²(π/8) cos²(π/8)] + 3
= 16/sin²(π/4)  + 3
35an²(π/16) + tan²(7π/16) 

= [tan(π/16) + tan(7π/16)]² − 2 tan(π/16)tan(7π/16)
= [tan(π/16) + tan(7π/16)]² − 2 
 
Similarly,
tan²(3π/16) + tan²(5π/16)  = [tan(3π/16) + tan(5π/16)]² − 2 
tan²(2π/16) + tan²(6π/16)  = [tan(2π/16) + tan(6π/16)]² − 2      

Now,  
[tan(π/16) + tan(7π/16)]² 
= [sin(7π/16 + π/16) / ((cos(π/16) cos(7π/16))]² 
=  1/((sin(7π/16) cos(7π/16))²
=  4/sin²(π/8)


Similarly,
[tan(2π/16) + tan(6π/16)]² = 4/sin²(π/4)
[tan(3π/16) + tan(5π/16)]² = 4/sin²(6π/16)

Therefore,
tan²(π/16) + tan²(2π/16) + tan²(3π/16) + tan²(4π/16) + tan²(5π/16) + tan²(6π/16) + tan²(7π/16)
= 4/sin²(π/8) − 2 + 4/sin²(π/4) − 2 + 4/sin²(6π/16) − 2 + tan²π/4
= 4/sin²(π/8) + 4/sin²(6π/16) + 3
= 4[1/sin²(π/8) + 1/sin²(3π/8)] + 3
= 4[sin²(π/8) + cos²(π/8] / [sin²(π/8) cos²(π/8)] + 3
= 16/sin²(π/4)  + 3
35

Mohammed Tameem Mohiuddin
10 Points
5 years ago
hey thanx bro

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